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

Optimal. Leaf size=261 \[ \frac{b \left (11 a^2-15 b^2\right ) \sin (c+d x)}{3 a^5 d}-\frac{\left (a^2-b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{a^2 b d (a \cos (c+d x)+b)}+\frac{\left (3 a^2-5 b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 a^3 b d}-\frac{\left (13 a^2-20 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 a^4 d}-\frac{2 b \sqrt{a-b} \sqrt{a+b} \left (2 a^2-5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^6 d}+\frac{x \left (-36 a^2 b^2+3 a^4+40 b^4\right )}{8 a^6}+\frac{\sin (c+d x) \cos ^3(c+d x)}{4 a^2 d} \]

[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]))

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Rubi [A]  time = 0.825278, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3872, 2892, 3049, 3023, 2735, 2659, 208} \[ \frac{b \left (11 a^2-15 b^2\right ) \sin (c+d x)}{3 a^5 d}-\frac{\left (a^2-b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{a^2 b d (a \cos (c+d x)+b)}+\frac{\left (3 a^2-5 b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 a^3 b d}-\frac{\left (13 a^2-20 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 a^4 d}-\frac{2 b \sqrt{a-b} \sqrt{a+b} \left (2 a^2-5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^6 d}+\frac{x \left (-36 a^2 b^2+3 a^4+40 b^4\right )}{8 a^6}+\frac{\sin (c+d x) \cos ^3(c+d x)}{4 a^2 d} \]

Antiderivative was successfully verified.

[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 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[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]

Rule 2892

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 3049

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 3023

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 2735

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 2659

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 208

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

Rubi steps

\begin{align*} \int \frac{\sin ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\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 ) \operatorname{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 ) \tanh ^{-1}\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]  time = 3.09277, size = 282, normalized size = 1.08 \[ \frac{\frac{40 a^3 b^2 \sin (3 (c+d x))-240 a^2 b^3 \sin (2 (c+d x))-24 a \left (-31 a^2 b^2+a^4+40 b^4\right ) \sin (c+d x)+24 a \left (-36 a^2 b^2+3 a^4+40 b^4\right ) (c+d x) \cos (c+d x)-864 a^2 b^3 c-864 a^2 b^3 d x+176 a^4 b \sin (2 (c+d x))-10 a^4 b \sin (4 (c+d x))+72 a^4 b c+72 a^4 b d x-21 a^5 \sin (3 (c+d x))+3 a^5 \sin (5 (c+d x))+960 b^5 c+960 b^5 d x}{a \cos (c+d x)+b}+\frac{384 b \left (-7 a^2 b^2+2 a^4+5 b^4\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}}{192 a^6 d} \]

Antiderivative was successfully verified.

[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)

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Maple [B]  time = 0.081, size = 883, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7+4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c
)^7*b-3/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*b^2-8/d/a^5/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d
*x+1/2*c)^7*b^3+52/3/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*b-3/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^
4*tan(1/2*d*x+1/2*c)^5*b^2-24/d/a^5/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*b^3+11/4/d/a^2/(1+tan(1/2*
d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5-11/4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3+3/d/a^4/(1+tan
(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*b^2+52/3/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*b-24/
d/a^5/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*b^3+4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c
)*b-8/d/a^5/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)*b^3-3/4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x
+1/2*c)+3/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)*b^2-9/d/a^4*arctan(tan(1/2*d*x+1/2*c))*b^2+10/d/
a^6*arctan(tan(1/2*d*x+1/2*c))*b^4+3/4/d/a^2*arctan(tan(1/2*d*x+1/2*c))-2/d*b^2/a^3*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/d*b^4/a^5*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)-4/d*b/a^2/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))+14/d*b^3
/a^4/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-10/d*b^5/a^6/((a+b)*(a-b))^(1/2
)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.18792, size = 1347, normalized size = 5.16 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin ^{4}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.28715, size = 651, normalized size = 2.49 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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