3.542 \(\int \frac{(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=194 \[ -\frac{a^{5/2} \left (3 c^2+10 c d+19 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{4 d^{5/2} f (c+d)^{5/2}}+\frac{3 a^3 (c-d) (c+3 d) \cos (e+f x)}{4 d^2 f (c+d)^2 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))}+\frac{a^2 (c-d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{2 d f (c+d) (c+d \sin (e+f x))^2} \]

[Out]

-(a^(5/2)*(3*c^2 + 10*c*d + 19*d^2)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x
]])])/(4*d^(5/2)*(c + d)^(5/2)*f) + (a^2*(c - d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(2*d*(c + d)*f*(c + d*
Sin[e + f*x])^2) + (3*a^3*(c - d)*(c + 3*d)*Cos[e + f*x])/(4*d^2*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*S
in[e + f*x]))

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Rubi [A]  time = 0.442532, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2762, 2980, 2773, 208} \[ -\frac{a^{5/2} \left (3 c^2+10 c d+19 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{4 d^{5/2} f (c+d)^{5/2}}+\frac{3 a^3 (c-d) (c+3 d) \cos (e+f x)}{4 d^2 f (c+d)^2 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))}+\frac{a^2 (c-d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{2 d f (c+d) (c+d \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^3,x]

[Out]

-(a^(5/2)*(3*c^2 + 10*c*d + 19*d^2)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x
]])])/(4*d^(5/2)*(c + d)^(5/2)*f) + (a^2*(c - d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(2*d*(c + d)*f*(c + d*
Sin[e + f*x])^2) + (3*a^3*(c - d)*(c + 3*d)*Cos[e + f*x])/(4*d^2*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*S
in[e + f*x]))

Rule 2762

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c
+ a*d)), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*
Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], 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] && GtQ[m, 1] && LtQ[n, -1]
&& (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2980

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((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]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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] && 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}}{(c+d \sin (e+f x))^3} \, dx &=\frac{a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac{a \int \frac{\sqrt{a+a \sin (e+f x)} \left (\frac{1}{2} a (c-9 d)-\frac{1}{2} a (3 c+5 d) \sin (e+f x)\right )}{(c+d \sin (e+f x))^2} \, dx}{2 d (c+d)}\\ &=\frac{a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{2 d (c+d) f (c+d \sin (e+f x))^2}+\frac{3 a^3 (c-d) (c+3 d) \cos (e+f x)}{4 d^2 (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}+\frac{\left (a^2 \left (3 c^2+10 c d+19 d^2\right )\right ) \int \frac{\sqrt{a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{8 d^2 (c+d)^2}\\ &=\frac{a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{2 d (c+d) f (c+d \sin (e+f x))^2}+\frac{3 a^3 (c-d) (c+3 d) \cos (e+f x)}{4 d^2 (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}-\frac{\left (a^3 \left (3 c^2+10 c d+19 d^2\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 )}{4 d^2 (c+d)^2 f}\\ &=-\frac{a^{5/2} \left (3 c^2+10 c d+19 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a+a \sin (e+f x)}}\right )}{4 d^{5/2} (c+d)^{5/2} f}+\frac{a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{2 d (c+d) f (c+d \sin (e+f x))^2}+\frac{3 a^3 (c-d) (c+3 d) \cos (e+f x)}{4 d^2 (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}\\ \end{align*}

Mathematica [A]  time = 4.723, size = 379, normalized size = 1.95 \[ \frac{(a (\sin (e+f x)+1))^{5/2} \left (-\frac{4 \sqrt{d} \left (-5 c^2-6 c d+11 d^2\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{(c+d)^2 (c+d \sin (e+f x))}+\frac{\left (3 c^2+10 c d+19 d^2\right ) \left (2 \log \left (\sqrt{d} \sqrt{c+d} \left (\tan ^2\left (\frac{1}{4} (e+f x)\right )+2 \tan \left (\frac{1}{4} (e+f x)\right )-1\right )+(c+d) \sec ^2\left (\frac{1}{4} (e+f x)\right )\right )-2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right )\right )+e+f x\right )}{(c+d)^{5/2}}-\frac{\left (3 c^2+10 c d+19 d^2\right ) \left (2 \log \left (-\sec ^2\left (\frac{1}{4} (e+f x)\right ) \left (-\sqrt{d} \sqrt{c+d} \sin \left (\frac{1}{2} (e+f x)\right )+\sqrt{d} \sqrt{c+d} \cos \left (\frac{1}{2} (e+f x)\right )+c+d\right )\right )-2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right )\right )+e+f x\right )}{(c+d)^{5/2}}-\frac{8 \sqrt{d} (c-d)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{(c+d) (c+d \sin (e+f x))^2}\right )}{16 d^{5/2} f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^3,x]

[Out]

((a*(1 + Sin[e + f*x]))^(5/2)*(-(((3*c^2 + 10*c*d + 19*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)^(5/2)) + ((3*c^2 + 10*c*d + 19*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)^(5/2) - (8*(c - d)^2*Sqrt[d]
*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c + d)*(c + d*Sin[e + f*x])^2) - (4*Sqrt[d]*(-5*c^2 - 6*c*d + 11*d^2
)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c + d)^2*(c + d*Sin[e + f*x]))))/(16*d^(5/2)*f*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^5)

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Maple [B]  time = 1.523, size = 567, normalized size = 2.9 \begin{align*} -{\frac{a \left ( 1+\sin \left ( fx+e \right ) \right ) }{4\, \left ( c+d\sin \left ( fx+e \right ) \right ) ^{2} \left ( c+d \right ) ^{2}{d}^{2}\cos \left ( fx+e \right ) f} \left ( 2\,\sin \left ( fx+e \right ){\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ){a}^{2}cd \left ( 3\,{c}^{2}+10\,cd+19\,{d}^{2} \right ) -{\it Artanh} \left ({d\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{acd+a{d}^{2}}}}} \right ){a}^{2}{d}^{2} \left ( 3\,{c}^{2}+10\,cd+19\,{d}^{2} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+5\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{a \left ( c+d \right ) d}{c}^{2}d+6\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{a \left ( c+d \right ) d}c{d}^{2}-11\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{a \left ( c+d \right ) d}{d}^{3}+3\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ){a}^{2}{c}^{4}+10\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ){a}^{2}{c}^{3}d+22\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ){a}^{2}{c}^{2}{d}^{2}+10\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ){a}^{2}c{d}^{3}+19\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ){a}^{2}{d}^{4}-3\,\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}a{c}^{3}-13\,\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}a{c}^{2}d+3\,\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}ac{d}^{2}+13\,\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}a{d}^{3} \right ) \sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \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)/(c+d*sin(f*x+e))^3,x)

[Out]

-1/4*a*(2*sin(f*x+e)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c*d*(3*c^2+10*c*d+19*d^2)-arcta
nh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*d^2*(3*c^2+10*c*d+19*d^2)*cos(f*x+e)^2+5*(a-a*sin(f*x+e))
^(3/2)*(a*(c+d)*d)^(1/2)*c^2*d+6*(a-a*sin(f*x+e))^(3/2)*(a*(c+d)*d)^(1/2)*c*d^2-11*(a-a*sin(f*x+e))^(3/2)*(a*(
c+d)*d)^(1/2)*d^3+3*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^4+10*arctanh((a-a*sin(f*x+e))^
(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^3*d+22*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^2*d^2+10
*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c*d^3+19*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*
d^2)^(1/2))*a^2*d^4-3*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*c^3-13*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/
2)*a*c^2*d+3*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*c*d^2+13*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*d^
3)*(-a*(-1+sin(f*x+e)))^(1/2)*(1+sin(f*x+e))/(a*(c+d)*d)^(1/2)/(c+d*sin(f*x+e))^2/(c+d)^2/d^2/cos(f*x+e)/(a+a*
sin(f*x+e))^(1/2)/f

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Maxima [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)/(c+d*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 3.4697, size = 4378, normalized size = 22.57 \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)/(c+d*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

[-1/16*((3*a^2*c^4 + 16*a^2*c^3*d + 42*a^2*c^2*d^2 + 48*a^2*c*d^3 + 19*a^2*d^4 - (3*a^2*c^2*d^2 + 10*a^2*c*d^3
+ 19*a^2*d^4)*cos(f*x + e)^3 - (6*a^2*c^3*d + 23*a^2*c^2*d^2 + 48*a^2*c*d^3 + 19*a^2*d^4)*cos(f*x + e)^2 + (3
*a^2*c^4 + 10*a^2*c^3*d + 22*a^2*c^2*d^2 + 10*a^2*c*d^3 + 19*a^2*d^4)*cos(f*x + e) + (3*a^2*c^4 + 16*a^2*c^3*d
+ 42*a^2*c^2*d^2 + 48*a^2*c*d^3 + 19*a^2*d^4 - (3*a^2*c^2*d^2 + 10*a^2*c*d^3 + 19*a^2*d^4)*cos(f*x + e)^2 + 2
*(3*a^2*c^3*d + 10*a^2*c^2*d^2 + 19*a^2*c*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)*co
s(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*a^2*c^3 + 3*a^2*c^2*d - 15*a^2*c*d^2 + 9
*a^2*d^3 + (5*a^2*c^2*d + 6*a^2*c*d^2 - 11*a^2*d^3)*cos(f*x + e)^2 + (3*a^2*c^3 + 8*a^2*c^2*d - 9*a^2*c*d^2 -
2*a^2*d^3)*cos(f*x + e) - (3*a^2*c^3 + 3*a^2*c^2*d - 15*a^2*c*d^2 + 9*a^2*d^3 - (5*a^2*c^2*d + 6*a^2*c*d^2 - 1
1*a^2*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e)^3 +
(2*c^3*d^3 + 5*c^2*d^4 + 4*c*d^5 + d^6)*f*cos(f*x + e)^2 - (c^4*d^2 + 2*c^3*d^3 + 2*c^2*d^4 + 2*c*d^5 + d^6)*
f*cos(f*x + e) - (c^4*d^2 + 4*c^3*d^3 + 6*c^2*d^4 + 4*c*d^5 + d^6)*f + ((c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x +
e)^2 - 2*(c^3*d^3 + 2*c^2*d^4 + c*d^5)*f*cos(f*x + e) - (c^4*d^2 + 4*c^3*d^3 + 6*c^2*d^4 + 4*c*d^5 + d^6)*f)*s
in(f*x + e)), 1/8*((3*a^2*c^4 + 16*a^2*c^3*d + 42*a^2*c^2*d^2 + 48*a^2*c*d^3 + 19*a^2*d^4 - (3*a^2*c^2*d^2 + 1
0*a^2*c*d^3 + 19*a^2*d^4)*cos(f*x + e)^3 - (6*a^2*c^3*d + 23*a^2*c^2*d^2 + 48*a^2*c*d^3 + 19*a^2*d^4)*cos(f*x
+ e)^2 + (3*a^2*c^4 + 10*a^2*c^3*d + 22*a^2*c^2*d^2 + 10*a^2*c*d^3 + 19*a^2*d^4)*cos(f*x + e) + (3*a^2*c^4 + 1
6*a^2*c^3*d + 42*a^2*c^2*d^2 + 48*a^2*c*d^3 + 19*a^2*d^4 - (3*a^2*c^2*d^2 + 10*a^2*c*d^3 + 19*a^2*d^4)*cos(f*x
+ e)^2 + 2*(3*a^2*c^3*d + 10*a^2*c^2*d^2 + 19*a^2*c*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(-a/(c*d + d^2))*arc
tan(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*a^2*
c^3 + 3*a^2*c^2*d - 15*a^2*c*d^2 + 9*a^2*d^3 + (5*a^2*c^2*d + 6*a^2*c*d^2 - 11*a^2*d^3)*cos(f*x + e)^2 + (3*a^
2*c^3 + 8*a^2*c^2*d - 9*a^2*c*d^2 - 2*a^2*d^3)*cos(f*x + e) - (3*a^2*c^3 + 3*a^2*c^2*d - 15*a^2*c*d^2 + 9*a^2*
d^3 - (5*a^2*c^2*d + 6*a^2*c*d^2 - 11*a^2*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((c^2*d^4
+ 2*c*d^5 + d^6)*f*cos(f*x + e)^3 + (2*c^3*d^3 + 5*c^2*d^4 + 4*c*d^5 + d^6)*f*cos(f*x + e)^2 - (c^4*d^2 + 2*c
^3*d^3 + 2*c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e) - (c^4*d^2 + 4*c^3*d^3 + 6*c^2*d^4 + 4*c*d^5 + d^6)*f + ((c
^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e)^2 - 2*(c^3*d^3 + 2*c^2*d^4 + c*d^5)*f*cos(f*x + e) - (c^4*d^2 + 4*c^3*d
^3 + 6*c^2*d^4 + 4*c*d^5 + d^6)*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))**(5/2)/(c+d*sin(f*x+e))**3,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))^(5/2)/(c+d*sin(f*x+e))^3,x, algorithm="giac")

[Out]

Exception raised: TypeError