\(\int \frac {\sin ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) [217]

Optimal result

Integrand size = 21, antiderivative size = 261 \[ \int \frac {\sin ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\left (3 a^4-36 a^2 b^2+40 b^4\right ) x}{8 a^6}-\frac {2 \sqrt {a-b} b \sqrt {a+b} \left (2 a^2-5 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^6 d}+\frac {b \left (11 a^2-15 b^2\right ) \sin (c+d x)}{3 a^5 d}-\frac {\left (13 a^2-20 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 a^4 d}+\frac {\left (3 a^2-5 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^3 b d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a^2 d}-\frac {\left (a^2-b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{a^2 b d (b+a \cos (c+d x))} \]

[Out]

1/8*(3*a^4-36*a^2*b^2+40*b^4)*x/a^6+1/3*b*(11*a^2-15*b^2)*sin(d*x+c)/a^5/d-1/8*(13*a^2-20*b^2)*cos(d*x+c)*sin(
d*x+c)/a^4/d+1/3*(3*a^2-5*b^2)*cos(d*x+c)^2*sin(d*x+c)/a^3/b/d+1/4*cos(d*x+c)^3*sin(d*x+c)/a^2/d-(a^2-b^2)*cos
(d*x+c)^3*sin(d*x+c)/a^2/b/d/(b+a*cos(d*x+c))-2*b*(2*a^2-5*b^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(
1/2))*(a-b)^(1/2)*(a+b)^(1/2)/a^6/d

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3957, 2971, 3128, 3102, 2814, 2738, 214} \[ \int \frac {\sin ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\left (a^2-b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{a^2 b d (a \cos (c+d x)+b)}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 a^2 d}-\frac {2 b \sqrt {a-b} \sqrt {a+b} \left (2 a^2-5 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^6 d}+\frac {b \left (11 a^2-15 b^2\right ) \sin (c+d x)}{3 a^5 d}-\frac {\left (13 a^2-20 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 a^4 d}+\frac {\left (3 a^2-5 b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 a^3 b d}+\frac {x \left (3 a^4-36 a^2 b^2+40 b^4\right )}{8 a^6} \]

[In]

Int[Sin[c + d*x]^4/(a + b*Sec[c + d*x])^2,x]

[Out]

((3*a^4 - 36*a^2*b^2 + 40*b^4)*x)/(8*a^6) - (2*Sqrt[a - b]*b*Sqrt[a + b]*(2*a^2 - 5*b^2)*ArcTanh[(Sqrt[a - b]*
Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^6*d) + (b*(11*a^2 - 15*b^2)*Sin[c + d*x])/(3*a^5*d) - ((13*a^2 - 20*b^2)*Co
s[c + d*x]*Sin[c + d*x])/(8*a^4*d) + ((3*a^2 - 5*b^2)*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^3*b*d) + (Cos[c + d*x]
^3*Sin[c + d*x])/(4*a^2*d) - ((a^2 - b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(a^2*b*d*(b + a*Cos[c + d*x]))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2971

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(a^2 - b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((d*Sin[e + f*x])^(n + 1)/(a*b^2*d*f
*(m + 1))), x] + (-Dist[1/(a*b^2*(m + 1)*(m + n + 4)), Int[(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^n*Sim
p[a^2*(n + 1)*(n + 3) - b^2*(m + n + 2)*(m + n + 4) + a*b*(m + 1)*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*(m
+ n + 3)*(m + n + 4))*Sin[e + f*x]^2, x], x], x] - Simp[Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 2)*((d*Sin[e +
f*x])^(n + 1)/(b^2*d*f*(m + n + 4))), x]) /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2
*m, 2*n] && LtQ[m, -1] &&  !LtQ[n, -1] && NeQ[m + n + 4, 0]

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3128

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e
+ f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*
x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n +
2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d
^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cos ^2(c+d x) \sin ^4(c+d x)}{(-b-a \cos (c+d x))^2} \, dx \\ & = \frac {\cos ^3(c+d x) \sin (c+d x)}{4 a^2 d}-\frac {\left (a^2-b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{a^2 b d (b+a \cos (c+d x))}-\frac {\int \frac {\cos ^2(c+d x) \left (-8 a^2+15 b^2-a b \cos (c+d x)+4 \left (3 a^2-5 b^2\right ) \cos ^2(c+d x)\right )}{-b-a \cos (c+d x)} \, dx}{4 a^2 b} \\ & = \frac {\left (3 a^2-5 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^3 b d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a^2 d}-\frac {\left (a^2-b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{a^2 b d (b+a \cos (c+d x))}+\frac {\int \frac {\cos (c+d x) \left (-8 b \left (3 a^2-5 b^2\right )-5 a b^2 \cos (c+d x)+3 b \left (13 a^2-20 b^2\right ) \cos ^2(c+d x)\right )}{-b-a \cos (c+d x)} \, dx}{12 a^3 b} \\ & = -\frac {\left (13 a^2-20 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 a^4 d}+\frac {\left (3 a^2-5 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^3 b d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a^2 d}-\frac {\left (a^2-b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{a^2 b d (b+a \cos (c+d x))}-\frac {\int \frac {-3 b^2 \left (13 a^2-20 b^2\right )+a b \left (9 a^2-20 b^2\right ) \cos (c+d x)+8 b^2 \left (11 a^2-15 b^2\right ) \cos ^2(c+d x)}{-b-a \cos (c+d x)} \, dx}{24 a^4 b} \\ & = \frac {b \left (11 a^2-15 b^2\right ) \sin (c+d x)}{3 a^5 d}-\frac {\left (13 a^2-20 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 a^4 d}+\frac {\left (3 a^2-5 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^3 b d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a^2 d}-\frac {\left (a^2-b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{a^2 b d (b+a \cos (c+d x))}+\frac {\int \frac {3 a b^2 \left (13 a^2-20 b^2\right )-3 b \left (3 a^4-36 a^2 b^2+40 b^4\right ) \cos (c+d x)}{-b-a \cos (c+d x)} \, dx}{24 a^5 b} \\ & = \frac {\left (3 a^4-36 a^2 b^2+40 b^4\right ) x}{8 a^6}+\frac {b \left (11 a^2-15 b^2\right ) \sin (c+d x)}{3 a^5 d}-\frac {\left (13 a^2-20 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 a^4 d}+\frac {\left (3 a^2-5 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^3 b d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a^2 d}-\frac {\left (a^2-b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{a^2 b d (b+a \cos (c+d x))}+\frac {\left (b \left (2 a^4-7 a^2 b^2+5 b^4\right )\right ) \int \frac {1}{-b-a \cos (c+d x)} \, dx}{a^6} \\ & = \frac {\left (3 a^4-36 a^2 b^2+40 b^4\right ) x}{8 a^6}+\frac {b \left (11 a^2-15 b^2\right ) \sin (c+d x)}{3 a^5 d}-\frac {\left (13 a^2-20 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 a^4 d}+\frac {\left (3 a^2-5 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^3 b d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a^2 d}-\frac {\left (a^2-b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{a^2 b d (b+a \cos (c+d x))}+\frac {\left (2 b \left (2 a^4-7 a^2 b^2+5 b^4\right )\right ) \text {Subst}\left (\int \frac {1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^6 d} \\ & = \frac {\left (3 a^4-36 a^2 b^2+40 b^4\right ) x}{8 a^6}-\frac {2 \sqrt {a-b} b \sqrt {a+b} \left (2 a^2-5 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^6 d}+\frac {b \left (11 a^2-15 b^2\right ) \sin (c+d x)}{3 a^5 d}-\frac {\left (13 a^2-20 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 a^4 d}+\frac {\left (3 a^2-5 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^3 b d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a^2 d}-\frac {\left (a^2-b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{a^2 b d (b+a \cos (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.06 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.08 \[ \int \frac {\sin ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {384 b \left (2 a^4-7 a^2 b^2+5 b^4\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {72 a^4 b c-864 a^2 b^3 c+960 b^5 c+72 a^4 b d x-864 a^2 b^3 d x+960 b^5 d x+24 a \left (3 a^4-36 a^2 b^2+40 b^4\right ) (c+d x) \cos (c+d x)-24 a \left (a^4-31 a^2 b^2+40 b^4\right ) \sin (c+d x)+176 a^4 b \sin (2 (c+d x))-240 a^2 b^3 \sin (2 (c+d x))-21 a^5 \sin (3 (c+d x))+40 a^3 b^2 \sin (3 (c+d x))-10 a^4 b \sin (4 (c+d x))+3 a^5 \sin (5 (c+d x))}{b+a \cos (c+d x)}}{192 a^6 d} \]

[In]

Integrate[Sin[c + d*x]^4/(a + b*Sec[c + d*x])^2,x]

[Out]

((384*b*(2*a^4 - 7*a^2*b^2 + 5*b^4)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (7
2*a^4*b*c - 864*a^2*b^3*c + 960*b^5*c + 72*a^4*b*d*x - 864*a^2*b^3*d*x + 960*b^5*d*x + 24*a*(3*a^4 - 36*a^2*b^
2 + 40*b^4)*(c + d*x)*Cos[c + d*x] - 24*a*(a^4 - 31*a^2*b^2 + 40*b^4)*Sin[c + d*x] + 176*a^4*b*Sin[2*(c + d*x)
] - 240*a^2*b^3*Sin[2*(c + d*x)] - 21*a^5*Sin[3*(c + d*x)] + 40*a^3*b^2*Sin[3*(c + d*x)] - 10*a^4*b*Sin[4*(c +
d*x)] + 3*a^5*Sin[5*(c + d*x)])/(b + a*Cos[c + d*x]))/(192*a^6*d)

Maple [A] (verified)

Time = 1.64 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.25

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a -b \right ) \left (a +b \right ) b \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b}-\frac {\left (2 a^{2}-5 b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{6}}+\frac {\frac {2 \left (\left (\frac {3}{8} a^{4}+2 a^{3} b -\frac {3}{2} a^{2} b^{2}-4 a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+\left (\frac {26}{3} a^{3} b -\frac {3}{2} a^{2} b^{2}-12 a \,b^{3}+\frac {11}{8} a^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {11}{8} a^{4}+\frac {3}{2} a^{2} b^{2}+\frac {26}{3} a^{3} b -12 a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (2 a^{3} b -4 a \,b^{3}-\frac {3}{8} a^{4}+\frac {3}{2} a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {\left (3 a^{4}-36 a^{2} b^{2}+40 b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{a^{6}}}{d}\)	\(325\)
default	\(\frac {\frac {2 \left (a -b \right ) \left (a +b \right ) b \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b}-\frac {\left (2 a^{2}-5 b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{6}}+\frac {\frac {2 \left (\left (\frac {3}{8} a^{4}+2 a^{3} b -\frac {3}{2} a^{2} b^{2}-4 a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+\left (\frac {26}{3} a^{3} b -\frac {3}{2} a^{2} b^{2}-12 a \,b^{3}+\frac {11}{8} a^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {11}{8} a^{4}+\frac {3}{2} a^{2} b^{2}+\frac {26}{3} a^{3} b -12 a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (2 a^{3} b -4 a \,b^{3}-\frac {3}{8} a^{4}+\frac {3}{2} a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {\left (3 a^{4}-36 a^{2} b^{2}+40 b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{a^{6}}}{d}\)	\(325\)
risch	\(\frac {3 x}{8 a^{2}}-\frac {9 x \,b^{2}}{2 a^{4}}+\frac {5 x \,b^{4}}{a^{6}}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{2} d}-\frac {5 i b \,{\mathrm e}^{i \left (d x +c \right )}}{4 a^{3} d}-\frac {2 i b^{3} {\mathrm e}^{-i \left (d x +c \right )}}{a^{5} d}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 a^{2} d}+\frac {2 i b^{3} {\mathrm e}^{i \left (d x +c \right )}}{a^{5} d}-\frac {3 i {\mathrm e}^{2 i \left (d x +c \right )} b^{2}}{8 a^{4} d}+\frac {5 i b \,{\mathrm e}^{-i \left (d x +c \right )}}{4 a^{3} d}+\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )} b^{2}}{8 a^{4} d}+\frac {2 i \left (a^{2}-b^{2}\right ) b^{2} \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{a^{6} d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {2 \sqrt {a^{2}-b^{2}}\, b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \sqrt {a^{2}-b^{2}}-b}{a}\right )}{d \,a^{4}}-\frac {5 \sqrt {a^{2}-b^{2}}\, b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \sqrt {a^{2}-b^{2}}-b}{a}\right )}{d \,a^{6}}-\frac {2 \sqrt {a^{2}-b^{2}}\, b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {b +i \sqrt {a^{2}-b^{2}}}{a}\right )}{d \,a^{4}}+\frac {5 \sqrt {a^{2}-b^{2}}\, b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {b +i \sqrt {a^{2}-b^{2}}}{a}\right )}{d \,a^{6}}+\frac {\sin \left (4 d x +4 c \right )}{32 a^{2} d}-\frac {b \sin \left (3 d x +3 c \right )}{6 d \,a^{3}}\)	\(494\)

[In]

int(sin(d*x+c)^4/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(2*(a-b)*(a+b)*b/a^6*(-a*b*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)-(2*a^2-5
*b^2)/((a-b)*(a+b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))+2/a^6*(((3/8*a^4+2*a^3*b-3/2*
a^2*b^2-4*a*b^3)*tan(1/2*d*x+1/2*c)^7+(26/3*a^3*b-3/2*a^2*b^2-12*a*b^3+11/8*a^4)*tan(1/2*d*x+1/2*c)^5+(-11/8*a
^4+3/2*a^2*b^2+26/3*a^3*b-12*a*b^3)*tan(1/2*d*x+1/2*c)^3+(2*a^3*b-4*a*b^3-3/8*a^4+3/2*a^2*b^2)*tan(1/2*d*x+1/2
*c))/(1+tan(1/2*d*x+1/2*c)^2)^4+1/8*(3*a^4-36*a^2*b^2+40*b^4)*arctan(tan(1/2*d*x+1/2*c))))

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 581, normalized size of antiderivative = 2.23 \[ \int \frac {\sin ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\left [\frac {3 \, {\left (3 \, a^{5} - 36 \, a^{3} b^{2} + 40 \, a b^{4}\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (3 \, a^{4} b - 36 \, a^{2} b^{3} + 40 \, b^{5}\right )} d x - 12 \, {\left (2 \, a^{2} b^{2} - 5 \, b^{4} + {\left (2 \, a^{3} b - 5 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (6 \, a^{5} \cos \left (d x + c\right )^{4} - 10 \, a^{4} b \cos \left (d x + c\right )^{3} + 88 \, a^{3} b^{2} - 120 \, a b^{4} - 5 \, {\left (3 \, a^{5} - 4 \, a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (49 \, a^{4} b - 60 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{7} d \cos \left (d x + c\right ) + a^{6} b d\right )}}, \frac {3 \, {\left (3 \, a^{5} - 36 \, a^{3} b^{2} + 40 \, a b^{4}\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (3 \, a^{4} b - 36 \, a^{2} b^{3} + 40 \, b^{5}\right )} d x - 24 \, {\left (2 \, a^{2} b^{2} - 5 \, b^{4} + {\left (2 \, a^{3} b - 5 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (6 \, a^{5} \cos \left (d x + c\right )^{4} - 10 \, a^{4} b \cos \left (d x + c\right )^{3} + 88 \, a^{3} b^{2} - 120 \, a b^{4} - 5 \, {\left (3 \, a^{5} - 4 \, a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (49 \, a^{4} b - 60 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{7} d \cos \left (d x + c\right ) + a^{6} b d\right )}}\right ] \]

[In]

integrate(sin(d*x+c)^4/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/24*(3*(3*a^5 - 36*a^3*b^2 + 40*a*b^4)*d*x*cos(d*x + c) + 3*(3*a^4*b - 36*a^2*b^3 + 40*b^5)*d*x - 12*(2*a^2*
b^2 - 5*b^4 + (2*a^3*b - 5*a*b^3)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*
x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(
d*x + c) + b^2)) + (6*a^5*cos(d*x + c)^4 - 10*a^4*b*cos(d*x + c)^3 + 88*a^3*b^2 - 120*a*b^4 - 5*(3*a^5 - 4*a^3
*b^2)*cos(d*x + c)^2 + (49*a^4*b - 60*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/(a^7*d*cos(d*x + c) + a^6*b*d), 1/2
4*(3*(3*a^5 - 36*a^3*b^2 + 40*a*b^4)*d*x*cos(d*x + c) + 3*(3*a^4*b - 36*a^2*b^3 + 40*b^5)*d*x - 24*(2*a^2*b^2
- 5*b^4 + (2*a^3*b - 5*a*b^3)*cos(d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a
^2 - b^2)*sin(d*x + c))) + (6*a^5*cos(d*x + c)^4 - 10*a^4*b*cos(d*x + c)^3 + 88*a^3*b^2 - 120*a*b^4 - 5*(3*a^5
- 4*a^3*b^2)*cos(d*x + c)^2 + (49*a^4*b - 60*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/(a^7*d*cos(d*x + c) + a^6*b
*d)]

Sympy [F]

\[ \int \frac {\sin ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\sin ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]

[In]

integrate(sin(d*x+c)**4/(a+b*sec(d*x+c))**2,x)

[Out]

Integral(sin(c + d*x)**4/(a + b*sec(c + d*x))**2, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sin ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(sin(d*x+c)^4/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 482, normalized size of antiderivative = 1.85 \[ \int \frac {\sin ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {3 \, {\left (3 \, a^{4} - 36 \, a^{2} b^{2} + 40 \, b^{4}\right )} {\left (d x + c\right )}}{a^{6}} - \frac {48 \, {\left (2 \, a^{4} b - 7 \, a^{2} b^{3} + 5 \, b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{6}} - \frac {48 \, {\left (a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )} a^{5}} + \frac {2 \, {\left (9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 48 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 33 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 208 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 288 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 33 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 208 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 288 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 48 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 96 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a^{5}}}{24 \, d} \]

[In]

integrate(sin(d*x+c)^4/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(3*(3*a^4 - 36*a^2*b^2 + 40*b^4)*(d*x + c)/a^6 - 48*(2*a^4*b - 7*a^2*b^3 + 5*b^5)*(pi*floor(1/2*(d*x + c)
/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/(sqr
t(-a^2 + b^2)*a^6) - 48*(a^2*b^2*tan(1/2*d*x + 1/2*c) - b^4*tan(1/2*d*x + 1/2*c))/((a*tan(1/2*d*x + 1/2*c)^2 -
b*tan(1/2*d*x + 1/2*c)^2 - a - b)*a^5) + 2*(9*a^3*tan(1/2*d*x + 1/2*c)^7 + 48*a^2*b*tan(1/2*d*x + 1/2*c)^7 -
36*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 96*b^3*tan(1/2*d*x + 1/2*c)^7 + 33*a^3*tan(1/2*d*x + 1/2*c)^5 + 208*a^2*b*ta
n(1/2*d*x + 1/2*c)^5 - 36*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 288*b^3*tan(1/2*d*x + 1/2*c)^5 - 33*a^3*tan(1/2*d*x +
1/2*c)^3 + 208*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 36*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 288*b^3*tan(1/2*d*x + 1/2*c)^
3 - 9*a^3*tan(1/2*d*x + 1/2*c) + 48*a^2*b*tan(1/2*d*x + 1/2*c) + 36*a*b^2*tan(1/2*d*x + 1/2*c) - 96*b^3*tan(1/
2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a^5))/d

Mupad [B] (verification not implemented)

Time = 15.42 (sec) , antiderivative size = 2804, normalized size of antiderivative = 10.74 \[ \int \frac {\sin ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

int(sin(c + d*x)^4/(a + b/cos(c + d*x))^2,x)

[Out]

((tan(c/2 + (d*x)/2)^5*(33*a^4 - 360*b^4 + 244*a^2*b^2))/(6*a^5) - (tan(c/2 + (d*x)/2)^3*(60*a*b^3 - 59*a^3*b
+ 12*a^4 + 240*b^4 - 176*a^2*b^2))/(6*a^5) - (tan(c/2 + (d*x)/2)^7*(59*a^3*b - 60*a*b^3 + 12*a^4 + 240*b^4 - 1
76*a^2*b^2))/(6*a^5) + (tan(c/2 + (d*x)/2)^9*(a - b)*(20*a*b^2 - 16*a^2*b - 3*a^3 + 40*b^3))/(4*a^5) + (tan(c/
2 + (d*x)/2)*(a + b)*(20*a*b^2 + 16*a^2*b - 3*a^3 - 40*b^3))/(4*a^5))/(d*(a + b - tan(c/2 + (d*x)/2)^10*(a - b
) + tan(c/2 + (d*x)/2)^2*(3*a + 5*b) + tan(c/2 + (d*x)/2)^4*(2*a + 10*b) - tan(c/2 + (d*x)/2)^8*(3*a - 5*b) -
tan(c/2 + (d*x)/2)^6*(2*a - 10*b))) + (atan(((((((76*a^17*b - 12*a^18 - 160*a^12*b^6 + 240*a^13*b^5 + 144*a^14
*b^4 - 316*a^15*b^3 + 28*a^16*b^2)/a^15 - (tan(c/2 + (d*x)/2)*(a^4*3i + b^4*40i - a^2*b^2*36i)*(128*a^14*b + 1
28*a^12*b^3 - 256*a^13*b^2))/(16*a^16))*(a^4*3i + b^4*40i - a^2*b^2*36i))/(8*a^6) + (tan(c/2 + (d*x)/2)*(6400*
a*b^10 - 27*a^10*b + 9*a^11 - 3200*b^11 + 2560*a^2*b^9 - 11520*a^3*b^8 + 2688*a^4*b^7 + 6144*a^5*b^6 - 2600*a^
6*b^5 - 904*a^7*b^4 + 383*a^8*b^3 + 67*a^9*b^2))/(2*a^10))*(a^4*3i + b^4*40i - a^2*b^2*36i)*1i)/(8*a^6) - ((((
(76*a^17*b - 12*a^18 - 160*a^12*b^6 + 240*a^13*b^5 + 144*a^14*b^4 - 316*a^15*b^3 + 28*a^16*b^2)/a^15 + (tan(c/
2 + (d*x)/2)*(a^4*3i + b^4*40i - a^2*b^2*36i)*(128*a^14*b + 128*a^12*b^3 - 256*a^13*b^2))/(16*a^16))*(a^4*3i +
b^4*40i - a^2*b^2*36i))/(8*a^6) - (tan(c/2 + (d*x)/2)*(6400*a*b^10 - 27*a^10*b + 9*a^11 - 3200*b^11 + 2560*a^
2*b^9 - 11520*a^3*b^8 + 2688*a^4*b^7 + 6144*a^5*b^6 - 2600*a^6*b^5 - 904*a^7*b^4 + 383*a^8*b^3 + 67*a^9*b^2))/
(2*a^10))*(a^4*3i + b^4*40i - a^2*b^2*36i)*1i)/(8*a^6))/((12000*a*b^13 + 18*a^13*b - 8000*b^14 + 21600*a^2*b^1
2 - 37400*a^3*b^11 - 19240*a^4*b^10 + 43960*a^5*b^9 + 4672*a^6*b^8 - 23963*a^7*b^7 + 1742*a^8*b^6 + 5958*a^9*b
^5 - 834*a^10*b^4 - 573*a^11*b^3 + 60*a^12*b^2)/a^15 + (((((76*a^17*b - 12*a^18 - 160*a^12*b^6 + 240*a^13*b^5
+ 144*a^14*b^4 - 316*a^15*b^3 + 28*a^16*b^2)/a^15 - (tan(c/2 + (d*x)/2)*(a^4*3i + b^4*40i - a^2*b^2*36i)*(128*
a^14*b + 128*a^12*b^3 - 256*a^13*b^2))/(16*a^16))*(a^4*3i + b^4*40i - a^2*b^2*36i))/(8*a^6) + (tan(c/2 + (d*x)
/2)*(6400*a*b^10 - 27*a^10*b + 9*a^11 - 3200*b^11 + 2560*a^2*b^9 - 11520*a^3*b^8 + 2688*a^4*b^7 + 6144*a^5*b^6
- 2600*a^6*b^5 - 904*a^7*b^4 + 383*a^8*b^3 + 67*a^9*b^2))/(2*a^10))*(a^4*3i + b^4*40i - a^2*b^2*36i))/(8*a^6)
+ (((((76*a^17*b - 12*a^18 - 160*a^12*b^6 + 240*a^13*b^5 + 144*a^14*b^4 - 316*a^15*b^3 + 28*a^16*b^2)/a^15 +
(tan(c/2 + (d*x)/2)*(a^4*3i + b^4*40i - a^2*b^2*36i)*(128*a^14*b + 128*a^12*b^3 - 256*a^13*b^2))/(16*a^16))*(a
^4*3i + b^4*40i - a^2*b^2*36i))/(8*a^6) - (tan(c/2 + (d*x)/2)*(6400*a*b^10 - 27*a^10*b + 9*a^11 - 3200*b^11 +
2560*a^2*b^9 - 11520*a^3*b^8 + 2688*a^4*b^7 + 6144*a^5*b^6 - 2600*a^6*b^5 - 904*a^7*b^4 + 383*a^8*b^3 + 67*a^9
*b^2))/(2*a^10))*(a^4*3i + b^4*40i - a^2*b^2*36i))/(8*a^6)))*(a^4*3i + b^4*40i - a^2*b^2*36i)*1i)/(4*a^6*d) +
(b*atan(((b*(a^2 - b^2)^(1/2)*(2*a^2 - 5*b^2)*((tan(c/2 + (d*x)/2)*(6400*a*b^10 - 27*a^10*b + 9*a^11 - 3200*b^
11 + 2560*a^2*b^9 - 11520*a^3*b^8 + 2688*a^4*b^7 + 6144*a^5*b^6 - 2600*a^6*b^5 - 904*a^7*b^4 + 383*a^8*b^3 + 6
7*a^9*b^2))/(2*a^10) + (b*((76*a^17*b - 12*a^18 - 160*a^12*b^6 + 240*a^13*b^5 + 144*a^14*b^4 - 316*a^15*b^3 +
28*a^16*b^2)/a^15 - (b*tan(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)*(2*a^2 - 5*b^2)*(128*a^14*b + 128*a^12*b^3 - 256*a
^13*b^2))/(2*a^16))*(a^2 - b^2)^(1/2)*(2*a^2 - 5*b^2))/a^6)*1i)/a^6 + (b*(a^2 - b^2)^(1/2)*(2*a^2 - 5*b^2)*((t
an(c/2 + (d*x)/2)*(6400*a*b^10 - 27*a^10*b + 9*a^11 - 3200*b^11 + 2560*a^2*b^9 - 11520*a^3*b^8 + 2688*a^4*b^7
+ 6144*a^5*b^6 - 2600*a^6*b^5 - 904*a^7*b^4 + 383*a^8*b^3 + 67*a^9*b^2))/(2*a^10) - (b*((76*a^17*b - 12*a^18 -
160*a^12*b^6 + 240*a^13*b^5 + 144*a^14*b^4 - 316*a^15*b^3 + 28*a^16*b^2)/a^15 + (b*tan(c/2 + (d*x)/2)*(a^2 -
b^2)^(1/2)*(2*a^2 - 5*b^2)*(128*a^14*b + 128*a^12*b^3 - 256*a^13*b^2))/(2*a^16))*(a^2 - b^2)^(1/2)*(2*a^2 - 5*
b^2))/a^6)*1i)/a^6)/((12000*a*b^13 + 18*a^13*b - 8000*b^14 + 21600*a^2*b^12 - 37400*a^3*b^11 - 19240*a^4*b^10
+ 43960*a^5*b^9 + 4672*a^6*b^8 - 23963*a^7*b^7 + 1742*a^8*b^6 + 5958*a^9*b^5 - 834*a^10*b^4 - 573*a^11*b^3 + 6
0*a^12*b^2)/a^15 + (b*(a^2 - b^2)^(1/2)*(2*a^2 - 5*b^2)*((tan(c/2 + (d*x)/2)*(6400*a*b^10 - 27*a^10*b + 9*a^11
- 3200*b^11 + 2560*a^2*b^9 - 11520*a^3*b^8 + 2688*a^4*b^7 + 6144*a^5*b^6 - 2600*a^6*b^5 - 904*a^7*b^4 + 383*a
^8*b^3 + 67*a^9*b^2))/(2*a^10) + (b*((76*a^17*b - 12*a^18 - 160*a^12*b^6 + 240*a^13*b^5 + 144*a^14*b^4 - 316*a
^15*b^3 + 28*a^16*b^2)/a^15 - (b*tan(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)*(2*a^2 - 5*b^2)*(128*a^14*b + 128*a^12*b
^3 - 256*a^13*b^2))/(2*a^16))*(a^2 - b^2)^(1/2)*(2*a^2 - 5*b^2))/a^6))/a^6 - (b*(a^2 - b^2)^(1/2)*(2*a^2 - 5*b
^2)*((tan(c/2 + (d*x)/2)*(6400*a*b^10 - 27*a^10*b + 9*a^11 - 3200*b^11 + 2560*a^2*b^9 - 11520*a^3*b^8 + 2688*a
^4*b^7 + 6144*a^5*b^6 - 2600*a^6*b^5 - 904*a^7*b^4 + 383*a^8*b^3 + 67*a^9*b^2))/(2*a^10) - (b*((76*a^17*b - 12
*a^18 - 160*a^12*b^6 + 240*a^13*b^5 + 144*a^14*b^4 - 316*a^15*b^3 + 28*a^16*b^2)/a^15 + (b*tan(c/2 + (d*x)/2)*
(a^2 - b^2)^(1/2)*(2*a^2 - 5*b^2)*(128*a^14*b + 128*a^12*b^3 - 256*a^13*b^2))/(2*a^16))*(a^2 - b^2)^(1/2)*(2*a
^2 - 5*b^2))/a^6))/a^6))*(a^2 - b^2)^(1/2)*(2*a^2 - 5*b^2)*2i)/(a^6*d)