\(\int \frac {\sec ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [448]

Optimal result

Integrand size = 21, antiderivative size = 193 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {10 a b^4 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2} d}+\frac {b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {\sec (c+d x) \left (15 a b^3+\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d} \]

[Out]

10*a*b^4*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(7/2)/d+b*sec(d*x+c)^3/(a^2-b^2)/d/(a+b*si
n(d*x+c))-1/3*sec(d*x+c)^3*(5*a*b-(a^2+4*b^2)*sin(d*x+c))/(a^2-b^2)^2/d+1/3*sec(d*x+c)*(15*a*b^3+(2*a^4-9*a^2*
b^2-8*b^4)*sin(d*x+c))/(a^2-b^2)^3/d

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2773, 2945, 12, 2739, 632, 210} \[ \int \frac {\sec ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {10 a b^4 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac {b \sec ^3(c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 d \left (a^2-b^2\right )^2}+\frac {\sec (c+d x) \left (\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)+15 a b^3\right )}{3 d \left (a^2-b^2\right )^3} \]

[In]

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

[Out]

(10*a*b^4*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2)*d) + (b*Sec[c + d*x]^3)/((a^2 -
b^2)*d*(a + b*Sin[c + d*x])) - (Sec[c + d*x]^3*(5*a*b - (a^2 + 4*b^2)*Sin[c + d*x]))/(3*(a^2 - b^2)^2*d) + (S
ec[c + d*x]*(15*a*b^3 + (2*a^4 - 9*a^2*b^2 - 8*b^4)*Sin[c + d*x]))/(3*(a^2 - b^2)^3*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*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 2773

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

Rule 2945

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c - a*d - (a*c -
b*d)*Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

Rubi steps \begin{align*} \text {integral}& = \frac {b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\int \frac {\sec ^4(c+d x) (-a+4 b \sin (c+d x))}{a+b \sin (c+d x)} \, dx}{-a^2+b^2} \\ & = \frac {b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {\int \frac {\sec ^2(c+d x) \left (a \left (2 a^2-7 b^2\right )+2 b \left (a^2+4 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{3 \left (a^2-b^2\right )^2} \\ & = \frac {b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {\sec (c+d x) \left (15 a b^3+\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}-\frac {\int -\frac {15 a b^4}{a+b \sin (c+d x)} \, dx}{3 \left (a^2-b^2\right )^3} \\ & = \frac {b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {\sec (c+d x) \left (15 a b^3+\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}+\frac {\left (5 a b^4\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^3} \\ & = \frac {b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {\sec (c+d x) \left (15 a b^3+\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}+\frac {\left (10 a b^4\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^3 d} \\ & = \frac {b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {\sec (c+d x) \left (15 a b^3+\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}-\frac {\left (20 a b^4\right ) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^3 d} \\ & = \frac {10 a b^4 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2} d}+\frac {b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {\sec (c+d x) \left (15 a b^3+\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.63 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.74 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {120 a b^4 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac {1}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {4 (2 a+5 b) \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {1}{(a-b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (2 a-5 b) \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {12 b^5 \cos (c+d x)}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))}}{12 d} \]

[In]

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

[Out]

((120*a*b^4*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(7/2) + 1/((a + b)^2*(Cos[(c + d*x)/
2] - Sin[(c + d*x)/2])^2) + (2*Sin[(c + d*x)/2])/((a + b)^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) + (4*(2*a
+ 5*b)*Sin[(c + d*x)/2])/((a + b)^3*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (2*Sin[(c + d*x)/2])/((a - b)^2*
(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) - 1/((a - b)^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (4*(2*a - 5
*b)*Sin[(c + d*x)/2])/((a - b)^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (12*b^5*Cos[c + d*x])/((a - b)^3*(a
+ b)^3*(a + b*Sin[c + d*x])))/(12*d)

Maple [A] (verified)

Time = 2.80 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.31

method	result	size
derivativedivides	\(\frac {-\frac {1}{3 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a +2 b}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 b^{4} \left (\frac {\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}+b}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {5 a \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}-\frac {1}{3 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {a -2 b}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\)	\(252\)
default	\(\frac {-\frac {1}{3 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a +2 b}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 b^{4} \left (\frac {\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}+b}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {5 a \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}-\frac {1}{3 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {a -2 b}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\)	\(252\)
risch	\(\frac {-\frac {8 i a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}}{3}+10 a \,b^{4} {\mathrm e}^{7 i \left (d x +c \right )}+\frac {70 a \,b^{4} {\mathrm e}^{5 i \left (d x +c \right )}}{3}+\frac {20 a^{3} b^{2} {\mathrm e}^{5 i \left (d x +c \right )}}{3}+\frac {88 a^{3} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}}{3}+\frac {26 a \,b^{4} {\mathrm e}^{3 i \left (d x +c \right )}}{3}+\frac {16 i b^{5}}{3}+\frac {2 a \,b^{4} {\mathrm e}^{i \left (d x +c \right )}}{3}+12 a^{3} b^{2} {\mathrm e}^{i \left (d x +c \right )}+22 i a^{2} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+\frac {32 i b^{5} {\mathrm e}^{2 i \left (d x +c \right )}}{3}-\frac {8 a^{5} {\mathrm e}^{i \left (d x +c \right )}}{3}-8 a^{5} {\mathrm e}^{3 i \left (d x +c \right )}+6 i a^{2} b^{3}+10 i a^{2} b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-\frac {20 i a^{4} b \,{\mathrm e}^{4 i \left (d x +c \right )}}{3}+\frac {110 i a^{2} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}}{3}-\frac {4 i a^{4} b}{3}}{\left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{3} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right ) d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \left (a^{2}-b^{2}\right )}-\frac {5 b^{4} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {5 b^{4} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}\)	\(504\)

[In]

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

[Out]

1/d*(-1/3/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)^3-1/2/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)^2-1/(a+b)^3*(a+2*b)/(tan(1/2*d*x
+1/2*c)-1)+2*b^4/(a-b)^3/(a+b)^3*((b^2/a*tan(1/2*d*x+1/2*c)+b)/(a*tan(1/2*d*x+1/2*c)^2+2*b*tan(1/2*d*x+1/2*c)+
a)+5*a/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2)))-1/3/(a-b)^2/(tan(1/2*d*x+1/2*
c)+1)^3+1/2/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)^2-1/(a-b)^3*(a-2*b)/(tan(1/2*d*x+1/2*c)+1))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 782, normalized size of antiderivative = 4.05 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\left [-\frac {2 \, a^{6} b - 6 \, a^{4} b^{3} + 6 \, a^{2} b^{5} - 2 \, b^{7} + 2 \, {\left (2 \, a^{6} b - 11 \, a^{4} b^{3} + a^{2} b^{5} + 8 \, b^{7}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{6} b + 2 \, a^{4} b^{3} - 7 \, a^{2} b^{5} + 4 \, b^{7}\right )} \cos \left (d x + c\right )^{2} - 15 \, {\left (a b^{5} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a^{2} b^{4} \cos \left (d x + c\right )^{3}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6} + {\left (2 \, a^{7} - 11 \, a^{5} b^{2} + 16 \, a^{3} b^{4} - 7 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left ({\left (a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + {\left (a^{9} - 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} - 4 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right )^{3}\right )}}, -\frac {a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7} + {\left (2 \, a^{6} b - 11 \, a^{4} b^{3} + a^{2} b^{5} + 8 \, b^{7}\right )} \cos \left (d x + c\right )^{4} - {\left (a^{6} b + 2 \, a^{4} b^{3} - 7 \, a^{2} b^{5} + 4 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (a b^{5} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a^{2} b^{4} \cos \left (d x + c\right )^{3}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6} + {\left (2 \, a^{7} - 11 \, a^{5} b^{2} + 16 \, a^{3} b^{4} - 7 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left ({\left (a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + {\left (a^{9} - 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} - 4 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right )^{3}\right )}}\right ] \]

[In]

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

[Out]

[-1/6*(2*a^6*b - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 + 2*(2*a^6*b - 11*a^4*b^3 + a^2*b^5 + 8*b^7)*cos(d*x + c)^4 - 2
*(a^6*b + 2*a^4*b^3 - 7*a^2*b^5 + 4*b^7)*cos(d*x + c)^2 - 15*(a*b^5*cos(d*x + c)^3*sin(d*x + c) + a^2*b^4*cos(
d*x + c)^3)*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*
x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2))
- 2*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 + (2*a^7 - 11*a^5*b^2 + 16*a^3*b^4 - 7*a*b^6)*cos(d*x + c)^2)*sin(d*
x + c))/((a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7 + b^9)*d*cos(d*x + c)^3*sin(d*x + c) + (a^9 - 4*a^7*b^2 +
6*a^5*b^4 - 4*a^3*b^6 + a*b^8)*d*cos(d*x + c)^3), -1/3*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7 + (2*a^6*b - 11*a^
4*b^3 + a^2*b^5 + 8*b^7)*cos(d*x + c)^4 - (a^6*b + 2*a^4*b^3 - 7*a^2*b^5 + 4*b^7)*cos(d*x + c)^2 + 15*(a*b^5*c
os(d*x + c)^3*sin(d*x + c) + a^2*b^4*cos(d*x + c)^3)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 -
b^2)*cos(d*x + c))) - (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 + (2*a^7 - 11*a^5*b^2 + 16*a^3*b^4 - 7*a*b^6)*cos(d
*x + c)^2)*sin(d*x + c))/((a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7 + b^9)*d*cos(d*x + c)^3*sin(d*x + c) + (a
^9 - 4*a^7*b^2 + 6*a^5*b^4 - 4*a^3*b^6 + a*b^8)*d*cos(d*x + c)^3)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate(sec(d*x+c)^4/(a+b*sin(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*b^2-4*a^2>0)', see `assume?`
for more de

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 427 vs. \(2 (184) = 368\).

Time = 0.35 (sec) , antiderivative size = 427, normalized size of antiderivative = 2.21 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 \, {\left (\frac {15 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a b^{4}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, {\left (b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a b^{5}\right )}}{{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}} - \frac {3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 18 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a^{3} b + 14 \, a b^{3}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )}}{3 \, d} \]

[In]

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

[Out]

2/3*(15*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))*a*b^4
/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(a^2 - b^2)) + 3*(b^6*tan(1/2*d*x + 1/2*c) + a*b^5)/((a^7 - 3*a^5*b^
2 + 3*a^3*b^4 - a*b^6)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)) - (3*a^4*tan(1/2*d*x + 1/2*c
)^5 - 9*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*b^4*tan(1/2*d*x + 1/2*c)^5 - 6*a^3*b*tan(1/2*d*x + 1/2*c)^4 + 18*a*
b^3*tan(1/2*d*x + 1/2*c)^4 - 2*a^4*tan(1/2*d*x + 1/2*c)^3 + 18*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 8*b^4*tan(1/2*
d*x + 1/2*c)^3 - 24*a*b^3*tan(1/2*d*x + 1/2*c)^2 + 3*a^4*tan(1/2*d*x + 1/2*c) - 9*a^2*b^2*tan(1/2*d*x + 1/2*c)
- 6*b^4*tan(1/2*d*x + 1/2*c) - 2*a^3*b + 14*a*b^3)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(tan(1/2*d*x + 1/2*c)
^2 - 1)^3))/d

Mupad [B] (verification not implemented)

Time = 7.93 (sec) , antiderivative size = 727, normalized size of antiderivative = 3.77 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {2\,\left (-2\,a^4\,b+14\,a^2\,b^3+3\,b^5\right )}{3\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {10\,b^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-4\,a^4\,b+28\,a^2\,b^3+21\,b^5\right )}{3\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^6-13\,a^4\,b^2+22\,a^2\,b^4+3\,b^6\right )}{3\,a\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (-a^6+3\,a^4\,b^2+2\,a^2\,b^4+b^6\right )}{a\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (a^6-3\,a^4\,b^2+38\,a^2\,b^4+9\,b^6\right )}{3\,a\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^6+9\,a^4\,b^2-46\,a^2\,b^4-9\,b^6\right )}{3\,a\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {10\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-2\,a^4+6\,a^2\,b^2+5\,b^4\right )}{3\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-6\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}+\frac {10\,a\,b^4\,\mathrm {atan}\left (\frac {\frac {5\,a\,b^4\,\left (2\,a^6\,b-6\,a^4\,b^3+6\,a^2\,b^5-2\,b^7\right )}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}+\frac {10\,a^2\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}}{10\,a\,b^4}\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]

[In]

int(1/(cos(c + d*x)^4*(a + b*sin(c + d*x))^2),x)

[Out]

((2*(3*b^5 - 2*a^4*b + 14*a^2*b^3))/(3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (10*b^5*tan(c/2 + (d*x)/2)^6)/(a
^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (2*tan(c/2 + (d*x)/2)^2*(21*b^5 - 4*a^4*b + 28*a^2*b^3))/(3*(a^6 - b^6 + 3
*a^2*b^4 - 3*a^4*b^2)) + (2*tan(c/2 + (d*x)/2)*(3*a^6 + 3*b^6 + 22*a^2*b^4 - 13*a^4*b^2))/(3*a*(a^6 - b^6 + 3*
a^2*b^4 - 3*a^4*b^2)) - (2*tan(c/2 + (d*x)/2)^7*(b^6 - a^6 + 2*a^2*b^4 + 3*a^4*b^2))/(a*(a^6 - b^6 + 3*a^2*b^4
- 3*a^4*b^2)) + (2*tan(c/2 + (d*x)/2)^5*(a^6 + 9*b^6 + 38*a^2*b^4 - 3*a^4*b^2))/(3*a*(a^6 - b^6 + 3*a^2*b^4 -
3*a^4*b^2)) + (2*tan(c/2 + (d*x)/2)^3*(a^6 - 9*b^6 - 46*a^2*b^4 + 9*a^4*b^2))/(3*a*(a^6 - b^6 + 3*a^2*b^4 - 3
*a^4*b^2)) + (10*b*tan(c/2 + (d*x)/2)^4*(5*b^4 - 2*a^4 + 6*a^2*b^2))/(3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)))/
(d*(a + 2*b*tan(c/2 + (d*x)/2) - 2*a*tan(c/2 + (d*x)/2)^2 + 2*a*tan(c/2 + (d*x)/2)^6 - a*tan(c/2 + (d*x)/2)^8
- 6*b*tan(c/2 + (d*x)/2)^3 + 6*b*tan(c/2 + (d*x)/2)^5 - 2*b*tan(c/2 + (d*x)/2)^7)) + (10*a*b^4*atan(((5*a*b^4*
(2*a^6*b - 2*b^7 + 6*a^2*b^5 - 6*a^4*b^3))/((a + b)^(7/2)*(a - b)^(7/2)) + (10*a^2*b^4*tan(c/2 + (d*x)/2)*(a^6
- b^6 + 3*a^2*b^4 - 3*a^4*b^2))/((a + b)^(7/2)*(a - b)^(7/2)))/(10*a*b^4)))/(d*(a + b)^(7/2)*(a - b)^(7/2))