3.1352 \(\int \frac {\tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=177 \[ \frac {a \tan ^3(c+d x)}{3 d \left (a^2-b^2\right )}-\frac {b \sec ^3(c+d x)}{3 d \left (a^2-b^2\right )}+\frac {a^2 b \sec (c+d x)}{d \left (a^2-b^2\right )^2}+\frac {b \sec (c+d x)}{d \left (a^2-b^2\right )}+\frac {2 a^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}-\frac {a^3 \tan (c+d x)}{d \left (a^2-b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2727, 2607, 30, 2606, 3767, 8, 2660, 618, 204} \[ \frac {2 a^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}+\frac {a \tan ^3(c+d x)}{3 d \left (a^2-b^2\right )}-\frac {a^3 \tan (c+d x)}{d \left (a^2-b^2\right )^2}-\frac {b \sec ^3(c+d x)}{3 d \left (a^2-b^2\right )}+\frac {a^2 b \sec (c+d x)}{d \left (a^2-b^2\right )^2}+\frac {b \sec (c+d x)}{d \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^4*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(5/2)*d) + (a^2*b*Sec[c + d*x])/((a^2 -
b^2)^2*d) + (b*Sec[c + d*x])/((a^2 - b^2)*d) - (b*Sec[c + d*x]^3)/(3*(a^2 - b^2)*d) - (a^3*Tan[c + d*x])/((a^2
- b^2)^2*d) + (a*Tan[c + d*x]^3)/(3*(a^2 - b^2)*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 204

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

Rule 618

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 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2660

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 2727

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \frac {\tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {a \int \sec ^2(c+d x) \tan ^2(c+d x) \, dx}{a^2-b^2}-\frac {a^2 \int \frac {\tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{a^2-b^2}-\frac {b \int \sec (c+d x) \tan ^3(c+d x) \, dx}{a^2-b^2}\\ &=-\frac {a^3 \int \sec ^2(c+d x) \, dx}{\left (a^2-b^2\right )^2}+\frac {a^4 \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2}+\frac {\left (a^2 b\right ) \int \sec (c+d x) \tan (c+d x) \, dx}{\left (a^2-b^2\right )^2}+\frac {a \operatorname {Subst}\left (\int x^2 \, dx,x,\tan (c+d x)\right )}{\left (a^2-b^2\right ) d}-\frac {b \operatorname {Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{\left (a^2-b^2\right ) d}\\ &=\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d}-\frac {b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {a \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {a^3 \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {\left (2 a^4\right ) \operatorname {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 )^2 d}+\frac {\left (a^2 b\right ) \operatorname {Subst}(\int 1 \, dx,x,\sec (c+d x))}{\left (a^2-b^2\right )^2 d}\\ &=\frac {a^2 b \sec (c+d x)}{\left (a^2-b^2\right )^2 d}+\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d}-\frac {b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {a^3 \tan (c+d x)}{\left (a^2-b^2\right )^2 d}+\frac {a \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {\left (4 a^4\right ) \operatorname {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 )^2 d}\\ &=\frac {2 a^4 \tan ^{-1}\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 )^{5/2} d}+\frac {a^2 b \sec (c+d x)}{\left (a^2-b^2\right )^2 d}+\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d}-\frac {b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {a^3 \tan (c+d x)}{\left (a^2-b^2\right )^2 d}+\frac {a \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d}\\ \end {align*}

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Mathematica [A]  time = 1.42, size = 195, normalized size = 1.10 \[ \frac {\frac {48 a^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {\sec ^3(c+d x) \left (8 a^3 \sin (3 (c+d x))+3 b \left (11 a^2-5 b^2\right ) \cos (c+d x)+12 b \left (b^2-2 a^2\right ) \cos (2 (c+d x))+11 a^2 b \cos (3 (c+d x))-16 a^2 b+6 a b^2 \sin (c+d x)-2 a b^2 \sin (3 (c+d x))-5 b^3 \cos (3 (c+d x))+4 b^3\right )}{(a-b)^2 (a+b)^2}}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((48*a^4*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) - (Sec[c + d*x]^3*(-16*a^2*b + 4*
b^3 + 3*b*(11*a^2 - 5*b^2)*Cos[c + d*x] + 12*b*(-2*a^2 + b^2)*Cos[2*(c + d*x)] + 11*a^2*b*Cos[3*(c + d*x)] - 5
*b^3*Cos[3*(c + d*x)] + 6*a*b^2*Sin[c + d*x] + 8*a^3*Sin[3*(c + d*x)] - 2*a*b^2*Sin[3*(c + d*x)]))/((a - b)^2*
(a + b)^2))/(24*d)

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fricas [A]  time = 0.74, size = 476, normalized size = 2.69 \[ \left [-\frac {3 \, \sqrt {-a^{2} + b^{2}} a^{4} \cos \left (d x + c\right )^{3} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} - 6 \, {\left (2 \, a^{4} b - 3 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} - {\left (4 \, a^{5} - 5 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}, -\frac {3 \, \sqrt {a^{2} - b^{2}} a^{4} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} + a^{4} b - 2 \, a^{2} b^{3} + b^{5} - 3 \, {\left (2 \, a^{4} b - 3 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} - {\left (4 \, a^{5} - 5 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(3*sqrt(-a^2 + b^2)*a^4*cos(d*x + c)^3*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^4*b - 4*a^2*b^3 + 2*b^5 - 6*(2*a^4*b - 3*a^2*b^3 + b^5)*cos(d*x + c)^2 - 2*(a^5 - 2*a^3*
b^2 + a*b^4 - (4*a^5 - 5*a^3*b^2 + a*b^4)*cos(d*x + c)^2)*sin(d*x + c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d
*cos(d*x + c)^3), -1/3*(3*sqrt(a^2 - b^2)*a^4*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*cos
(d*x + c)^3 + a^4*b - 2*a^2*b^3 + b^5 - 3*(2*a^4*b - 3*a^2*b^3 + b^5)*cos(d*x + c)^2 - (a^5 - 2*a^3*b^2 + a*b^
4 - (4*a^5 - 5*a^3*b^2 + a*b^4)*cos(d*x + c)^2)*sin(d*x + c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x +
c)^3)]

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giac [A]  time = 0.27, size = 241, normalized size = 1.36 \[ \frac {2 \, {\left (\frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{4}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 10 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a^{2} b + 2 \, b^{3}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*(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^4/((
a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + (3*a^3*tan(1/2*d*x + 1/2*c)^5 - 3*a^2*b*tan(1/2*d*x + 1/2*c)^4 - 10*
a^3*tan(1/2*d*x + 1/2*c)^3 + 4*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 12*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 6*b^3*tan(1/2*
d*x + 1/2*c)^2 + 3*a^3*tan(1/2*d*x + 1/2*c) - 5*a^2*b + 2*b^3)/((a^4 - 2*a^2*b^2 + b^4)*(tan(1/2*d*x + 1/2*c)^
2 - 1)^3))/d

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maple [A]  time = 0.39, size = 269, normalized size = 1.52 \[ -\frac {32}{3 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (32 a +32 b \right )}-\frac {16}{d \left (32 a +32 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a}{d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {b}{2 d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {32}{3 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (32 a -32 b \right )}+\frac {16}{d \left (32 a -32 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {a}{d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {b}{2 d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 a^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-32/3/d/(tan(1/2*d*x+1/2*c)-1)^3/(32*a+32*b)-16/d/(32*a+32*b)/(tan(1/2*d*x+1/2*c)-1)^2+1/d/(a+b)^2/(tan(1/2*d*
x+1/2*c)-1)*a+1/2/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)*b-32/3/d/(tan(1/2*d*x+1/2*c)+1)^3/(32*a-32*b)+16/d/(32*a-32
*b)/(tan(1/2*d*x+1/2*c)+1)^2+1/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)*a-1/2/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)*b+2/d*a
^4/(a-b)^2/(a+b)^2/(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))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^4/(a+b*sin(d*x+c)),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 details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 17.43, size = 372, normalized size = 2.10 \[ \frac {\frac {2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,\left (5\,a^2\,b-2\,b^3\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{a^4-2\,a^2\,b^2+b^4}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a\,b^2-5\,a^3\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2\,b-b^3\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a^4-2\,a^2\,b^2+b^4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}+\frac {2\,a^4\,\mathrm {atan}\left (\frac {\frac {a^4\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}+\frac {2\,a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}}{2\,a^4}\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*a^3*tan(c/2 + (d*x)/2))/(a^4 + b^4 - 2*a^2*b^2) - (2*(5*a^2*b - 2*b^3))/(3*(a^4 + b^4 - 2*a^2*b^2)) + (2*a
^3*tan(c/2 + (d*x)/2)^5)/(a^4 + b^4 - 2*a^2*b^2) + (4*tan(c/2 + (d*x)/2)^3*(2*a*b^2 - 5*a^3))/(3*(a^4 + b^4 -
2*a^2*b^2)) + (4*tan(c/2 + (d*x)/2)^2*(2*a^2*b - b^3))/(a^4 + b^4 - 2*a^2*b^2) - (2*a^2*b*tan(c/2 + (d*x)/2)^4
)/(a^4 + b^4 - 2*a^2*b^2))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1)) +
(2*a^4*atan(((a^4*(2*a^4*b + 2*b^5 - 4*a^2*b^3))/((a + b)^(5/2)*(a - b)^(5/2)) + (2*a^5*tan(c/2 + (d*x)/2)*(a^
4 + b^4 - 2*a^2*b^2))/((a + b)^(5/2)*(a - b)^(5/2)))/(2*a^4)))/(d*(a + b)^(5/2)*(a - b)^(5/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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