3.724 \(\int (a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx\)
Optimal. Leaf size=235 \[ -\frac{2 \left (c^2-d^2\right ) (5 a d+3 b c) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d f \sqrt{c+d \sin (e+f x)}}+\frac{2 \left (20 a c d+3 b \left (c^2+3 d^2\right )\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 (5 a d+3 b c) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 f}-\frac{2 b \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f} \]
[Out]
(-2*(3*b*c + 5*a*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(15*f) - (2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/
2))/(5*f) + (2*(20*a*c*d + 3*b*(c^2 + 3*d^2))*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e +
f*x]])/(15*d*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) - (2*(3*b*c + 5*a*d)*(c^2 - d^2)*EllipticF[(e - Pi/2 + f*x)
/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(15*d*f*Sqrt[c + d*Sin[e + f*x]])
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Rubi [A] time = 0.358819, antiderivative size = 235, 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 =
{2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 \left (c^2-d^2\right ) (5 a d+3 b c) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d f \sqrt{c+d \sin (e+f x)}}+\frac{2 \left (20 a c d+3 b \left (c^2+3 d^2\right )\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 (5 a d+3 b c) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 f}-\frac{2 b \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^(3/2),x]
[Out]
(-2*(3*b*c + 5*a*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(15*f) - (2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/
2))/(5*f) + (2*(20*a*c*d + 3*b*(c^2 + 3*d^2))*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e +
f*x]])/(15*d*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) - (2*(3*b*c + 5*a*d)*(c^2 - d^2)*EllipticF[(e - Pi/2 + f*x)
/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(15*d*f*Sqrt[c + d*Sin[e + f*x]])
Rule 2753
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*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] && GtQ[m, 0] && 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 (a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx &=-\frac{2 b \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f}+\frac{2}{5} \int \sqrt{c+d \sin (e+f x)} \left (\frac{1}{2} (5 a c+3 b d)+\frac{1}{2} (3 b c+5 a d) \sin (e+f x)\right ) \, dx\\ &=-\frac{2 (3 b c+5 a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 f}-\frac{2 b \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f}+\frac{4}{15} \int \frac{\frac{1}{4} \left (12 b c d+5 a \left (3 c^2+d^2\right )\right )+\frac{1}{4} \left (20 a c d+3 b \left (c^2+3 d^2\right )\right ) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx\\ &=-\frac{2 (3 b c+5 a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 f}-\frac{2 b \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f}-\frac{\left ((3 b c+5 a d) \left (c^2-d^2\right )\right ) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{15 d}+\frac{\left (20 a c d+3 b \left (c^2+3 d^2\right )\right ) \int \sqrt{c+d \sin (e+f x)} \, dx}{15 d}\\ &=-\frac{2 (3 b c+5 a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 f}-\frac{2 b \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f}+\frac{\left (\left (20 a c d+3 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}{15 d \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{\left ((3 b c+5 a d) \left (c^2-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 \sqrt{c+d \sin (e+f x)}}\\ &=-\frac{2 (3 b c+5 a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 f}-\frac{2 b \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f}+\frac{2 \left (20 a c d+3 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)}}{15 d f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 (3 b c+5 a d) \left (c^2-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 f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.728198, size = 218, normalized size = 0.93 \[ \frac{-2 d \left (5 a \left (3 c^2+d^2\right )+12 b c d\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-2 \left (20 a c d+3 b \left (c^2+3 d^2\right )\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \left ((c+d) E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-c F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )-2 d \cos (e+f x) (c+d \sin (e+f x)) (5 a d+6 b c+3 b d \sin (e+f x))}{15 d f \sqrt{c+d \sin (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^(3/2),x]
[Out]
(-2*d*(12*b*c*d + 5*a*(3*c^2 + d^2))*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])
/(c + d)] - 2*(20*a*c*d + 3*b*(c^2 + 3*d^2))*((c + d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - c*Elli
pticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)])*Sqrt[(c + d*Sin[e + f*x])/(c + d)] - 2*d*Cos[e + f*x]*(c + d*Sin[
e + f*x])*(6*b*c + 5*a*d + 3*b*d*Sin[e + f*x]))/(15*d*f*Sqrt[c + d*Sin[e + f*x]])
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Maple [B] time = 1.217, size = 1449, normalized size = 6.2 \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))^(3/2),x)
[Out]
2/15*(15*c^3*a*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)
*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*d+5*c^2*a*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+
sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+
d))^(1/2))*d^2-15*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1
/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*a*c*d^3-5*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(
-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/
(c+d))^(1/2))*a*d^4+3*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d)
)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*b*c^3*d+9*((c+d*sin(f*x+e))/(c-d))^(1/2)
*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c
-d)/(c+d))^(1/2))*b*c^2*d^2-3*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e
))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*b*c*d^3-9*((c+d*sin(f*x+e))/(c-d
))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(
1/2),((c-d)/(c+d))^(1/2))*b*d^4-20*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(
f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*a*c^3*d+20*((c+d*sin(f*x+e)
)/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c
-d))^(1/2),((c-d)/(c+d))^(1/2))*a*c*d^3-3*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*
(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*b*c^4-6*((c+d*sin(f*
x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e)
)/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*b*c^2*d^2+9*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2
)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*b*d^4+3*b*d^4*
sin(f*x+e)^4+5*a*d^4*sin(f*x+e)^3+9*b*c*d^3*sin(f*x+e)^3+5*a*c*d^3*sin(f*x+e)^2+6*b*c^2*d^2*sin(f*x+e)^2-3*b*d
^4*sin(f*x+e)^2-5*a*d^4*sin(f*x+e)-9*b*c*d^3*sin(f*x+e)-5*a*c*d^3-6*b*c^2*d^2)/d^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{\left (b \sin \left (f x + e\right ) + a\right )}{\left (d \sin \left (f x + e\right ) + c\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))*(c+d*sin(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
integrate((b*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(3/2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (b d \cos \left (f x + e\right )^{2} - a c - b d -{\left (b c + a d\right )} \sin \left (f x + e\right )\right )} \sqrt{d \sin \left (f x + e\right ) + c}, 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))^(3/2),x, algorithm="fricas")
[Out]
integral(-(b*d*cos(f*x + e)^2 - a*c - b*d - (b*c + a*d)*sin(f*x + e))*sqrt(d*sin(f*x + e) + c), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sin{\left (e + f x \right )}\right ) \left (c + d \sin{\left (e + f x \right )}\right )^{\frac{3}{2}}\, dx \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))**(3/2),x)
[Out]
Integral((a + b*sin(e + f*x))*(c + d*sin(e + f*x))**(3/2), x)
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Giac [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))^(3/2),x, algorithm="giac")
[Out]
Timed out