3.4.12 \(\int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{3/2}} \, dx\) [312]

Optimal. Leaf size=150 \[ -\frac {10 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{c^{3/2} f}+\frac {a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^{7/2}}+\frac {5 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}+\frac {10 a^3 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}} \]

[Out]

a^3*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(7/2)+5/3*a^3*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)-10*a^3*arctanh(1/2
*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))*2^(1/2)/c^(3/2)/f+10*a^3*cos(f*x+e)/c/f/(c-c*sin(f*x+e))^(
1/2)

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Rubi [A]
time = 0.22, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2815, 2759, 2758, 2728, 212} \begin {gather*} -\frac {10 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{c^{3/2} f}+\frac {a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^{7/2}}+\frac {5 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}+\frac {10 a^3 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(3/2),x]

[Out]

(-10*Sqrt[2]*a^3*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(c^(3/2)*f) + (a^3*c^2*Co
s[e + f*x]^5)/(f*(c - c*Sin[e + f*x])^(7/2)) + (5*a^3*Cos[e + f*x]^3)/(3*f*(c - c*Sin[e + f*x])^(3/2)) + (10*a
^3*Cos[e + f*x])/(c*f*Sqrt[c - c*Sin[e + f*x]])

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 2758

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + p))), x] + Dist[g^2*((p - 1)/(a*(m + p))), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
0] && IntegersQ[2*m, 2*p]

Rule 2759

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

Rule 2815

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

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{3/2}} \, dx &=\left (a^3 c^3\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{9/2}} \, dx\\ &=\frac {a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^{7/2}}-\frac {1}{2} \left (5 a^3 c\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx\\ &=\frac {a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^{7/2}}+\frac {5 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\left (5 a^3\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=\frac {a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^{7/2}}+\frac {5 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}+\frac {10 a^3 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}-\frac {\left (10 a^3\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{c}\\ &=\frac {a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^{7/2}}+\frac {5 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}+\frac {10 a^3 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}+\frac {\left (20 a^3\right ) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{c f}\\ &=-\frac {10 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{c^{3/2} f}+\frac {a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^{7/2}}+\frac {5 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}+\frac {10 a^3 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.50, size = 173, normalized size = 1.15 \begin {gather*} -\frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (50 \cos \left (\frac {1}{2} (e+f x)\right )+25 \cos \left (\frac {3}{2} (e+f x)\right )+\cos \left (\frac {5}{2} (e+f x)\right )+50 \sin \left (\frac {1}{2} (e+f x)\right )-(120+120 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) (-1+\sin (e+f x))-25 \sin \left (\frac {3}{2} (e+f x)\right )+\sin \left (\frac {5}{2} (e+f x)\right )\right )}{6 c f (-1+\sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(3/2),x]

[Out]

-1/6*(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(50*Cos[(e + f*x)/2] + 25*Cos[(3*(e + f*x))/2] + Cos[(5*(e + f
*x))/2] + 50*Sin[(e + f*x)/2] - (120 + 120*I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]
*(-1 + Sin[e + f*x]) - 25*Sin[(3*(e + f*x))/2] + Sin[(5*(e + f*x))/2]))/(c*f*(-1 + Sin[e + f*x])*Sqrt[c - c*Si
n[e + f*x]])

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Maple [A]
time = 1.88, size = 189, normalized size = 1.26

method	result	size
default	\(\frac {2 a^{3} \left (15 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c^{2}-\left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {c}\, \sin \left (f x +e \right )-12 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}} \sin \left (f x +e \right )-15 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+\left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {c}+18 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{3 c^{\frac {7}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(189\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/3*a^3*(15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)*c^2-(c*(1+sin(f*x+e)))^(3
/2)*c^(1/2)*sin(f*x+e)-12*(c*(1+sin(f*x+e)))^(1/2)*c^(3/2)*sin(f*x+e)-15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e))
)^(1/2)*2^(1/2)/c^(1/2))*c^2+(c*(1+sin(f*x+e)))^(3/2)*c^(1/2)+18*(c*(1+sin(f*x+e)))^(1/2)*c^(3/2))*(c*(1+sin(f
*x+e)))^(1/2)/c^(7/2)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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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((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^3/(-c*sin(f*x + e) + c)^(3/2), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 352 vs. \(2 (141) = 282\).
time = 0.36, size = 352, normalized size = 2.35 \begin {gather*} \frac {\frac {15 \, \sqrt {2} {\left (a^{3} c \cos \left (f x + e\right )^{2} - a^{3} c \cos \left (f x + e\right ) - 2 \, a^{3} c + {\left (a^{3} c \cos \left (f x + e\right ) + 2 \, a^{3} c\right )} \sin \left (f x + e\right )\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) - \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} - 2 \, {\left (a^{3} \cos \left (f x + e\right )^{3} + 13 \, a^{3} \cos \left (f x + e\right )^{2} + 18 \, a^{3} \cos \left (f x + e\right ) + 6 \, a^{3} + {\left (a^{3} \cos \left (f x + e\right )^{2} - 12 \, a^{3} \cos \left (f x + e\right ) + 6 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/3*(15*sqrt(2)*(a^3*c*cos(f*x + e)^2 - a^3*c*cos(f*x + e) - 2*a^3*c + (a^3*c*cos(f*x + e) + 2*a^3*c)*sin(f*x
+ e))*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2)*sin(f*x + e) - 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*(cos(f*x +
e) + sin(f*x + e) + 1)/sqrt(c) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f
*x + e) - 2))/sqrt(c) - 2*(a^3*cos(f*x + e)^3 + 13*a^3*cos(f*x + e)^2 + 18*a^3*cos(f*x + e) + 6*a^3 + (a^3*cos
(f*x + e)^2 - 12*a^3*cos(f*x + e) + 6*a^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^2*f*cos(f*x + e)^2 - c^
2*f*cos(f*x + e) - 2*c^2*f + (c^2*f*cos(f*x + e) + 2*c^2*f)*sin(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**(3/2),x)

[Out]

a**3*(Integral(3*sin(e + f*x)/(-c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) + c*sqrt(-c*sin(e + f*x) + c)), x) +
Integral(3*sin(e + f*x)**2/(-c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) + c*sqrt(-c*sin(e + f*x) + c)), x) + Int
egral(sin(e + f*x)**3/(-c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) + c*sqrt(-c*sin(e + f*x) + c)), x) + Integral
(1/(-c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) + c*sqrt(-c*sin(e + f*x) + c)), x))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 376 vs. \(2 (141) = 282\).
time = 0.57, size = 376, normalized size = 2.51 \begin {gather*} -\frac {\frac {30 \, \sqrt {2} a^{3} \log \left (-\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )}{c^{\frac {3}{2}} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {3 \, \sqrt {2} a^{3} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{c^{\frac {3}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {3 \, \sqrt {2} {\left (a^{3} + \frac {10 \, a^{3} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}}{c^{\frac {3}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {16 \, \sqrt {2} {\left (7 \, a^{3} \sqrt {c} - \frac {12 \, a^{3} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {9 \, a^{3} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}\right )}}{c^{2} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - 1\right )}^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{6 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

-1/6*(30*sqrt(2)*a^3*log(-(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1))/(c^(3/2)*
sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + 3*sqrt(2)*a^3*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(c^(3/2)*(cos(-1/4*p
i + 1/2*f*x + 1/2*e) + 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 3*sqrt(2)*(a^3 + 10*a^3*(cos(-1/4*pi + 1/2*f*
x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1))*(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(c^(3/2)*(cos(-1/4*
pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 16*sqrt(2)*(7*a^3*sqrt(c) - 12*a^3*sqrt(c)*(
cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 9*a^3*sqrt(c)*(cos(-1/4*pi + 1/2*f*
x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)/(c^2*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4
*pi + 1/2*f*x + 1/2*e) + 1) - 1)^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^3/(c - c*sin(e + f*x))^(3/2),x)

[Out]

int((a + a*sin(e + f*x))^3/(c - c*sin(e + f*x))^(3/2), x)

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