3.33 \(\int \frac{(a+b \sin (e+f x)) (A+B \sin (e+f x)+C \sin ^2(e+f x))}{\sin ^{\frac{3}{2}}(e+f x)} \, dx\)

Optimal. Leaf size=117 \[ \frac{2 F\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) (3 a B+3 A b+b C)}{3 f}+\frac{2 E\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) (b B-a (A-C))}{f}-\frac{2 a A \cos (e+f x)}{f \sqrt{\sin (e+f x)}}-\frac{2 b C \sqrt{\sin (e+f x)} \cos (e+f x)}{3 f} \]

[Out]

(2*(b*B - a*(A - C))*EllipticE[(e - Pi/2 + f*x)/2, 2])/f + (2*(3*A*b + 3*a*B + b*C)*EllipticF[(e - Pi/2 + f*x)
/2, 2])/(3*f) - (2*a*A*Cos[e + f*x])/(f*Sqrt[Sin[e + f*x]]) - (2*b*C*Cos[e + f*x]*Sqrt[Sin[e + f*x]])/(3*f)

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Rubi [A]  time = 0.219201, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {3031, 3023, 2748, 2641, 2639} \[ \frac{2 F\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) (3 a B+3 A b+b C)}{3 f}+\frac{2 E\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) (b B-a (A-C))}{f}-\frac{2 a A \cos (e+f x)}{f \sqrt{\sin (e+f x)}}-\frac{2 b C \sqrt{\sin (e+f x)} \cos (e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Sin[e + f*x])*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2))/Sin[e + f*x]^(3/2),x]

[Out]

(2*(b*B - a*(A - C))*EllipticE[(e - Pi/2 + f*x)/2, 2])/f + (2*(3*A*b + 3*a*B + b*C)*EllipticF[(e - Pi/2 + f*x)
/2, 2])/(3*f) - (2*a*A*Cos[e + f*x])/(f*Sqrt[Sin[e + f*x]]) - (2*b*C*Cos[e + f*x]*Sqrt[Sin[e + f*x]])/(3*f)

Rule 3031

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

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{(a+b \sin (e+f x)) \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right )}{\sin ^{\frac{3}{2}}(e+f x)} \, dx &=-\frac{2 a A \cos (e+f x)}{f \sqrt{\sin (e+f x)}}-2 \int \frac{\frac{1}{2} (-A b-a B)-\frac{1}{2} (b B-a (A-C)) \sin (e+f x)-\frac{1}{2} b C \sin ^2(e+f x)}{\sqrt{\sin (e+f x)}} \, dx\\ &=-\frac{2 a A \cos (e+f x)}{f \sqrt{\sin (e+f x)}}-\frac{2 b C \cos (e+f x) \sqrt{\sin (e+f x)}}{3 f}-\frac{4}{3} \int \frac{\frac{1}{4} (-3 A b-3 a B-b C)-\frac{3}{4} (b B-a (A-C)) \sin (e+f x)}{\sqrt{\sin (e+f x)}} \, dx\\ &=-\frac{2 a A \cos (e+f x)}{f \sqrt{\sin (e+f x)}}-\frac{2 b C \cos (e+f x) \sqrt{\sin (e+f x)}}{3 f}-(-b B+a (A-C)) \int \sqrt{\sin (e+f x)} \, dx-\frac{1}{3} (-3 A b-3 a B-b C) \int \frac{1}{\sqrt{\sin (e+f x)}} \, dx\\ &=\frac{2 (b B-a (A-C)) E\left (\left .\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )\right |2\right )}{f}+\frac{2 (3 A b+3 a B+b C) F\left (\left .\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )\right |2\right )}{3 f}-\frac{2 a A \cos (e+f x)}{f \sqrt{\sin (e+f x)}}-\frac{2 b C \cos (e+f x) \sqrt{\sin (e+f x)}}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.748548, size = 97, normalized size = 0.83 \[ -\frac{2 F\left (\left .\frac{1}{4} (-2 e-2 f x+\pi )\right |2\right ) (3 a B+3 A b+b C)+6 E\left (\left .\frac{1}{4} (-2 e-2 f x+\pi )\right |2\right ) (a (C-A)+b B)+\frac{2 \cos (e+f x) (3 a A+b C \sin (e+f x))}{\sqrt{\sin (e+f x)}}}{3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Sin[e + f*x])*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2))/Sin[e + f*x]^(3/2),x]

[Out]

-(6*(b*B + a*(-A + C))*EllipticE[(-2*e + Pi - 2*f*x)/4, 2] + 2*(3*A*b + 3*a*B + b*C)*EllipticF[(-2*e + Pi - 2*
f*x)/4, 2] + (2*Cos[e + f*x]*(3*a*A + b*C*Sin[e + f*x]))/Sqrt[Sin[e + f*x]])/(3*f)

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Maple [B]  time = 1.063, size = 516, normalized size = 4.4 \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))*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/sin(f*x+e)^(3/2),x)

[Out]

(-A*(1+sin(f*x+e))^(1/2)*(2-2*sin(f*x+e))^(1/2)*(-sin(f*x+e))^(1/2)*EllipticF((1+sin(f*x+e))^(1/2),1/2*2^(1/2)
)*a+A*b*(1+sin(f*x+e))^(1/2)*(2-2*sin(f*x+e))^(1/2)*(-sin(f*x+e))^(1/2)*EllipticF((1+sin(f*x+e))^(1/2),1/2*2^(
1/2))+2*A*(1+sin(f*x+e))^(1/2)*(2-2*sin(f*x+e))^(1/2)*(-sin(f*x+e))^(1/2)*EllipticE((1+sin(f*x+e))^(1/2),1/2*2
^(1/2))*a+B*a*(1+sin(f*x+e))^(1/2)*(2-2*sin(f*x+e))^(1/2)*(-sin(f*x+e))^(1/2)*EllipticF((1+sin(f*x+e))^(1/2),1
/2*2^(1/2))+B*b*(1+sin(f*x+e))^(1/2)*(2-2*sin(f*x+e))^(1/2)*(-sin(f*x+e))^(1/2)*EllipticF((1+sin(f*x+e))^(1/2)
,1/2*2^(1/2))-2*B*b*(1+sin(f*x+e))^(1/2)*(2-2*sin(f*x+e))^(1/2)*(-sin(f*x+e))^(1/2)*EllipticE((1+sin(f*x+e))^(
1/2),1/2*2^(1/2))+a*C*(1+sin(f*x+e))^(1/2)*(2-2*sin(f*x+e))^(1/2)*(-sin(f*x+e))^(1/2)*EllipticF((1+sin(f*x+e))
^(1/2),1/2*2^(1/2))+1/3*C*(1+sin(f*x+e))^(1/2)*(2-2*sin(f*x+e))^(1/2)*(-sin(f*x+e))^(1/2)*EllipticF((1+sin(f*x
+e))^(1/2),1/2*2^(1/2))*b-2*a*C*(1+sin(f*x+e))^(1/2)*(2-2*sin(f*x+e))^(1/2)*(-sin(f*x+e))^(1/2)*EllipticE((1+s
in(f*x+e))^(1/2),1/2*2^(1/2))-2/3*C*cos(f*x+e)^2*sin(f*x+e)*b-2*A*a*cos(f*x+e)^2)/cos(f*x+e)/sin(f*x+e)^(1/2)/
f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )}{\left (b \sin \left (f x + e\right ) + a\right )}}{\sin \left (f x + e\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/sin(f*x+e)^(3/2),x, algorithm="maxima")

[Out]

integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(b*sin(f*x + e) + a)/sin(f*x + e)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left ({\left (C a + B b\right )} \cos \left (f x + e\right )^{2} -{\left (A + C\right )} a - B b +{\left (C b \cos \left (f x + e\right )^{2} - B a -{\left (A + C\right )} b\right )} \sin \left (f x + e\right )\right )} \sqrt{\sin \left (f x + e\right )}}{\cos \left (f x + e\right )^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/sin(f*x+e)^(3/2),x, algorithm="fricas")

[Out]

integral(((C*a + B*b)*cos(f*x + e)^2 - (A + C)*a - B*b + (C*b*cos(f*x + e)^2 - B*a - (A + C)*b)*sin(f*x + e))*
sqrt(sin(f*x + e))/(cos(f*x + e)^2 - 1), 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))*(A+B*sin(f*x+e)+C*sin(f*x+e)**2)/sin(f*x+e)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )}{\left (b \sin \left (f x + e\right ) + a\right )}}{\sin \left (f x + e\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/sin(f*x+e)^(3/2),x, algorithm="giac")

[Out]

integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(b*sin(f*x + e) + a)/sin(f*x + e)^(3/2), x)