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

Optimal. Leaf size=153 \[ -\frac {2 a^{3/2} (c-d) (B c-A d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{d^{5/2} f \sqrt {c+d}}+\frac {2 a^2 (-3 A d+3 B c-4 B d) \cos (e+f x)}{3 d^2 f \sqrt {a \sin (e+f x)+a}}-\frac {2 a B \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f} \]

[Out]

-2*a^(3/2)*(c-d)*(-A*d+B*c)*arctanh(cos(f*x+e)*a^(1/2)*d^(1/2)/(c+d)^(1/2)/(a+a*sin(f*x+e))^(1/2))/d^(5/2)/f/(
c+d)^(1/2)+2/3*a^2*(-3*A*d+3*B*c-4*B*d)*cos(f*x+e)/d^2/f/(a+a*sin(f*x+e))^(1/2)-2/3*a*B*cos(f*x+e)*(a+a*sin(f*
x+e))^(1/2)/d/f

________________________________________________________________________________________

Rubi [A]  time = 0.50, antiderivative size = 153, normalized size of antiderivative = 1.00, 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 = {2976, 2981, 2773, 208} \[ \frac {2 a^2 (-3 A d+3 B c-4 B d) \cos (e+f x)}{3 d^2 f \sqrt {a \sin (e+f x)+a}}-\frac {2 a^{3/2} (c-d) (B c-A d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{d^{5/2} f \sqrt {c+d}}-\frac {2 a B \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f} \]

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]),x]

[Out]

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

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]

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 2976

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

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)} \, dx &=-\frac {2 a B \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{3 d f}+\frac {2 \int \frac {\sqrt {a+a \sin (e+f x)} \left (\frac {1}{2} a (B c+3 A d)-\frac {1}{2} a (3 B c-3 A d-4 B d) \sin (e+f x)\right )}{c+d \sin (e+f x)} \, dx}{3 d}\\ &=\frac {2 a^2 (3 B c-3 A d-4 B d) \cos (e+f x)}{3 d^2 f \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{3 d f}+\frac {(a (c-d) (B c-A d)) \int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{d^2}\\ &=\frac {2 a^2 (3 B c-3 A d-4 B d) \cos (e+f x)}{3 d^2 f \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{3 d f}-\frac {\left (2 a^2 (c-d) (B c-A d)\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 f}\\ &=-\frac {2 a^{3/2} (c-d) (B c-A d) \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} \sqrt {c+d} f}+\frac {2 a^2 (3 B c-3 A d-4 B d) \cos (e+f x)}{3 d^2 f \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{3 d f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 3.54, size = 356, normalized size = 2.33 \[ \frac {(a (\sin (e+f x)+1))^{3/2} \left (6 \sqrt {d} (2 A d-2 B c+3 B d) \sin \left (\frac {1}{2} (e+f x)\right )-6 \sqrt {d} (2 A d-2 B c+3 B d) \cos \left (\frac {1}{2} (e+f x)\right )+\frac {3 (c-d) (B c-A d) \left (2 \log \left (\sqrt {d} \sqrt {c+d} \left (\tan ^2\left (\frac {1}{4} (e+f x)\right )+2 \tan \left (\frac {1}{4} (e+f x)\right )-1\right )+(c+d) \sec ^2\left (\frac {1}{4} (e+f x)\right )\right )-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+e+f x\right )}{\sqrt {c+d}}-\frac {3 (c-d) (B c-A d) \left (2 \log \left (-\sec ^2\left (\frac {1}{4} (e+f x)\right ) \left (-\sqrt {d} \sqrt {c+d} \sin \left (\frac {1}{2} (e+f x)\right )+\sqrt {d} \sqrt {c+d} \cos \left (\frac {1}{2} (e+f x)\right )+c+d\right )\right )-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+e+f x\right )}{\sqrt {c+d}}-2 B d^{3/2} \sin \left (\frac {3}{2} (e+f x)\right )-2 B d^{3/2} \cos \left (\frac {3}{2} (e+f x)\right )\right )}{6 d^{5/2} f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

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]),x]

[Out]

((a*(1 + Sin[e + f*x]))^(3/2)*(-6*Sqrt[d]*(-2*B*c + 2*A*d + 3*B*d)*Cos[(e + f*x)/2] - 2*B*d^(3/2)*Cos[(3*(e +
f*x))/2] - (3*(c - d)*(B*c - A*d)*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2] + 2*Log[-(Sec[(e + f*x)/4]^2*(c + d + S
qrt[d]*Sqrt[c + d]*Cos[(e + f*x)/2] - Sqrt[d]*Sqrt[c + d]*Sin[(e + f*x)/2]))]))/Sqrt[c + d] + (3*(c - d)*(B*c
- A*d)*(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*T
an[(e + f*x)/4] + Tan[(e + f*x)/4]^2)]))/Sqrt[c + d] + 6*Sqrt[d]*(-2*B*c + 2*A*d + 3*B*d)*Sin[(e + f*x)/2] - 2
*B*d^(3/2)*Sin[(3*(e + f*x))/2]))/(6*d^(5/2)*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)

________________________________________________________________________________________

fricas [B]  time = 1.47, size = 880, normalized size = 5.75 \[ \left [-\frac {3 \, {\left (B a c^{2} - {\left (A + B\right )} a c d + A a d^{2} + {\left (B a c^{2} - {\left (A + B\right )} a c d + A a d^{2}\right )} \cos \left (f x + e\right ) + {\left (B a c^{2} - {\left (A + B\right )} a c d + A a d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {a}{c d + d^{2}}} \log \left (\frac {a d^{2} \cos \left (f x + e\right )^{3} - a c^{2} - 2 \, a c d - a d^{2} - {\left (6 \, a c d + 7 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left (c^{2} d + 4 \, c d^{2} + 3 \, d^{3} - {\left (c d^{2} + d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (c^{2} d + 3 \, c d^{2} + 2 \, d^{3}\right )} \cos \left (f x + e\right ) - {\left (c^{2} d + 4 \, c d^{2} + 3 \, d^{3} + {\left (c d^{2} + d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {\frac {a}{c d + d^{2}}} - {\left (a c^{2} + 8 \, a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right ) + {\left (a d^{2} \cos \left (f x + e\right )^{2} - a c^{2} - 2 \, a c d - a d^{2} + 2 \, {\left (3 \, a c d + 4 \, a d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{d^{2} \cos \left (f x + e\right )^{3} + {\left (2 \, c d + d^{2}\right )} \cos \left (f x + e\right )^{2} - c^{2} - 2 \, c d - d^{2} - {\left (c^{2} + d^{2}\right )} \cos \left (f x + e\right ) + {\left (d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \cos \left (f x + e\right ) - c^{2} - 2 \, c d - d^{2}\right )} \sin \left (f x + e\right )}\right ) + 4 \, {\left (B a d \cos \left (f x + e\right )^{2} - 3 \, B a c + {\left (3 \, A + 4 \, B\right )} a d - {\left (3 \, B a c - {\left (3 \, A + 5 \, B\right )} a d\right )} \cos \left (f x + e\right ) + {\left (B a d \cos \left (f x + e\right ) + 3 \, B a c - {\left (3 \, A + 4 \, B\right )} a d\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{6 \, {\left (d^{2} f \cos \left (f x + e\right ) + d^{2} f \sin \left (f x + e\right ) + d^{2} f\right )}}, -\frac {3 \, {\left (B a c^{2} - {\left (A + B\right )} a c d + A a d^{2} + {\left (B a c^{2} - {\left (A + B\right )} a c d + A a d^{2}\right )} \cos \left (f x + e\right ) + {\left (B a c^{2} - {\left (A + B\right )} a c d + A a d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-\frac {a}{c d + d^{2}}} \arctan \left (\frac {\sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) - c - 2 \, d\right )} \sqrt {-\frac {a}{c d + d^{2}}}}{2 \, a \cos \left (f x + e\right )}\right ) + 2 \, {\left (B a d \cos \left (f x + e\right )^{2} - 3 \, B a c + {\left (3 \, A + 4 \, B\right )} a d - {\left (3 \, B a c - {\left (3 \, A + 5 \, B\right )} a d\right )} \cos \left (f x + e\right ) + {\left (B a d \cos \left (f x + e\right ) + 3 \, B a c - {\left (3 \, A + 4 \, B\right )} a d\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{3 \, {\left (d^{2} f \cos \left (f x + e\right ) + d^{2} f \sin \left (f x + e\right ) + d^{2} f\right )}}\right ] \]

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)),x, algorithm="fricas")

[Out]

[-1/6*(3*(B*a*c^2 - (A + B)*a*c*d + A*a*d^2 + (B*a*c^2 - (A + B)*a*c*d + A*a*d^2)*cos(f*x + e) + (B*a*c^2 - (A
+ B)*a*c*d + A*a*d^2)*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*(B*a*d*cos(f*x + e)^2 - 3*B*a*c + (3*A + 4*B)*a*d - (3*B*a*c - (3*A + 5*B)*a*d)*c
os(f*x + e) + (B*a*d*cos(f*x + e) + 3*B*a*c - (3*A + 4*B)*a*d)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(d^2*f*
cos(f*x + e) + d^2*f*sin(f*x + e) + d^2*f), -1/3*(3*(B*a*c^2 - (A + B)*a*c*d + A*a*d^2 + (B*a*c^2 - (A + B)*a*
c*d + A*a*d^2)*cos(f*x + e) + (B*a*c^2 - (A + B)*a*c*d + A*a*d^2)*sin(f*x + e))*sqrt(-a/(c*d + d^2))*arctan(1/
2*sqrt(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*(B*a*d*cos(f*
x + e)^2 - 3*B*a*c + (3*A + 4*B)*a*d - (3*B*a*c - (3*A + 5*B)*a*d)*cos(f*x + e) + (B*a*d*cos(f*x + e) + 3*B*a*
c - (3*A + 4*B)*a*d)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(d^2*f*cos(f*x + e) + d^2*f*sin(f*x + e) + d^2*f)
]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

maple [B]  time = 1.73, size = 291, normalized size = 1.90 \[ \frac {2 \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (3 A \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a^{2} c d -3 A \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a^{2} d^{2}-3 B \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a^{2} c^{2}+3 B \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a^{2} c d +B \left (-a \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, d -3 A \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (c +d \right ) d}\, a d +3 B \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (c +d \right ) d}\, a c -6 B \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (c +d \right ) d}\, a d \right )}{3 d^{2} \sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

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)),x)

[Out]

2/3*(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)*(3*A*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^2*c
*d-3*A*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^2*d^2-3*B*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/
(a*(c+d)*d)^(1/2))*a^2*c^2+3*B*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^2*c*d+B*(-a*(sin(f*x+e
)-1))^(3/2)*(a*(c+d)*d)^(1/2)*d-3*A*(-a*(sin(f*x+e)-1))^(1/2)*(a*(c+d)*d)^(1/2)*a*d+3*B*(-a*(sin(f*x+e)-1))^(1
/2)*(a*(c+d)*d)^(1/2)*a*c-6*B*(-a*(sin(f*x+e)-1))^(1/2)*(a*(c+d)*d)^(1/2)*a*d)/d^2/(a*(c+d)*d)^(1/2)/cos(f*x+e
)/(a+a*sin(f*x+e))^(1/2)/f

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{d \sin \left (f x + e\right ) + c}\,{d x} \]

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)),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), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________