3.495 \(\int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=320 \[ -\frac {4 a^2 \left (c^2+5 c d-12 d^2\right ) \cos (e+f x)}{15 d f (c-d) (c+d)^3 \sqrt {c+d \sin (e+f x)}}-\frac {4 a^2 \left (c^2+5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d^2 f (c-d) (c+d)^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 a^2 (c+5 d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d^2 f (c+d)^2 \sqrt {c+d \sin (e+f x)}}-\frac {4 a^2 (c+5 d) \cos (e+f x)}{15 d f (c+d)^2 (c+d \sin (e+f x))^{3/2}}+\frac {2 a^2 (c-d) \cos (e+f x)}{5 d f (c+d) (c+d \sin (e+f x))^{5/2}} \]

[Out]

2/5*a^2*(c-d)*cos(f*x+e)/d/(c+d)/f/(c+d*sin(f*x+e))^(5/2)-4/15*a^2*(c+5*d)*cos(f*x+e)/d/(c+d)^2/f/(c+d*sin(f*x
+e))^(3/2)-4/15*a^2*(c^2+5*c*d-12*d^2)*cos(f*x+e)/(c-d)/d/(c+d)^3/f/(c+d*sin(f*x+e))^(1/2)+4/15*a^2*(c^2+5*c*d
-12*d^2)*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticE(cos(1/2*e+1/4*Pi+1/2*f*x),2^(
1/2)*(d/(c+d))^(1/2))*(c+d*sin(f*x+e))^(1/2)/(c-d)/d^2/(c+d)^3/f/((c+d*sin(f*x+e))/(c+d))^(1/2)-4/15*a^2*(c+5*
d)*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticF(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)*(
d/(c+d))^(1/2))*((c+d*sin(f*x+e))/(c+d))^(1/2)/d^2/(c+d)^2/f/(c+d*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.58, antiderivative size = 320, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2762, 2754, 2752, 2663, 2661, 2655, 2653} \[ -\frac {4 a^2 \left (c^2+5 c d-12 d^2\right ) \cos (e+f x)}{15 d f (c-d) (c+d)^3 \sqrt {c+d \sin (e+f x)}}-\frac {4 a^2 \left (c^2+5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d^2 f (c-d) (c+d)^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 a^2 (c+5 d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d^2 f (c+d)^2 \sqrt {c+d \sin (e+f x)}}-\frac {4 a^2 (c+5 d) \cos (e+f x)}{15 d f (c+d)^2 (c+d \sin (e+f x))^{3/2}}+\frac {2 a^2 (c-d) \cos (e+f x)}{5 d f (c+d) (c+d \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^(7/2),x]

[Out]

(2*a^2*(c - d)*Cos[e + f*x])/(5*d*(c + d)*f*(c + d*Sin[e + f*x])^(5/2)) - (4*a^2*(c + 5*d)*Cos[e + f*x])/(15*d
*(c + d)^2*f*(c + d*Sin[e + f*x])^(3/2)) - (4*a^2*(c^2 + 5*c*d - 12*d^2)*Cos[e + f*x])/(15*(c - d)*d*(c + d)^3
*f*Sqrt[c + d*Sin[e + f*x]]) - (4*a^2*(c^2 + 5*c*d - 12*d^2)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt
[c + d*Sin[e + f*x]])/(15*(c - d)*d^2*(c + d)^3*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (4*a^2*(c + 5*d)*Ellip
ticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(15*d^2*(c + d)^2*f*Sqrt[c + d*Sin
[e + f*x]])

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2754

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

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]))

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{7/2}} \, dx &=\frac {2 a^2 (c-d) \cos (e+f x)}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {(2 a) \int \frac {-5 a d-a (c+4 d) \sin (e+f x)}{(c+d \sin (e+f x))^{5/2}} \, dx}{5 d (c+d)}\\ &=\frac {2 a^2 (c-d) \cos (e+f x)}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {4 a^2 (c+5 d) \cos (e+f x)}{15 d (c+d)^2 f (c+d \sin (e+f x))^{3/2}}+\frac {(4 a) \int \frac {6 a (c-d) d+\frac {1}{2} a (c-d) (c+5 d) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{15 (c-d) d (c+d)^2}\\ &=\frac {2 a^2 (c-d) \cos (e+f x)}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {4 a^2 (c+5 d) \cos (e+f x)}{15 d (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^2 \left (c^2+5 c d-12 d^2\right ) \cos (e+f x)}{15 (c-d) d (c+d)^3 f \sqrt {c+d \sin (e+f x)}}-\frac {(8 a) \int \frac {-\frac {1}{4} a (11 c-5 d) (c-d) d+\frac {1}{4} a (c-d) \left (c^2+5 c d-12 d^2\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 (c-d)^2 d (c+d)^3}\\ &=\frac {2 a^2 (c-d) \cos (e+f x)}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {4 a^2 (c+5 d) \cos (e+f x)}{15 d (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^2 \left (c^2+5 c d-12 d^2\right ) \cos (e+f x)}{15 (c-d) d (c+d)^3 f \sqrt {c+d \sin (e+f x)}}+\frac {\left (2 a^2 (c+5 d)\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 d^2 (c+d)^2}-\frac {\left (2 a^2 \left (c^2+5 c d-12 d^2\right )\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{15 (c-d) d^2 (c+d)^3}\\ &=\frac {2 a^2 (c-d) \cos (e+f x)}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {4 a^2 (c+5 d) \cos (e+f x)}{15 d (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^2 \left (c^2+5 c d-12 d^2\right ) \cos (e+f x)}{15 (c-d) d (c+d)^3 f \sqrt {c+d \sin (e+f x)}}-\frac {\left (2 a^2 \left (c^2+5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{15 (c-d) d^2 (c+d)^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (2 a^2 (c+5 d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right ) \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}} \, dx}{15 d^2 (c+d)^2 \sqrt {c+d \sin (e+f x)}}\\ &=\frac {2 a^2 (c-d) \cos (e+f x)}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {4 a^2 (c+5 d) \cos (e+f x)}{15 d (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^2 \left (c^2+5 c d-12 d^2\right ) \cos (e+f x)}{15 (c-d) d (c+d)^3 f \sqrt {c+d \sin (e+f x)}}-\frac {4 a^2 \left (c^2+5 c d-12 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{15 (c-d) d^2 (c+d)^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 a^2 (c+5 d) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{15 d^2 (c+d)^2 f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 2.03, size = 283, normalized size = 0.88 \[ \frac {2 a^2 (\sin (e+f x)+1)^2 \left (d \cos (e+f x) \left (-2 \left (c^2+5 c d-12 d^2\right ) (c+d \sin (e+f x))^2-2 (c-d) (c+5 d) (c+d) (c+d \sin (e+f x))+3 (c-d)^2 (c+d)^2\right )-2 (c+d \sin (e+f x))^2 \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \left (d^2 (11 c-5 d) F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )-\left (c^2+5 c d-12 d^2\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )-c F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )\right )\right )\right )}{15 d^2 f (c-d) (c+d)^3 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4 (c+d \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^(7/2),x]

[Out]

(2*a^2*(1 + Sin[e + f*x])^2*(-2*((11*c - 5*d)*d^2*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - (c^2 + 5*c
*d - 12*d^2)*((c + d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - c*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*
d)/(c + d)]))*(c + d*Sin[e + f*x])^2*Sqrt[(c + d*Sin[e + f*x])/(c + d)] + d*Cos[e + f*x]*(3*(c - d)^2*(c + d)^
2 - 2*(c - d)*(c + d)*(c + 5*d)*(c + d*Sin[e + f*x]) - 2*(c^2 + 5*c*d - 12*d^2)*(c + d*Sin[e + f*x])^2)))/(15*
(c - d)*d^2*(c + d)^3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*(c + d*Sin[e + f*x])^(5/2))

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}\right )} \sqrt {d \sin \left (f x + e\right ) + c}}{d^{4} \cos \left (f x + e\right )^{4} + c^{4} + 6 \, c^{2} d^{2} + d^{4} - 2 \, {\left (3 \, c^{2} d^{2} + d^{4}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left (c d^{3} \cos \left (f x + e\right )^{2} - c^{3} d - c d^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

integral(-(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2)*sqrt(d*sin(f*x + e) + c)/(d^4*cos(f*x + e)^4 + c^4
+ 6*c^2*d^2 + d^4 - 2*(3*c^2*d^2 + d^4)*cos(f*x + e)^2 - 4*(c*d^3*cos(f*x + e)^2 - c^3*d - c*d^3)*sin(f*x + e
)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^2/(d*sin(f*x + e) + c)^(7/2), x)

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maple [B]  time = 6.39, size = 1436, normalized size = 4.49 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(7/2),x)

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*a^2*(1/d^2*(2*d*cos(f*x+e)^2/(c^2-d^2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^
2)^(1/2)+2*c/(c^2-d^2)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*
d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^
(1/2))+2/(c^2-d^2)*d*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/
(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)
/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))))+2*(-c+d)/d^2*(2/3/(c^2-d^2)/d*(
-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+c/d)^2+8/3*d*cos(f*x+e)^2/(c^2-d^2)^2*c/(-(-d*sin(f*x+e)-c)
*cos(f*x+e)^2)^(1/2)+2*(3*c^2+d^2)/(3*c^4-6*c^2*d^2+3*d^4)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*
x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin
(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+8/3*c*d/(c^2-d^2)^2*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-si
n(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*Ellip
ticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d)
)^(1/2))))+(c^2-2*c*d+d^2)/d^2*(2/5/(c^2-d^2)/d^2*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+c/d)^3+1
6/15*c/(c^2-d^2)^2/d*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+c/d)^2+2/15*d*cos(f*x+e)^2/(c^2-d^2)^
3*(23*c^2+9*d^2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+2*(15*c^3+17*c*d^2)/(15*c^6-45*c^4*d^2+45*c^2*d^4-15*
d^6)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(
-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+2/15*d*(23*
c^2+9*d^2)/(c^2-d^2)^3*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*
d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-
d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2)))))/cos(f*x+e)/(c+d*sin(f*x+e))^
(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^2/(d*sin(f*x + e) + c)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^2/(c + d*sin(e + f*x))^(7/2),x)

[Out]

int((a + a*sin(e + f*x))^2/(c + d*sin(e + f*x))^(7/2), x)

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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))**2/(c+d*sin(f*x+e))**(7/2),x)

[Out]

Timed out

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