3.298 \(\int \frac{(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=191 \[ -\frac{a^{3/2} \left (A d (c+3 d)-B \left (3 c^2+3 c d-2 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{d^{5/2} f (c+d)^{3/2}}-\frac{a^2 (-A d+3 B c+2 B d) \cos (e+f x)}{d^2 f (c+d) \sqrt{a \sin (e+f x)+a}}+\frac{a (B c-A d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{d f (c+d) (c+d \sin (e+f x))} \]

[Out]

-((a^(3/2)*(A*d*(c + 3*d) - B*(3*c^2 + 3*c*d - 2*d^2))*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqr
t[a + a*Sin[e + f*x]])])/(d^(5/2)*(c + d)^(3/2)*f)) - (a^2*(3*B*c - A*d + 2*B*d)*Cos[e + f*x])/(d^2*(c + d)*f*
Sqrt[a + a*Sin[e + f*x]]) + (a*(B*c - A*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(d*(c + d)*f*(c + d*Sin[e +
f*x]))

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Rubi [A]  time = 0.552218, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used = {2975, 2981, 2773, 208} \[ -\frac{a^{3/2} \left (A d (c+3 d)-B \left (3 c^2+3 c d-2 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{d^{5/2} f (c+d)^{3/2}}-\frac{a^2 (-A d+3 B c+2 B d) \cos (e+f x)}{d^2 f (c+d) \sqrt{a \sin (e+f x)+a}}+\frac{a (B c-A d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{d f (c+d) (c+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^(3/2)*(A*d*(c + 3*d) - B*(3*c^2 + 3*c*d - 2*d^2))*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqr
t[a + a*Sin[e + f*x]])])/(d^(5/2)*(c + d)^(3/2)*f)) - (a^2*(3*B*c - A*d + 2*B*d)*Cos[e + f*x])/(d^2*(c + d)*f*
Sqrt[a + a*Sin[e + f*x]]) + (a*(B*c - A*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(d*(c + d)*f*(c + d*Sin[e +
f*x]))

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 2981

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

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 208

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

Rubi steps

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

Mathematica [A]  time = 4.88958, size = 381, normalized size = 1.99 \[ \frac{(a (\sin (e+f x)+1))^{3/2} \left (\frac{\left (A d (c+3 d)+B \left (-3 c^2-3 c d+2 d^2\right )\right ) \left (2 \log \left (\sqrt{d} \sqrt{c+d} \left (\tan ^2\left (\frac{1}{4} (e+f x)\right )+2 \tan \left (\frac{1}{4} (e+f x)\right )-1\right )+(c+d) \sec ^2\left (\frac{1}{4} (e+f x)\right )\right )-2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right )\right )+e+f x\right )}{(c+d)^{3/2}}+\frac{\left (B \left (3 c^2+3 c d-2 d^2\right )-A d (c+3 d)\right ) \left (2 \log \left (-\sec ^2\left (\frac{1}{4} (e+f x)\right ) \left (-\sqrt{d} \sqrt{c+d} \sin \left (\frac{1}{2} (e+f x)\right )+\sqrt{d} \sqrt{c+d} \cos \left (\frac{1}{2} (e+f x)\right )+c+d\right )\right )-2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right )\right )+e+f x\right )}{(c+d)^{3/2}}-\frac{4 \sqrt{d} (d-c) (A d-B c) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{(c+d) (c+d \sin (e+f x))}+8 B \sqrt{d} \sin \left (\frac{1}{2} (e+f x)\right )-8 B \sqrt{d} \cos \left (\frac{1}{2} (e+f x)\right )\right )}{4 d^{5/2} f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[e + f*x]))^(3/2)*(-8*B*Sqrt[d]*Cos[(e + f*x)/2] + ((-(A*d*(c + 3*d)) + B*(3*c^2 + 3*c*d - 2*d^2))
*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2] + 2*Log[-(Sec[(e + f*x)/4]^2*(c + d + Sqrt[d]*Sqrt[c + d]*Cos[(e + f*x)/
2] - Sqrt[d]*Sqrt[c + d]*Sin[(e + f*x)/2]))]))/(c + d)^(3/2) + ((A*d*(c + 3*d) + B*(-3*c^2 - 3*c*d + 2*d^2))*(
e + f*x - 2*Log[Sec[(e + f*x)/4]^2] + 2*Log[(c + d)*Sec[(e + f*x)/4]^2 + Sqrt[d]*Sqrt[c + d]*(-1 + 2*Tan[(e +
f*x)/4] + Tan[(e + f*x)/4]^2)]))/(c + d)^(3/2) + 8*B*Sqrt[d]*Sin[(e + f*x)/2] - (4*Sqrt[d]*(-c + d)*(-(B*c) +
A*d)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c + d)*(c + d*Sin[e + f*x]))))/(4*d^(5/2)*f*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^3)

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Maple [B]  time = 1.839, size = 592, normalized size = 3.1 \begin{align*}{\frac{a \left ( 1+\sin \left ( fx+e \right ) \right ) }{ \left ( c+d \right ){d}^{2} \left ( c+d\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f}\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) } \left ( -\sin \left ( fx+e \right ) d \left ( A{\it Artanh} \left ({d\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{acd+a{d}^{2}}}}} \right ) acd+3\,A{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ) a{d}^{2}-3\,B{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ) a{c}^{2}-3\,B{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ) acd+2\,B{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ) a{d}^{2}+2\,B\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}c+2\,B\sqrt{a \left ( c+d \right ) d}\sqrt{a-a\sin \left ( fx+e \right ) }d \right ) -A{\it Artanh} \left ({d\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{acd+a{d}^{2}}}}} \right ) a{c}^{2}d-3\,A{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ) ac{d}^{2}+3\,B{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ) a{c}^{3}+3\,B{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ) a{c}^{2}d-2\,B{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ) ac{d}^{2}+A\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}cd-A\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}{d}^{2}-3\,B\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}{c}^{2}-B\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}cd \right ){\frac{1}{\sqrt{a \left ( c+d \right ) d}}}{\frac{1}{\sqrt{a+a\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))^(3/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2,x)

[Out]

a*(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(-sin(f*x+e)*d*(A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(
1/2))*a*c*d+3*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*d^2-3*B*arctanh((a-a*sin(f*x+e))^(1/2)
*d/(a*c*d+a*d^2)^(1/2))*a*c^2-3*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*c*d+2*B*arctanh((a-a
*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*d^2+2*B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*c+2*B*(a*(c+d)*d)
^(1/2)*(a-a*sin(f*x+e))^(1/2)*d)-A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*c^2*d-3*A*arctanh((
a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*c*d^2+3*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))
*a*c^3+3*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*c^2*d-2*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/
(a*c*d+a*d^2)^(1/2))*a*c*d^2+A*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*c*d-A*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*
d)^(1/2)*d^2-3*B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*c^2-B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*c*d)/
d^2/(c+d)/(c+d*sin(f*x+e))/(a*(c+d)*d)^(1/2)/cos(f*x+e)/(a+a*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 )}^{\frac{3}{2}}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 11.553, size = 3245, normalized size = 16.99 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((3*B*a*c^3 - (A - 6*B)*a*c^2*d - (4*A - B)*a*c*d^2 - (3*A + 2*B)*a*d^3 - (3*B*a*c^2*d - (A - 3*B)*a*c*d^
2 - (3*A + 2*B)*a*d^3)*cos(f*x + e)^2 + (3*B*a*c^3 - (A - 3*B)*a*c^2*d - (3*A + 2*B)*a*c*d^2)*cos(f*x + e) + (
3*B*a*c^3 - (A - 6*B)*a*c^2*d - (4*A - B)*a*c*d^2 - (3*A + 2*B)*a*d^3 + (3*B*a*c^2*d - (A - 3*B)*a*c*d^2 - (3*
A + 2*B)*a*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a/(c*d + d^2))*log((a*d^2*cos(f*x + e)^3 - a*c^2 - 2*a*c*d -
a*d^2 - (6*a*c*d + 7*a*d^2)*cos(f*x + e)^2 + 4*(c^2*d + 4*c*d^2 + 3*d^3 - (c*d^2 + d^3)*cos(f*x + e)^2 + (c^2*
d + 3*c*d^2 + 2*d^3)*cos(f*x + e) - (c^2*d + 4*c*d^2 + 3*d^3 + (c*d^2 + d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(
a*sin(f*x + e) + a)*sqrt(a/(c*d + d^2)) - (a*c^2 + 8*a*c*d + 9*a*d^2)*cos(f*x + e) + (a*d^2*cos(f*x + e)^2 - a
*c^2 - 2*a*c*d - a*d^2 + 2*(3*a*c*d + 4*a*d^2)*cos(f*x + e))*sin(f*x + e))/(d^2*cos(f*x + e)^3 + (2*c*d + d^2)
*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 + d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^
2 - 2*c*d - d^2)*sin(f*x + e))) + 4*(3*B*a*c^2 - (A + B)*a*c*d + (A - 2*B)*a*d^2 + 2*(B*a*c*d + B*a*d^2)*cos(f
*x + e)^2 + (3*B*a*c^2 - (A - B)*a*c*d + A*a*d^2)*cos(f*x + e) - (3*B*a*c^2 - (A + B)*a*c*d + (A - 2*B)*a*d^2
- 2*(B*a*c*d + B*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((c*d^3 + d^4)*f*cos(f*x + e)^2
- (c^2*d^2 + c*d^3)*f*cos(f*x + e) - (c^2*d^2 + 2*c*d^3 + d^4)*f - ((c*d^3 + d^4)*f*cos(f*x + e) + (c^2*d^2 +
2*c*d^3 + d^4)*f)*sin(f*x + e)), -1/2*((3*B*a*c^3 - (A - 6*B)*a*c^2*d - (4*A - B)*a*c*d^2 - (3*A + 2*B)*a*d^3
- (3*B*a*c^2*d - (A - 3*B)*a*c*d^2 - (3*A + 2*B)*a*d^3)*cos(f*x + e)^2 + (3*B*a*c^3 - (A - 3*B)*a*c^2*d - (3*A
+ 2*B)*a*c*d^2)*cos(f*x + e) + (3*B*a*c^3 - (A - 6*B)*a*c^2*d - (4*A - B)*a*c*d^2 - (3*A + 2*B)*a*d^3 + (3*B*
a*c^2*d - (A - 3*B)*a*c*d^2 - (3*A + 2*B)*a*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(-a/(c*d + d^2))*arctan(1/2*s
qrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) - c - 2*d)*sqrt(-a/(c*d + d^2))/(a*cos(f*x + e))) - 2*(3*B*a*c^2 - (A
+ B)*a*c*d + (A - 2*B)*a*d^2 + 2*(B*a*c*d + B*a*d^2)*cos(f*x + e)^2 + (3*B*a*c^2 - (A - B)*a*c*d + A*a*d^2)*co
s(f*x + e) - (3*B*a*c^2 - (A + B)*a*c*d + (A - 2*B)*a*d^2 - 2*(B*a*c*d + B*a*d^2)*cos(f*x + e))*sin(f*x + e))*
sqrt(a*sin(f*x + e) + a))/((c*d^3 + d^4)*f*cos(f*x + e)^2 - (c^2*d^2 + c*d^3)*f*cos(f*x + e) - (c^2*d^2 + 2*c*
d^3 + d^4)*f - ((c*d^3 + d^4)*f*cos(f*x + e) + (c^2*d^2 + 2*c*d^3 + d^4)*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))**(3/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError