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

Optimal. Leaf size=369 \[ -\frac {2 \left (-8 a c d+3 b c^2+5 b d^2\right ) \cos (e+f x)}{15 f \left (c^2-d^2\right )^2 (c+d \sin (e+f x))^{3/2}}-\frac {2 (b c-a d) \cos (e+f x)}{5 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^{5/2}}+\frac {2 \left (-8 a c d+3 b c^2+5 b d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d f \left (c^2-d^2\right )^2 \sqrt {c+d \sin (e+f x)}}-\frac {2 \left (-23 a c^2 d-9 a d^3+3 b c^3+29 b c d^2\right ) \cos (e+f x)}{15 f \left (c^2-d^2\right )^3 \sqrt {c+d \sin (e+f x)}}-\frac {2 \left (-23 a c^2 d-9 a d^3+3 b c^3+29 b c d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d f \left (c^2-d^2\right )^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}} \]

[Out]

-2/5*(-a*d+b*c)*cos(f*x+e)/(c^2-d^2)/f/(c+d*sin(f*x+e))^(5/2)-2/15*(-8*a*c*d+3*b*c^2+5*b*d^2)*cos(f*x+e)/(c^2-
d^2)^2/f/(c+d*sin(f*x+e))^(3/2)-2/15*(-23*a*c^2*d-9*a*d^3+3*b*c^3+29*b*c*d^2)*cos(f*x+e)/(c^2-d^2)^3/f/(c+d*si
n(f*x+e))^(1/2)+2/15*(-23*a*c^2*d-9*a*d^3+3*b*c^3+29*b*c*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)/d/(c^2-d^2)^
3/f/((c+d*sin(f*x+e))/(c+d))^(1/2)-2/15*(-8*a*c*d+3*b*c^2+5*b*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)*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/(c^2-d^2)^2/f/(c+d*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.53, antiderivative size = 369, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2754, 2752, 2663, 2661, 2655, 2653} \[ -\frac {2 \left (-23 a c^2 d-9 a d^3+3 b c^3+29 b c d^2\right ) \cos (e+f x)}{15 f \left (c^2-d^2\right )^3 \sqrt {c+d \sin (e+f x)}}-\frac {2 \left (-8 a c d+3 b c^2+5 b d^2\right ) \cos (e+f x)}{15 f \left (c^2-d^2\right )^2 (c+d \sin (e+f x))^{3/2}}-\frac {2 (b c-a d) \cos (e+f x)}{5 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^{5/2}}+\frac {2 \left (-8 a c d+3 b c^2+5 b d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d f \left (c^2-d^2\right )^2 \sqrt {c+d \sin (e+f x)}}-\frac {2 \left (-23 a c^2 d-9 a d^3+3 b c^3+29 b c d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d f \left (c^2-d^2\right )^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*c - a*d)*Cos[e + f*x])/(5*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^(5/2)) - (2*(3*b*c^2 - 8*a*c*d + 5*b*d^2)*
Cos[e + f*x])/(15*(c^2 - d^2)^2*f*(c + d*Sin[e + f*x])^(3/2)) - (2*(3*b*c^3 - 23*a*c^2*d + 29*b*c*d^2 - 9*a*d^
3)*Cos[e + f*x])/(15*(c^2 - d^2)^3*f*Sqrt[c + d*Sin[e + f*x]]) - (2*(3*b*c^3 - 23*a*c^2*d + 29*b*c*d^2 - 9*a*d
^3)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(15*d*(c^2 - d^2)^3*f*Sqrt[(c + d*S
in[e + f*x])/(c + d)]) + (2*(3*b*c^2 - 8*a*c*d + 5*b*d^2)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c
+ d*Sin[e + f*x])/(c + d)])/(15*d*(c^2 - d^2)^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]

Rubi steps

\begin {align*} \int \frac {a+b \sin (e+f x)}{(c+d \sin (e+f x))^{7/2}} \, dx &=-\frac {2 (b c-a d) \cos (e+f x)}{5 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {2 \int \frac {-\frac {5}{2} (a c-b d)-\frac {3}{2} (b c-a d) \sin (e+f x)}{(c+d \sin (e+f x))^{5/2}} \, dx}{5 \left (c^2-d^2\right )}\\ &=-\frac {2 (b c-a d) \cos (e+f x)}{5 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {2 \left (3 b c^2-8 a c d+5 b d^2\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))^{3/2}}+\frac {4 \int \frac {\frac {3}{4} \left (5 a c^2-8 b c d+3 a d^2\right )+\frac {1}{4} \left (3 b c^2-8 a c d+5 b d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{15 \left (c^2-d^2\right )^2}\\ &=-\frac {2 (b c-a d) \cos (e+f x)}{5 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {2 \left (3 b c^2-8 a c d+5 b d^2\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))^{3/2}}-\frac {2 \left (3 b c^3-23 a c^2 d+29 b c d^2-9 a d^3\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^3 f \sqrt {c+d \sin (e+f x)}}-\frac {8 \int \frac {\frac {1}{8} \left (-15 a c^3+27 b c^2 d-17 a c d^2+5 b d^3\right )+\frac {1}{8} \left (3 b c^3-23 a c^2 d+29 b c d^2-9 a d^3\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 \left (c^2-d^2\right )^3}\\ &=-\frac {2 (b c-a d) \cos (e+f x)}{5 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {2 \left (3 b c^2-8 a c d+5 b d^2\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))^{3/2}}-\frac {2 \left (3 b c^3-23 a c^2 d+29 b c d^2-9 a d^3\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^3 f \sqrt {c+d \sin (e+f x)}}+\frac {\left (3 b c^2-8 a c d+5 b d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 d \left (c^2-d^2\right )^2}-\frac {\left (3 b c^3-23 a c^2 d+29 b c d^2-9 a d^3\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{15 d \left (c^2-d^2\right )^3}\\ &=-\frac {2 (b c-a d) \cos (e+f x)}{5 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {2 \left (3 b c^2-8 a c d+5 b d^2\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))^{3/2}}-\frac {2 \left (3 b c^3-23 a c^2 d+29 b c d^2-9 a d^3\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^3 f \sqrt {c+d \sin (e+f x)}}-\frac {\left (\left (3 b c^3-23 a c^2 d+29 b c d^2-9 a d^3\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 d \left (c^2-d^2\right )^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (\left (3 b c^2-8 a c d+5 b d^2\right ) \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 \left (c^2-d^2\right )^2 \sqrt {c+d \sin (e+f x)}}\\ &=-\frac {2 (b c-a d) \cos (e+f x)}{5 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {2 \left (3 b c^2-8 a c d+5 b d^2\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))^{3/2}}-\frac {2 \left (3 b c^3-23 a c^2 d+29 b c d^2-9 a d^3\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^3 f \sqrt {c+d \sin (e+f x)}}-\frac {2 \left (3 b c^3-23 a c^2 d+29 b c d^2-9 a d^3\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 d \left (c^2-d^2\right )^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 \left (3 b c^2-8 a c d+5 b d^2\right ) 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 \left (c^2-d^2\right )^2 f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 3.06, size = 297, normalized size = 0.80 \[ \frac {2 \left (\frac {\left (\frac {c+d \sin (e+f x)}{c+d}\right )^{5/2} \left (\left (-23 a c^2 d-9 a d^3+3 b c^3+29 b c d^2\right ) E\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )-(c-d) \left (-8 a c d+3 b c^2+5 b d^2\right ) F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )\right )}{d (c-d)^3}+\frac {\cos (e+f x) \left (d^2 \left (23 a c^2 d+9 a d^3-3 b c^3-29 b c d^2\right ) \sin ^2(e+f x)+d \left (54 a c^3 d+10 a c d^3-9 b c^4-60 b c^2 d^2+5 b d^4\right ) \sin (e+f x)+a d \left (34 c^4-5 c^2 d^2+3 d^4\right )+b \left (-9 c^5-25 c^3 d^2+2 c d^4\right )\right )}{\left (c^2-d^2\right )^3}\right )}{15 f (c+d \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((((3*b*c^3 - 23*a*c^2*d + 29*b*c*d^2 - 9*a*d^3)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - (c - d)*
(3*b*c^2 - 8*a*c*d + 5*b*d^2)*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)])*((c + d*Sin[e + f*x])/(c + d))^
(5/2))/((c - d)^3*d) + (Cos[e + f*x]*(a*d*(34*c^4 - 5*c^2*d^2 + 3*d^4) + b*(-9*c^5 - 25*c^3*d^2 + 2*c*d^4) + d
*(-9*b*c^4 + 54*a*c^3*d - 60*b*c^2*d^2 + 10*a*c*d^3 + 5*b*d^4)*Sin[e + f*x] + d^2*(-3*b*c^3 + 23*a*c^2*d - 29*
b*c*d^2 + 9*a*d^3)*Sin[e + f*x]^2))/(c^2 - d^2)^3))/(15*f*(c + d*Sin[e + f*x])^(5/2))

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b \sin \left (f x + e\right ) + a\right )} \sqrt {d \sin \left (f x + e\right ) + c}}{d^{4} \cos \left (f x + e\right )^{4} + c^{4} + 6 \, c^{2} d^{2} + d^{4} - 2 \, {\left (3 \, c^{2} d^{2} + d^{4}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left (c d^{3} \cos \left (f x + e\right )^{2} - c^{3} d - c d^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*sin(f*x + e) + a)*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 \[ \int \frac {b \sin \left (f x + e\right ) + a}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 7.60, size = 1049, normalized size = 2.84 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((a*d-b*c)/d*(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+16/15*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*c
os(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))))+b/d*(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)*Elli
pticF(((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-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 \[ \int \frac {b \sin \left (f x + e\right ) + a}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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