3.1.68 \(\int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\) [68]

Optimal. Leaf size=183 \[ \frac {15 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac {31 \cos (c+d x)}{5 a d \sqrt {a+a \sin (c+d x)}}-\frac {9 \cos (c+d x) \sin ^2(c+d x)}{10 a d \sqrt {a+a \sin (c+d x)}}+\frac {13 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{10 a^2 d} \]

[Out]

1/2*cos(d*x+c)*sin(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)+15/4*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+
c))^(1/2))*2^(1/2)/a^(3/2)/d-31/5*cos(d*x+c)/a/d/(a+a*sin(d*x+c))^(1/2)-9/10*cos(d*x+c)*sin(d*x+c)^2/a/d/(a+a*
sin(d*x+c))^(1/2)+13/10*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/a^2/d

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Rubi [A]
time = 0.26, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2844, 3062, 3047, 3102, 2830, 2728, 212} \begin {gather*} \frac {15 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {13 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{10 a^2 d}+\frac {\sin ^3(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {9 \sin ^2(c+d x) \cos (c+d x)}{10 a d \sqrt {a \sin (c+d x)+a}}-\frac {31 \cos (c+d x)}{5 a d \sqrt {a \sin (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) + (Cos[c + d*x]*
Sin[c + d*x]^3)/(2*d*(a + a*Sin[c + d*x])^(3/2)) - (31*Cos[c + d*x])/(5*a*d*Sqrt[a + a*Sin[c + d*x]]) - (9*Cos
[c + d*x]*Sin[c + d*x]^2)/(10*a*d*Sqrt[a + a*Sin[c + d*x]]) + (13*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(10*a
^2*d)

Rule 212

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

Rule 2728

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 2830

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

Rule 2844

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Dist[1/
(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], 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[m, -1] && GtQ[n, 1] && (IntegersQ
[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))

Rule 3047

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

Rule 3062

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

Rule 3102

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]

Rubi steps

\begin {align*} \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac {\cos (c+d x) \sin ^3(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac {\int \frac {\sin ^2(c+d x) \left (3 a-\frac {9}{2} a \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{2 a^2}\\ &=\frac {\cos (c+d x) \sin ^3(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac {9 \cos (c+d x) \sin ^2(c+d x)}{10 a d \sqrt {a+a \sin (c+d x)}}-\frac {\int \frac {\sin (c+d x) \left (-9 a^2+\frac {39}{4} a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{5 a^3}\\ &=\frac {\cos (c+d x) \sin ^3(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac {9 \cos (c+d x) \sin ^2(c+d x)}{10 a d \sqrt {a+a \sin (c+d x)}}-\frac {\int \frac {-9 a^2 \sin (c+d x)+\frac {39}{4} a^2 \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{5 a^3}\\ &=\frac {\cos (c+d x) \sin ^3(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac {9 \cos (c+d x) \sin ^2(c+d x)}{10 a d \sqrt {a+a \sin (c+d x)}}+\frac {13 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{10 a^2 d}-\frac {2 \int \frac {\frac {39 a^3}{8}-\frac {93}{4} a^3 \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{15 a^4}\\ &=\frac {\cos (c+d x) \sin ^3(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac {31 \cos (c+d x)}{5 a d \sqrt {a+a \sin (c+d x)}}-\frac {9 \cos (c+d x) \sin ^2(c+d x)}{10 a d \sqrt {a+a \sin (c+d x)}}+\frac {13 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{10 a^2 d}-\frac {15 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{4 a}\\ &=\frac {\cos (c+d x) \sin ^3(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac {31 \cos (c+d x)}{5 a d \sqrt {a+a \sin (c+d x)}}-\frac {9 \cos (c+d x) \sin ^2(c+d x)}{10 a d \sqrt {a+a \sin (c+d x)}}+\frac {13 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{10 a^2 d}+\frac {15 \text {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 )}{2 a d}\\ &=\frac {15 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac {31 \cos (c+d x)}{5 a d \sqrt {a+a \sin (c+d x)}}-\frac {9 \cos (c+d x) \sin ^2(c+d x)}{10 a d \sqrt {a+a \sin (c+d x)}}+\frac {13 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{10 a^2 d}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.28, size = 178, normalized size = 0.97 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (-55 \cos \left (\frac {1}{2} (c+d x)\right )-41 \cos \left (\frac {3}{2} (c+d x)\right )-3 \cos \left (\frac {5}{2} (c+d x)\right )+\cos \left (\frac {7}{2} (c+d x)\right )+55 \sin \left (\frac {1}{2} (c+d x)\right )-(150+150 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right ) (1+\sin (c+d x))-41 \sin \left (\frac {3}{2} (c+d x)\right )+3 \sin \left (\frac {5}{2} (c+d x)\right )+\sin \left (\frac {7}{2} (c+d x)\right )\right )}{20 d (a (1+\sin (c+d x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-55*Cos[(c + d*x)/2] - 41*Cos[(3*(c + d*x))/2] - 3*Cos[(5*(c + d*x))/2
] + Cos[(7*(c + d*x))/2] + 55*Sin[(c + d*x)/2] - (150 + 150*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 +
Tan[(c + d*x)/4])]*(1 + Sin[c + d*x]) - 41*Sin[(3*(c + d*x))/2] + 3*Sin[(5*(c + d*x))/2] + Sin[(7*(c + d*x))/
2]))/(20*d*(a*(1 + Sin[c + d*x]))^(3/2))

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Maple [A]
time = 1.76, size = 183, normalized size = 1.00

method	result	size
default	\(\frac {\left (\sin \left (d x +c \right ) \left (-80 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {5}{2}}-8 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} \sqrt {a}+75 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}\right )-90 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {5}{2}}-8 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} \sqrt {a}+75 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{20 a^{\frac {9}{2}} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(183\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/20*(sin(d*x+c)*(-80*(a-a*sin(d*x+c))^(1/2)*a^(5/2)-8*(a-a*sin(d*x+c))^(5/2)*a^(1/2)+75*2^(1/2)*arctanh(1/2*(
a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^3)-90*(a-a*sin(d*x+c))^(1/2)*a^(5/2)-8*(a-a*sin(d*x+c))^(5/2)*a^(1/2)
+75*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^3)*(-a*(sin(d*x+c)-1))^(1/2)/a^(9/2)/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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (156) = 312\).
time = 0.38, size = 314, normalized size = 1.72 \begin {gather*} \frac {75 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) - 4 \, {\left (4 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{3} - 48 \, \cos \left (d x + c\right )^{2} + {\left (4 \, \cos \left (d x + c\right )^{3} + 8 \, \cos \left (d x + c\right )^{2} - 40 \, \cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right ) - 45 \, \cos \left (d x + c\right ) - 5\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{40 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \cos \left (d x + c\right ) - 2 \, a^{2} d - {\left (a^{2} d \cos \left (d x + c\right ) + 2 \, a^{2} d\right )} \sin \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(75*sqrt(2)*(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)*sqrt(a)*log(-(a*cos(d*x
+ c)^2 + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - sin(d*x + c) + 1) + 3*a*cos(d*x + c) - (a
*cos(d*x + c) - 2*a)*sin(d*x + c) + 2*a)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)
) - 4*(4*cos(d*x + c)^4 - 4*cos(d*x + c)^3 - 48*cos(d*x + c)^2 + (4*cos(d*x + c)^3 + 8*cos(d*x + c)^2 - 40*cos
(d*x + c) + 5)*sin(d*x + c) - 45*cos(d*x + c) - 5)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d*x + c)^2 - a^2*d*cos
(d*x + c) - 2*a^2*d - (a^2*d*cos(d*x + c) + 2*a^2*d)*sin(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin ^{4}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**4/(a+a*sin(d*x+c))**(3/2),x)

[Out]

Integral(sin(c + d*x)**4/(a*(sin(c + d*x) + 1))**(3/2), x)

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Giac [A]
time = 0.54, size = 197, normalized size = 1.08 \begin {gather*} -\frac {\frac {75 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {75 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {10 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {32 \, \sqrt {2} {\left (2 \, a^{\frac {17}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5 \, a^{\frac {17}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{10} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{40 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-1/40*(75*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^(3/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 75*s
qrt(2)*log(-sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^(3/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) + 10*sqrt(2)*sin
(-1/4*pi + 1/2*d*x + 1/2*c)/((sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)*a^(3/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)
)) - 32*sqrt(2)*(2*a^(17/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 + 5*a^(17/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c))/(a^1
0*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\sin \left (c+d\,x\right )}^4}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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