3.95 \(\int \frac{(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=175 \[ \frac{a^2 c^2 (A+B) \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac{3 a^2 (A+9 B) \cos (e+f x)}{8 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{3 a^2 (A+9 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{4 \sqrt{2} c^{5/2} f}-\frac{a^2 (A+9 B) \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{5/2}} \]

[Out]

(3*a^2*(A + 9*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(4*Sqrt[2]*c^(5/2)*f) + (
a^2*(A + B)*c^2*Cos[e + f*x]^5)/(4*f*(c - c*Sin[e + f*x])^(9/2)) - (a^2*(A + 9*B)*Cos[e + f*x]^3)/(8*f*(c - c*
Sin[e + f*x])^(5/2)) - (3*a^2*(A + 9*B)*Cos[e + f*x])/(8*c^2*f*Sqrt[c - c*Sin[e + f*x]])

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Rubi [A]  time = 0.478418, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2967, 2859, 2680, 2679, 2649, 206} \[ \frac{a^2 c^2 (A+B) \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac{3 a^2 (A+9 B) \cos (e+f x)}{8 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{3 a^2 (A+9 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{4 \sqrt{2} c^{5/2} f}-\frac{a^2 (A+9 B) \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

(3*a^2*(A + 9*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(4*Sqrt[2]*c^(5/2)*f) + (
a^2*(A + B)*c^2*Cos[e + f*x]^5)/(4*f*(c - c*Sin[e + f*x])^(9/2)) - (a^2*(A + 9*B)*Cos[e + f*x]^3)/(8*f*(c - c*
Sin[e + f*x])^(5/2)) - (3*a^2*(A + 9*B)*Cos[e + f*x])/(8*c^2*f*Sqrt[c - c*Sin[e + f*x]])

Rule 2967

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

Rule 2859

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

Rule 2680

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 2679

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[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 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])

Rubi steps

\begin{align*} \int \frac{(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx &=\left (a^2 c^2\right ) \int \frac{\cos ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx\\ &=\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac{1}{8} \left (a^2 (A+9 B) c\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^{7/2}} \, dx\\ &=\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac{a^2 (A+9 B) \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{5/2}}+\frac{\left (3 a^2 (A+9 B)\right ) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx}{16 c}\\ &=\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac{a^2 (A+9 B) \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{5/2}}-\frac{3 a^2 (A+9 B) \cos (e+f x)}{8 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{\left (3 a^2 (A+9 B)\right ) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{8 c^2}\\ &=\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac{a^2 (A+9 B) \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{5/2}}-\frac{3 a^2 (A+9 B) \cos (e+f x)}{8 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{\left (3 a^2 (A+9 B)\right ) \operatorname{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 )}{4 c^2 f}\\ &=\frac{3 a^2 (A+9 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{4 \sqrt{2} c^{5/2} f}+\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac{a^2 (A+9 B) \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{5/2}}-\frac{3 a^2 (A+9 B) \cos (e+f x)}{8 c^2 f \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.20048, size = 344, normalized size = 1.97 \[ \frac{a^2 (\sin (e+f x)+1)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (8 (A+B) \sin \left (\frac{1}{2} (e+f x)\right )-(5 A+13 B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3-2 (5 A+13 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+4 (A+B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+(-3-3 i) \sqrt [4]{-1} (A+9 B) \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4-8 B \cos \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4-8 B \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4\right )}{4 f (c-c \sin (e+f x))^{5/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(4*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - (5*A + 13*B)*(Co
s[(e + f*x)/2] - Sin[(e + f*x)/2])^3 - (3 + 3*I)*(-1)^(1/4)*(A + 9*B)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(
e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 - 8*B*Cos[(e + f*x)/2]*(Cos[(e + f*x)/2] - Sin[(e + f*x)
/2])^4 + 8*(A + B)*Sin[(e + f*x)/2] - 2*(5*A + 13*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2]
- 8*B*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2)/(4*f*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^4*(c - c*Sin[e + f*x])^(5/2))

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Maple [B]  time = 1.443, size = 386, normalized size = 2.2 \begin{align*} -{\frac{{a}^{2}}{ \left ( -8+8\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( 3\,A\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}{c}^{2}+27\,B\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}{c}^{2}-6\,A\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sin \left ( fx+e \right ){c}^{2}-16\,B\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{3/2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}-54\,B\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sin \left ( fx+e \right ){c}^{2}+10\,A \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}\sqrt{c}+3\,A\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{2}+26\,B \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}\sqrt{c}+32\,B{c}^{3/2}\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sin \left ( fx+e \right ) +27\,B\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{2}-12\,A\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{3/2}-60\,B\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{3/2} \right ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{-{\frac{9}{2}}}{\frac{1}{\sqrt{c-c\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x)

[Out]

-1/8/c^(9/2)*a^2*(3*A*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^2*c^2+27*B*2^(1
/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^2*c^2-6*A*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^2-16*B*(c*(1+sin(f*x+e)))^(1/2)*c^(3/2)*sin(f*x+e)^2-54*B*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+10*A*(c*(1+sin(f*x+e)))^(3/2)*c^(1/2)+3*A
*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^2+26*B*(c*(1+sin(f*x+e)))^(3/2)*c^(1/2)+32*B*
c^(3/2)*(c*(1+sin(f*x+e)))^(1/2)*sin(f*x+e)+27*B*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))
*c^2-12*A*(c*(1+sin(f*x+e)))^(1/2)*c^(3/2)-60*B*(c*(1+sin(f*x+e)))^(1/2)*c^(3/2))*(c*(1+sin(f*x+e)))^(1/2)/(-1
+sin(f*x+e))/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^2/(-c*sin(f*x + e) + c)^(5/2), x)

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Fricas [B]  time = 1.53259, size = 1143, normalized size = 6.53 \begin{align*} \frac{3 \, \sqrt{2}{\left ({\left (A + 9 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} + 3 \,{\left (A + 9 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \,{\left (A + 9 \, B\right )} a^{2} \cos \left (f x + e\right ) - 4 \,{\left (A + 9 \, B\right )} a^{2} -{\left ({\left (A + 9 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \,{\left (A + 9 \, B\right )} a^{2} \cos \left (f x + e\right ) - 4 \,{\left (A + 9 \, B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{c} \log \left (-\frac{c \cos \left (f x + e\right )^{2} + 2 \, \sqrt{2} \sqrt{-c \sin \left (f x + e\right ) + c} \sqrt{c}{\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) +{\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\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 ) - 4 \,{\left (8 \, B a^{2} \cos \left (f x + e\right )^{3} -{\left (5 \, A + 21 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} -{\left (A + 25 \, B\right )} a^{2} \cos \left (f x + e\right ) + 4 \,{\left (A + B\right )} a^{2} +{\left (8 \, B a^{2} \cos \left (f x + e\right )^{2} +{\left (5 \, A + 29 \, B\right )} a^{2} \cos \left (f x + e\right ) + 4 \,{\left (A + B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{16 \,{\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f -{\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/16*(3*sqrt(2)*((A + 9*B)*a^2*cos(f*x + e)^3 + 3*(A + 9*B)*a^2*cos(f*x + e)^2 - 2*(A + 9*B)*a^2*cos(f*x + e)
- 4*(A + 9*B)*a^2 - ((A + 9*B)*a^2*cos(f*x + e)^2 - 2*(A + 9*B)*a^2*cos(f*x + e) - 4*(A + 9*B)*a^2)*sin(f*x +
e))*sqrt(c)*log(-(c*cos(f*x + e)^2 + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f*x + e) + sin(f*x + e)
+ 1) + 3*c*cos(f*x + e) + (c*cos(f*x + e) - 2*c)*sin(f*x + e) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(
f*x + e) - cos(f*x + e) - 2)) - 4*(8*B*a^2*cos(f*x + e)^3 - (5*A + 21*B)*a^2*cos(f*x + e)^2 - (A + 25*B)*a^2*c
os(f*x + e) + 4*(A + B)*a^2 + (8*B*a^2*cos(f*x + e)^2 + (5*A + 29*B)*a^2*cos(f*x + e) + 4*(A + B)*a^2)*sin(f*x
+ e))*sqrt(-c*sin(f*x + e) + c))/(c^3*f*cos(f*x + e)^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^
3*f - (c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f)*sin(f*x + e))

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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((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

sage2