3.459 \(\int \cot ^4(c+d x) \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=205 \[ \frac {21 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{64 d}-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}+\frac {149 a^2 \cot (c+d x)}{64 d \sqrt {a \sin (c+d x)+a}}+\frac {19 a^2 \cot (c+d x) \csc (c+d x)}{32 d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x) \csc ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}-\frac {a \cot (c+d x) \csc ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{8 d} \]

[Out]

21/64*a^(3/2)*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/d-1/4*cot(d*x+c)*csc(d*x+c)^3*(a+a*sin(d*x+c)
)^(3/2)/d-2*a^2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+149/64*a^2*cot(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+19/32*a^2*c
ot(d*x+c)*csc(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-1/8*a*cot(d*x+c)*csc(d*x+c)^2*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.70, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {2881, 2763, 21, 2773, 206, 3044, 2975, 2980, 2772} \[ -\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}+\frac {149 a^2 \cot (c+d x)}{64 d \sqrt {a \sin (c+d x)+a}}+\frac {21 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{64 d}+\frac {19 a^2 \cot (c+d x) \csc (c+d x)}{32 d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x) \csc ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}-\frac {a \cot (c+d x) \csc ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*Csc[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(21*a^(3/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(64*d) - (2*a^2*Cos[c + d*x])/(d*Sqrt[a
+ a*Sin[c + d*x]]) + (149*a^2*Cot[c + d*x])/(64*d*Sqrt[a + a*Sin[c + d*x]]) + (19*a^2*Cot[c + d*x]*Csc[c + d*x
])/(32*d*Sqrt[a + a*Sin[c + d*x]]) - (a*Cot[c + d*x]*Csc[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]])/(8*d) - (Cot[c +
d*x]*Csc[c + d*x]^3*(a + a*Sin[c + d*x])^(3/2))/(4*d)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 2763

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

Rule 2772

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

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 2881

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

Rule 2975

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^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], 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] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])

Rule 2980

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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] && LtQ[n, -1]

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 \cot ^4(c+d x) \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=\int \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx+\int \csc ^5(c+d x) (a+a \sin (c+d x))^{3/2} \left (1-2 \sin ^2(c+d x)\right ) \, dx\\ &=-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}+2 \int \frac {\csc (c+d x) \left (\frac {a^2}{2}+\frac {1}{2} a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx+\frac {\int \csc ^4(c+d x) \left (\frac {3 a}{2}-\frac {13}{2} a \sin (c+d x)\right ) (a+a \sin (c+d x))^{3/2} \, dx}{4 a}\\ &=-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {a \cot (c+d x) \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{8 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}+\frac {\int \csc ^3(c+d x) \sqrt {a+a \sin (c+d x)} \left (-\frac {57 a^2}{4}-\frac {69}{4} a^2 \sin (c+d x)\right ) \, dx}{12 a}+a \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+\frac {19 a^2 \cot (c+d x) \csc (c+d x)}{32 d \sqrt {a+a \sin (c+d x)}}-\frac {a \cot (c+d x) \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{8 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}-\frac {1}{64} (149 a) \int \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx-\frac {\left (2 a^2\right ) \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 )}{d}\\ &=-\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+\frac {149 a^2 \cot (c+d x)}{64 d \sqrt {a+a \sin (c+d x)}}+\frac {19 a^2 \cot (c+d x) \csc (c+d x)}{32 d \sqrt {a+a \sin (c+d x)}}-\frac {a \cot (c+d x) \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{8 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}-\frac {1}{128} (149 a) \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+\frac {149 a^2 \cot (c+d x)}{64 d \sqrt {a+a \sin (c+d x)}}+\frac {19 a^2 \cot (c+d x) \csc (c+d x)}{32 d \sqrt {a+a \sin (c+d x)}}-\frac {a \cot (c+d x) \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{8 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}+\frac {\left (149 a^2\right ) \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 )}{64 d}\\ &=\frac {21 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{64 d}-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+\frac {149 a^2 \cot (c+d x)}{64 d \sqrt {a+a \sin (c+d x)}}+\frac {19 a^2 \cot (c+d x) \csc (c+d x)}{32 d \sqrt {a+a \sin (c+d x)}}-\frac {a \cot (c+d x) \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{8 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}\\ \end {align*}

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Mathematica [A]  time = 1.46, size = 392, normalized size = 1.91 \[ -\frac {a \csc ^{13}\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sin (c+d x)+1)} \left (-1486 \sin \left (\frac {1}{2} (c+d x)\right )-1030 \sin \left (\frac {3}{2} (c+d x)\right )+754 \sin \left (\frac {5}{2} (c+d x)\right )+426 \sin \left (\frac {7}{2} (c+d x)\right )-128 \sin \left (\frac {9}{2} (c+d x)\right )+1486 \cos \left (\frac {1}{2} (c+d x)\right )-1030 \cos \left (\frac {3}{2} (c+d x)\right )-754 \cos \left (\frac {5}{2} (c+d x)\right )+426 \cos \left (\frac {7}{2} (c+d x)\right )+128 \cos \left (\frac {9}{2} (c+d x)\right )+84 \cos (2 (c+d x)) \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )-21 \cos (4 (c+d x)) \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )-63 \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )-84 \cos (2 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )+21 \cos (4 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )+63 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )\right )}{64 d \left (\cot \left (\frac {1}{2} (c+d x)\right )+1\right ) \left (\csc ^2\left (\frac {1}{4} (c+d x)\right )-\sec ^2\left (\frac {1}{4} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*Csc[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

-1/64*(a*Csc[(c + d*x)/2]^13*Sqrt[a*(1 + Sin[c + d*x])]*(1486*Cos[(c + d*x)/2] - 1030*Cos[(3*(c + d*x))/2] - 7
54*Cos[(5*(c + d*x))/2] + 426*Cos[(7*(c + d*x))/2] + 128*Cos[(9*(c + d*x))/2] - 63*Log[1 + Cos[(c + d*x)/2] -
Sin[(c + d*x)/2]] + 84*Cos[2*(c + d*x)]*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 21*Cos[4*(c + d*x)]*Log
[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 63*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 84*Cos[2*(c + d*
x)]*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 21*Cos[4*(c + d*x)]*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x
)/2]] - 1486*Sin[(c + d*x)/2] - 1030*Sin[(3*(c + d*x))/2] + 754*Sin[(5*(c + d*x))/2] + 426*Sin[(7*(c + d*x))/2
] - 128*Sin[(9*(c + d*x))/2]))/(d*(1 + Cot[(c + d*x)/2])*(Csc[(c + d*x)/4]^2 - Sec[(c + d*x)/4]^2)^4)

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fricas [B]  time = 0.48, size = 460, normalized size = 2.24 \[ \frac {21 \, {\left (a \cos \left (d x + c\right )^{5} + a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{3} - 2 \, a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \sin \left (d x + c\right ) + a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (128 \, a \cos \left (d x + c\right )^{5} + 277 \, a \cos \left (d x + c\right )^{4} - 242 \, a \cos \left (d x + c\right )^{3} - 500 \, a \cos \left (d x + c\right )^{2} + 130 \, a \cos \left (d x + c\right ) - {\left (128 \, a \cos \left (d x + c\right )^{4} - 149 \, a \cos \left (d x + c\right )^{3} - 391 \, a \cos \left (d x + c\right )^{2} + 109 \, a \cos \left (d x + c\right ) + 239 \, a\right )} \sin \left (d x + c\right ) + 239 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{256 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{3} - 2 \, d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/256*(21*(a*cos(d*x + c)^5 + a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^3 - 2*a*cos(d*x + c)^2 + a*cos(d*x + c) + (a
*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 + a)*sin(d*x + c) + a)*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)) - 4*(128*a*cos(d*x + c)^5 + 277*a*cos(d*x + c)^4
- 242*a*cos(d*x + c)^3 - 500*a*cos(d*x + c)^2 + 130*a*cos(d*x + c) - (128*a*cos(d*x + c)^4 - 149*a*cos(d*x + c
)^3 - 391*a*cos(d*x + c)^2 + 109*a*cos(d*x + c) + 239*a)*sin(d*x + c) + 239*a)*sqrt(a*sin(d*x + c) + a))/(d*co
s(d*x + c)^5 + d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^3 - 2*d*cos(d*x + c)^2 + d*cos(d*x + c) + (d*cos(d*x + c)^4
- 2*d*cos(d*x + c)^2 + d)*sin(d*x + c) + d)

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giac [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(cos(d*x+c)^4*csc(d*x+c)^5*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 1.42, size = 188, normalized size = 0.92 \[ -\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (128 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {7}{2}} \left (\sin ^{4}\left (d x +c \right )\right )-21 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) a^{4} \left (\sin ^{4}\left (d x +c \right )\right )+149 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {7}{2}} \sqrt {a}-461 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {5}{2}} a^{\frac {3}{2}}+435 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} a^{\frac {5}{2}}-107 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {7}{2}}\right )}{64 a^{\frac {5}{2}} \sin \left (d x +c \right )^{4} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^5*(a+a*sin(d*x+c))^(3/2),x)

[Out]

-1/64*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)/a^(5/2)*(128*(-a*(sin(d*x+c)-1))^(1/2)*a^(7/2)*sin(d*x+c)^4-21*
arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*a^4*sin(d*x+c)^4+149*(-a*(sin(d*x+c)-1))^(7/2)*a^(1/2)-461*(-a*(sin
(d*x+c)-1))^(5/2)*a^(3/2)+435*(-a*(sin(d*x+c)-1))^(3/2)*a^(5/2)-107*(-a*(sin(d*x+c)-1))^(1/2)*a^(7/2))/sin(d*x
+c)^4/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{4} \csc \left (d x + c\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(3/2)*cos(d*x + c)^4*csc(d*x + c)^5, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}}{{\sin \left (c+d\,x\right )}^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*(a + a*sin(c + d*x))^(3/2))/sin(c + d*x)^5,x)

[Out]

int((cos(c + d*x)^4*(a + a*sin(c + d*x))^(3/2))/sin(c + d*x)^5, x)

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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(cos(d*x+c)**4*csc(d*x+c)**5*(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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