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

Optimal. Leaf size=218 \[ \frac{2 a^3 \left (5 A d (3 c-7 d)-B \left (15 c^2-35 c d+32 d^2\right )\right ) \cos (e+f x)}{15 d^3 f \sqrt{a \sin (e+f x)+a}}+\frac{2 a^2 (-5 A d+5 B c-8 B d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{15 d^2 f}+\frac{2 a^{5/2} (c-d)^2 (B c-A d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{d^{7/2} f \sqrt{c+d}}-\frac{2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 d f} \]

[Out]

(2*a^(5/2)*(c - d)^2*(B*c - A*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])
])/(d^(7/2)*Sqrt[c + d]*f) + (2*a^3*(5*A*(3*c - 7*d)*d - B*(15*c^2 - 35*c*d + 32*d^2))*Cos[e + f*x])/(15*d^3*f
*Sqrt[a + a*Sin[e + f*x]]) + (2*a^2*(5*B*c - 5*A*d - 8*B*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(15*d^2*f)
- (2*a*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(5*d*f)

________________________________________________________________________________________

Rubi [A]  time = 0.884646, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used = {2976, 2981, 2773, 208} \[ \frac{2 a^3 \left (5 A d (3 c-7 d)-B \left (15 c^2-35 c d+32 d^2\right )\right ) \cos (e+f x)}{15 d^3 f \sqrt{a \sin (e+f x)+a}}+\frac{2 a^2 (-5 A d+5 B c-8 B d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{15 d^2 f}+\frac{2 a^{5/2} (c-d)^2 (B c-A d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{d^{7/2} f \sqrt{c+d}}-\frac{2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(5/2)*(c - d)^2*(B*c - A*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])
])/(d^(7/2)*Sqrt[c + d]*f) + (2*a^3*(5*A*(3*c - 7*d)*d - B*(15*c^2 - 35*c*d + 32*d^2))*Cos[e + f*x])/(15*d^3*f
*Sqrt[a + a*Sin[e + f*x]]) + (2*a^2*(5*B*c - 5*A*d - 8*B*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(15*d^2*f)
- (2*a*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(5*d*f)

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]

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))^{5/2} (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx &=-\frac{2 a B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 d f}+\frac{2 \int \frac{(a+a \sin (e+f x))^{3/2} \left (\frac{1}{2} a (3 B c+5 A d)-\frac{1}{2} a (5 B c-5 A d-8 B d) \sin (e+f x)\right )}{c+d \sin (e+f x)} \, dx}{5 d}\\ &=\frac{2 a^2 (5 B c-5 A d-8 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 d^2 f}-\frac{2 a B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 d f}+\frac{4 \int \frac{\sqrt{a+a \sin (e+f x)} \left (-\frac{1}{4} a^2 (B c (5 c-17 d)-5 A d (c+3 d))-\frac{1}{4} a^2 \left (5 A (3 c-7 d) d-B \left (15 c^2-35 c d+32 d^2\right )\right ) \sin (e+f x)\right )}{c+d \sin (e+f x)} \, dx}{15 d^2}\\ &=\frac{2 a^3 \left (5 A (3 c-7 d) d-B \left (15 c^2-35 c d+32 d^2\right )\right ) \cos (e+f x)}{15 d^3 f \sqrt{a+a \sin (e+f x)}}+\frac{2 a^2 (5 B c-5 A d-8 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 d^2 f}-\frac{2 a B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 d f}-\frac{\left (a^2 (c-d)^2 (B c-A d)\right ) \int \frac{\sqrt{a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{d^3}\\ &=\frac{2 a^3 \left (5 A (3 c-7 d) d-B \left (15 c^2-35 c d+32 d^2\right )\right ) \cos (e+f x)}{15 d^3 f \sqrt{a+a \sin (e+f x)}}+\frac{2 a^2 (5 B c-5 A d-8 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 d^2 f}-\frac{2 a B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 d f}+\frac{\left (2 a^3 (c-d)^2 (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^3 f}\\ &=\frac{2 a^{5/2} (c-d)^2 (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^{7/2} \sqrt{c+d} f}+\frac{2 a^3 \left (5 A (3 c-7 d) d-B \left (15 c^2-35 c d+32 d^2\right )\right ) \cos (e+f x)}{15 d^3 f \sqrt{a+a \sin (e+f x)}}+\frac{2 a^2 (5 B c-5 A d-8 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 d^2 f}-\frac{2 a B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 d f}\\ \end{align*}

Mathematica [B]  time = 5.86886, size = 450, normalized size = 2.06 \[ \frac{(a (\sin (e+f x)+1))^{5/2} \left (30 \sqrt{d} \left (A d (5 d-2 c)+B \left (2 c^2-5 c d+5 d^2\right )\right ) \sin \left (\frac{1}{2} (e+f x)\right )-30 \sqrt{d} \left (A d (5 d-2 c)+B \left (2 c^2-5 c d+5 d^2\right )\right ) \cos \left (\frac{1}{2} (e+f x)\right )-5 d^{3/2} (2 A d-2 B c+5 B d) \sin \left (\frac{3}{2} (e+f x)\right )-5 d^{3/2} (2 A d-2 B c+5 B d) \cos \left (\frac{3}{2} (e+f x)\right )-\frac{15 (c-d)^2 (B c-A d) \left (2 \log \left (\sqrt{d} \sqrt{c+d} \left (\tan ^2\left (\frac{1}{4} (e+f x)\right )+2 \tan \left (\frac{1}{4} (e+f x)\right )-1\right )+(c+d) \sec ^2\left (\frac{1}{4} (e+f x)\right )\right )-2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right )\right )+e+f x\right )}{\sqrt{c+d}}+\frac{15 (c-d)^2 (B c-A d) \left (2 \log \left (-\sec ^2\left (\frac{1}{4} (e+f x)\right ) \left (-\sqrt{d} \sqrt{c+d} \sin \left (\frac{1}{2} (e+f x)\right )+\sqrt{d} \sqrt{c+d} \cos \left (\frac{1}{2} (e+f x)\right )+c+d\right )\right )-2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right )\right )+e+f x\right )}{\sqrt{c+d}}-3 B d^{5/2} \sin \left (\frac{5}{2} (e+f x)\right )+3 B d^{5/2} \cos \left (\frac{5}{2} (e+f x)\right )\right )}{30 d^{7/2} f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a*(1 + Sin[e + f*x]))^(5/2)*(-30*Sqrt[d]*(A*d*(-2*c + 5*d) + B*(2*c^2 - 5*c*d + 5*d^2))*Cos[(e + f*x)/2] - 5
*d^(3/2)*(-2*B*c + 2*A*d + 5*B*d)*Cos[(3*(e + f*x))/2] + 3*B*d^(5/2)*Cos[(5*(e + f*x))/2] + (15*(c - d)^2*(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 + Sqrt[d]*Sqrt[c + d]*Cos[(e
+ f*x)/2] - Sqrt[d]*Sqrt[c + d]*Sin[(e + f*x)/2]))]))/Sqrt[c + d] - (15*(c - d)^2*(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*Tan[(e + f*x)/4] + Tan[(
e + f*x)/4]^2)]))/Sqrt[c + d] + 30*Sqrt[d]*(A*d*(-2*c + 5*d) + B*(2*c^2 - 5*c*d + 5*d^2))*Sin[(e + f*x)/2] - 5
*d^(3/2)*(-2*B*c + 2*A*d + 5*B*d)*Sin[(3*(e + f*x))/2] - 3*B*d^(5/2)*Sin[(5*(e + f*x))/2]))/(30*d^(7/2)*f*(Cos
[(e + f*x)/2] + Sin[(e + f*x)/2])^5)

________________________________________________________________________________________

Maple [B]  time = 1.764, size = 543, normalized size = 2.5 \begin{align*}{\frac{2+2\,\sin \left ( fx+e \right ) }{15\,{d}^{3}\cos \left ( fx+e \right ) f}\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) } \left ( -3\,B \left ( a-a\sin \left ( fx+e \right ) \right ) ^{5/2}\sqrt{a \left ( c+d \right ) d}{d}^{2}+5\,A \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{a \left ( c+d \right ) d}a{d}^{2}-15\,A{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ){a}^{3}{c}^{2}d+30\,A{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ){a}^{3}c{d}^{2}-15\,A{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ){a}^{3}{d}^{3}-5\,B \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{a \left ( c+d \right ) d}acd+20\,B \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{a \left ( c+d \right ) d}a{d}^{2}+15\,B{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ){a}^{3}{c}^{3}-30\,B{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ){a}^{3}{c}^{2}d+15\,B{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ){a}^{3}c{d}^{2}+15\,A\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}{a}^{2}cd-45\,A\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}{a}^{2}{d}^{2}-15\,B\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}{a}^{2}{c}^{2}+45\,B\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}{a}^{2}cd-60\,B\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}{a}^{2}{d}^{2} \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))^(5/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x)

[Out]

2/15*(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(-3*B*(a-a*sin(f*x+e))^(5/2)*(a*(c+d)*d)^(1/2)*d^2+5*A*(a-a*sin
(f*x+e))^(3/2)*(a*(c+d)*d)^(1/2)*a*d^2-15*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^3*c^2*d+30
*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^3*c*d^2-15*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*
d+a*d^2)^(1/2))*a^3*d^3-5*B*(a-a*sin(f*x+e))^(3/2)*(a*(c+d)*d)^(1/2)*a*c*d+20*B*(a-a*sin(f*x+e))^(3/2)*(a*(c+d
)*d)^(1/2)*a*d^2+15*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^3*c^3-30*B*arctanh((a-a*sin(f*x+
e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^3*c^2*d+15*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^3*c*d^
2+15*A*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a^2*c*d-45*A*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a^2*d^2-
15*B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a^2*c^2+45*B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a^2*c*d-60
*B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a^2*d^2)/d^3/(a*(c+d)*d)^(1/2)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

________________________________________________________________________________________

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{5}{2}}}{d \sin \left (f x + e\right ) + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

Fricas [B]  time = 17.6338, size = 2961, normalized size = 13.58 \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))^(5/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

[1/30*(15*(B*a^2*c^3 - (A + 2*B)*a^2*c^2*d + (2*A + B)*a^2*c*d^2 - A*a^2*d^3 + (B*a^2*c^3 - (A + 2*B)*a^2*c^2*
d + (2*A + B)*a^2*c*d^2 - A*a^2*d^3)*cos(f*x + e) + (B*a^2*c^3 - (A + 2*B)*a^2*c^2*d + (2*A + B)*a^2*c*d^2 - A
*a^2*d^3)*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^2*d^2*cos(f*x + e)^3 - 15*B*a^2*c^2 + 5*(3*A + 7*B)*a^2*c*d - (35*A + 32*B)*a^2*d^2 + (
5*B*a^2*c*d - (5*A + 11*B)*a^2*d^2)*cos(f*x + e)^2 - (15*B*a^2*c^2 - 5*(3*A + 8*B)*a^2*c*d + 2*(20*A + 23*B)*a
^2*d^2)*cos(f*x + e) - (3*B*a^2*d^2*cos(f*x + e)^2 - 15*B*a^2*c^2 + 5*(3*A + 7*B)*a^2*c*d - (35*A + 32*B)*a^2*
d^2 - (5*B*a^2*c*d - (5*A + 14*B)*a^2*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(d^3*f*cos(f*
x + e) + d^3*f*sin(f*x + e) + d^3*f), 1/15*(15*(B*a^2*c^3 - (A + 2*B)*a^2*c^2*d + (2*A + B)*a^2*c*d^2 - A*a^2*
d^3 + (B*a^2*c^3 - (A + 2*B)*a^2*c^2*d + (2*A + B)*a^2*c*d^2 - A*a^2*d^3)*cos(f*x + e) + (B*a^2*c^3 - (A + 2*B
)*a^2*c^2*d + (2*A + B)*a^2*c*d^2 - A*a^2*d^3)*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*(3*B*a^2*d^2*cos(f*x + e)^3 - 15
*B*a^2*c^2 + 5*(3*A + 7*B)*a^2*c*d - (35*A + 32*B)*a^2*d^2 + (5*B*a^2*c*d - (5*A + 11*B)*a^2*d^2)*cos(f*x + e)
^2 - (15*B*a^2*c^2 - 5*(3*A + 8*B)*a^2*c*d + 2*(20*A + 23*B)*a^2*d^2)*cos(f*x + e) - (3*B*a^2*d^2*cos(f*x + e)
^2 - 15*B*a^2*c^2 + 5*(3*A + 7*B)*a^2*c*d - (35*A + 32*B)*a^2*d^2 - (5*B*a^2*c*d - (5*A + 14*B)*a^2*d^2)*cos(f
*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(d^3*f*cos(f*x + e) + d^3*f*sin(f*x + e) + d^3*f)]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [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))^(5/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

Timed out