3.488 \(\int \frac{\cot ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=191 \[ -\frac{19 \cot (c+d x)}{8 a^2 d \sqrt{a \sin (c+d x)+a}}+\frac{45 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{8 a^{5/2} d}-\frac{4 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{a^{5/2} d}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 a^2 d \sqrt{a \sin (c+d x)+a}}+\frac{13 \cot (c+d x) \csc (c+d x)}{12 a^2 d \sqrt{a \sin (c+d x)+a}} \]

[Out]

(45*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(8*a^(5/2)*d) - (4*Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[
c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(a^(5/2)*d) - (19*Cot[c + d*x])/(8*a^2*d*Sqrt[a + a*Sin[c + d*x
]]) + (13*Cot[c + d*x]*Csc[c + d*x])/(12*a^2*d*Sqrt[a + a*Sin[c + d*x]]) - (Cot[c + d*x]*Csc[c + d*x]^2)/(3*a^
2*d*Sqrt[a + a*Sin[c + d*x]])

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Rubi [A]  time = 0.947277, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {2717, 2779, 2984, 2985, 2649, 206, 2773, 3044} \[ -\frac{19 \cot (c+d x)}{8 a^2 d \sqrt{a \sin (c+d x)+a}}+\frac{45 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{8 a^{5/2} d}-\frac{4 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{a^{5/2} d}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 a^2 d \sqrt{a \sin (c+d x)+a}}+\frac{13 \cot (c+d x) \csc (c+d x)}{12 a^2 d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(45*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(8*a^(5/2)*d) - (4*Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[
c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(a^(5/2)*d) - (19*Cot[c + d*x])/(8*a^2*d*Sqrt[a + a*Sin[c + d*x
]]) + (13*Cot[c + d*x]*Csc[c + d*x])/(12*a^2*d*Sqrt[a + a*Sin[c + d*x]]) - (Cot[c + d*x]*Csc[c + d*x]^2)/(3*a^
2*d*Sqrt[a + a*Sin[c + d*x]])

Rule 2717

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^4, x_Symbol] :> Dist[-2/(a*b), Int[(a
+ b*Sin[e + f*x])^(m + 2)/Sin[e + f*x]^3, x], x] + Dist[1/a^2, Int[((a + b*Sin[e + f*x])^(m + 2)*(1 + Sin[e +
f*x]^2))/Sin[e + f*x]^4, x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m - 1/2] && LtQ[m,
-1]

Rule 2779

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

Rule 2984

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

Rule 2985

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

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

Rule 2773

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3044

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

Rubi steps

\begin{align*} \int \frac{\cot ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=\frac{\int \frac{\csc ^4(c+d x) \left (1+\sin ^2(c+d x)\right )}{\sqrt{a+a \sin (c+d x)}} \, dx}{a^2}-\frac{2 \int \frac{\csc ^3(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx}{a^2}\\ &=\frac{\cot (c+d x) \csc (c+d x)}{a^2 d \sqrt{a+a \sin (c+d x)}}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 a^2 d \sqrt{a+a \sin (c+d x)}}+\frac{\int \frac{\csc ^3(c+d x) \left (-\frac{a}{2}+\frac{11}{2} a \sin (c+d x)\right )}{\sqrt{a+a \sin (c+d x)}} \, dx}{3 a^3}+\frac{\int \frac{\csc ^2(c+d x) (a-3 a \sin (c+d x))}{\sqrt{a+a \sin (c+d x)}} \, dx}{2 a^3}\\ &=-\frac{\cot (c+d x)}{2 a^2 d \sqrt{a+a \sin (c+d x)}}+\frac{13 \cot (c+d x) \csc (c+d x)}{12 a^2 d \sqrt{a+a \sin (c+d x)}}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 a^2 d \sqrt{a+a \sin (c+d x)}}+\frac{\int \frac{\csc ^2(c+d x) \left (\frac{45 a^2}{4}-\frac{3}{4} a^2 \sin (c+d x)\right )}{\sqrt{a+a \sin (c+d x)}} \, dx}{6 a^4}+\frac{\int \frac{\csc (c+d x) \left (-\frac{7 a^2}{2}+\frac{1}{2} a^2 \sin (c+d x)\right )}{\sqrt{a+a \sin (c+d x)}} \, dx}{2 a^4}\\ &=-\frac{19 \cot (c+d x)}{8 a^2 d \sqrt{a+a \sin (c+d x)}}+\frac{13 \cot (c+d x) \csc (c+d x)}{12 a^2 d \sqrt{a+a \sin (c+d x)}}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 a^2 d \sqrt{a+a \sin (c+d x)}}+\frac{\int \frac{\csc (c+d x) \left (-\frac{51 a^3}{8}+\frac{45}{8} a^3 \sin (c+d x)\right )}{\sqrt{a+a \sin (c+d x)}} \, dx}{6 a^5}-\frac{7 \int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{4 a^3}+\frac{2 \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx}{a^2}\\ &=-\frac{19 \cot (c+d x)}{8 a^2 d \sqrt{a+a \sin (c+d x)}}+\frac{13 \cot (c+d x) \csc (c+d x)}{12 a^2 d \sqrt{a+a \sin (c+d x)}}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 a^2 d \sqrt{a+a \sin (c+d x)}}-\frac{17 \int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{16 a^3}+\frac{2 \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx}{a^2}+\frac{7 \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,\frac{a \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{2 a^2 d}-\frac{4 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{a^2 d}\\ &=\frac{7 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{2 a^{5/2} d}-\frac{2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{a^{5/2} d}-\frac{19 \cot (c+d x)}{8 a^2 d \sqrt{a+a \sin (c+d x)}}+\frac{13 \cot (c+d x) \csc (c+d x)}{12 a^2 d \sqrt{a+a \sin (c+d x)}}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 a^2 d \sqrt{a+a \sin (c+d x)}}+\frac{17 \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,\frac{a \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{8 a^2 d}-\frac{4 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{a^2 d}\\ &=\frac{45 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{8 a^{5/2} d}-\frac{4 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{a^{5/2} d}-\frac{19 \cot (c+d x)}{8 a^2 d \sqrt{a+a \sin (c+d x)}}+\frac{13 \cot (c+d x) \csc (c+d x)}{12 a^2 d \sqrt{a+a \sin (c+d x)}}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 a^2 d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 2.3391, size = 332, normalized size = 1.74 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5 \left (-\frac{8 \csc ^9\left (\frac{1}{2} (c+d x)\right ) \left (-396 \sin \left (\frac{1}{2} (c+d x)\right )-218 \sin \left (\frac{3}{2} (c+d x)\right )+114 \sin \left (\frac{5}{2} (c+d x)\right )+396 \cos \left (\frac{1}{2} (c+d x)\right )-218 \cos \left (\frac{3}{2} (c+d x)\right )-114 \cos \left (\frac{5}{2} (c+d x)\right )-405 \sin (c+d x) \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )+405 \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )+135 \sin (3 (c+d x)) \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-135 \sin (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )\right )}{\left (\csc ^2\left (\frac{1}{4} (c+d x)\right )-\sec ^2\left (\frac{1}{4} (c+d x)\right )\right )^3}+(1536+1536 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )\right )}{192 d (a (\sin (c+d x)+1))^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cot[c + d*x]^4/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5*((1536 + 1536*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[
(c + d*x)/4])] - (8*Csc[(c + d*x)/2]^9*(396*Cos[(c + d*x)/2] - 218*Cos[(3*(c + d*x))/2] - 114*Cos[(5*(c + d*x)
)/2] - 396*Sin[(c + d*x)/2] - 405*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] + 405*Log[1 - Cos[
(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] - 218*Sin[(3*(c + d*x))/2] + 114*Sin[(5*(c + d*x))/2] + 135*Log[
1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 135*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*S
in[3*(c + d*x)]))/(Csc[(c + d*x)/4]^2 - Sec[(c + d*x)/4]^2)^3))/(192*d*(a*(1 + Sin[c + d*x]))^(5/2))

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Maple [A]  time = 1.19, size = 182, normalized size = 1. \begin{align*} -{\frac{1+\sin \left ( dx+c \right ) }{24\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) d}\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) } \left ( -135\,{a}^{5}{\it Artanh} \left ({\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }}{\sqrt{a}}} \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}+57\, \left ( -a \left ( \sin \left ( dx+c \right ) -1 \right ) \right ) ^{5/2}{a}^{5/2}+96\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{3}-88\, \left ( -a \left ( \sin \left ( dx+c \right ) -1 \right ) \right ) ^{3/2}{a}^{7/2}+39\,\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{9/2} \right ){a}^{-{\frac{15}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x)

[Out]

-1/24/a^(15/2)*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(-135*a^5*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*s
in(d*x+c)^3+57*(-a*(sin(d*x+c)-1))^(5/2)*a^(5/2)+96*2^(1/2)*arctanh(1/2*(-a*(sin(d*x+c)-1))^(1/2)*2^(1/2)/a^(1
/2))*a^5*sin(d*x+c)^3-88*(-a*(sin(d*x+c)-1))^(3/2)*a^(7/2)+39*(-a*(sin(d*x+c)-1))^(1/2)*a^(9/2))/sin(d*x+c)^3/
cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.29704, size = 1504, normalized size = 7.87 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/96*(135*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 - (cos(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*sin(d*x +
c) + 1)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 + (cos(d*x + c) + 3)*sin(d*x +
c) - 2*cos(d*x + c) - 3)*sqrt(a*sin(d*x + c) + a)*sqrt(a) - 9*a*cos(d*x + c) + (a*cos(d*x + c)^2 + 8*a*cos(d*x
+ c) - a)*sin(d*x + c) - a)/(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x + c) - cos(d*x +
c) - 1)) + 192*sqrt(2)*(a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 - (a*cos(d*x + c)^3 + a*cos(d*x + c)^2 - a*cos(d
*x + c) - a)*sin(d*x + c) + a)*log(-(cos(d*x + c)^2 - (cos(d*x + c) - 2)*sin(d*x + c) - 2*sqrt(2)*sqrt(a*sin(d
*x + c) + a)*(cos(d*x + c) - sin(d*x + c) + 1)/sqrt(a) + 3*cos(d*x + c) + 2)/(cos(d*x + c)^2 - (cos(d*x + c) +
2)*sin(d*x + c) - cos(d*x + c) - 2))/sqrt(a) + 4*(57*cos(d*x + c)^3 + 83*cos(d*x + c)^2 - (57*cos(d*x + c)^2
- 26*cos(d*x + c) - 91)*sin(d*x + c) - 65*cos(d*x + c) - 91)*sqrt(a*sin(d*x + c) + a))/(a^3*d*cos(d*x + c)^4 -
2*a^3*d*cos(d*x + c)^2 + a^3*d - (a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2 - a^3*d*cos(d*x + c) - a^3*d)*s
in(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)**4/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 2.63869, size = 938, normalized size = 4.91 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/48*(sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*((2*tan(1/2*d*x + 1/2*c)/(a^3*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 15/(a^
3*sgn(tan(1/2*d*x + 1/2*c) + 1)))*tan(1/2*d*x + 1/2*c) + 74/(a^3*sgn(tan(1/2*d*x + 1/2*c) + 1))) + (1890*sqrt(
2)*sqrt(a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 3840*sqrt(2)*sqrt(a)*arctan(sqrt(a)/sqrt(-a)) - 945*
sqrt(2)*sqrt(-a)*log(sqrt(2)*sqrt(a) + sqrt(a)) + 2700*sqrt(a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) -
5376*sqrt(a)*arctan(sqrt(a)/sqrt(-a)) - 1350*sqrt(-a)*log(sqrt(2)*sqrt(a) + sqrt(a)) - 1302*sqrt(2)*sqrt(-a) -
1808*sqrt(-a))*sgn(tan(1/2*d*x + 1/2*c) + 1)/(7*sqrt(2)*sqrt(-a)*a^(5/2) + 10*sqrt(-a)*a^(5/2)) + 384*sqrt(2)
*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) + sqrt(a))/sqrt(-a))/(
sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 270*arctan(-(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x +
1/2*c)^2 + a))/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 135*log(abs(-sqrt(a)*tan(1/2*d*x + 1/
2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/(a^(5/2)*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 2*(15*(sqrt(a)*tan(1/2*d
*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^5 - 78*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x +
1/2*c)^2 + a))^4*sqrt(a) + 144*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a^(3/2) -
15*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*a^2 - 74*a^(5/2))/(((sqrt(a)*tan(1/2*d
*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a)^3*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d