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

Optimal. Leaf size=285 \[ \frac{2 \left (4 a c d-b \left (c^2+3 d^2\right )\right ) \cos (e+f x)}{3 f \left (c^2-d^2\right )^2 \sqrt{c+d \sin (e+f x)}}-\frac{2 (b c-a d) \cos (e+f x)}{3 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^{3/2}}+\frac{2 (b c-a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d f \left (c^2-d^2\right ) \sqrt{c+d \sin (e+f x)}}+\frac{2 \left (4 a c d-b \left (c^2+3 d^2\right )\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d f \left (c^2-d^2\right )^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]

[Out]

(-2*(b*c - a*d)*Cos[e + f*x])/(3*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^(3/2)) + (2*(4*a*c*d - b*(c^2 + 3*d^2))*Co
s[e + f*x])/(3*(c^2 - d^2)^2*f*Sqrt[c + d*Sin[e + f*x]]) + (2*(4*a*c*d - b*(c^2 + 3*d^2))*EllipticE[(e - Pi/2
+ f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(3*d*(c^2 - d^2)^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) +
(2*(b*c - a*d)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*d*(c^2 - d^
2)*f*Sqrt[c + d*Sin[e + f*x]])

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Rubi [A]  time = 0.394533, antiderivative size = 285, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2754, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \left (4 a c d-b \left (c^2+3 d^2\right )\right ) \cos (e+f x)}{3 f \left (c^2-d^2\right )^2 \sqrt{c+d \sin (e+f x)}}-\frac{2 (b c-a d) \cos (e+f x)}{3 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^{3/2}}+\frac{2 (b c-a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d f \left (c^2-d^2\right ) \sqrt{c+d \sin (e+f x)}}+\frac{2 \left (4 a c d-b \left (c^2+3 d^2\right )\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d f \left (c^2-d^2\right )^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*c - a*d)*Cos[e + f*x])/(3*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^(3/2)) + (2*(4*a*c*d - b*(c^2 + 3*d^2))*Co
s[e + f*x])/(3*(c^2 - d^2)^2*f*Sqrt[c + d*Sin[e + f*x]]) + (2*(4*a*c*d - b*(c^2 + 3*d^2))*EllipticE[(e - Pi/2
+ f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(3*d*(c^2 - d^2)^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) +
(2*(b*c - a*d)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*d*(c^2 - d^
2)*f*Sqrt[c + d*Sin[e + f*x]])

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 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 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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 1.48343, size = 199, normalized size = 0.7 \[ \frac{2 \left (\frac{\left (\frac{c+d \sin (e+f x)}{c+d}\right )^{3/2} \left (\left (b \left (c^2+3 d^2\right )-4 a c d\right ) E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-(c-d) (b c-a d) F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )}{d (c-d)^2}-\frac{\cos (e+f x) \left (d \left (b \left (c^2+3 d^2\right )-4 a c d\right ) \sin (e+f x)+a d \left (d^2-5 c^2\right )+2 b c \left (c^2+d^2\right )\right )}{\left (c^2-d^2\right )^2}\right )}{3 f (c+d \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 4.556, size = 887, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((a*d-b*c)/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)*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-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*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)
))))/cos(f*x+e)/(c+d*sin(f*x+e))^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \sin \left (f x + e\right ) + a}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))/(c+d*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (b \sin \left (f x + e\right ) + a\right )} \sqrt{d \sin \left (f x + e\right ) + c}}{3 \, c d^{2} \cos \left (f x + e\right )^{2} - c^{3} - 3 \, c d^{2} +{\left (d^{3} \cos \left (f x + e\right )^{2} - 3 \, c^{2} d - d^{3}\right )} \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))/(c+d*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \sin \left (f x + e\right ) + a}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))/(c+d*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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