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

Optimal. Leaf size=298 \[ -\frac{2 \left (56 a c d+15 b c^2+25 b d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{105 f}-\frac{2 \left (c^2-d^2\right ) \left (56 a c d+15 b c^2+25 b d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{105 d f \sqrt{c+d \sin (e+f x)}}+\frac{2 \left (161 a c^2 d+63 a d^3+15 b c^3+145 b c d^2\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{105 d f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 (7 a d+5 b c) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 f}-\frac{2 b \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f} \]

[Out]

(-2*(15*b*c^2 + 56*a*c*d + 25*b*d^2)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(105*f) - (2*(5*b*c + 7*a*d)*Cos[e
+ f*x]*(c + d*Sin[e + f*x])^(3/2))/(35*f) - (2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^(5/2))/(7*f) + (2*(15*b*c^
3 + 161*a*c^2*d + 145*b*c*d^2 + 63*a*d^3)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]
])/(105*d*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) - (2*(c^2 - d^2)*(15*b*c^2 + 56*a*c*d + 25*b*d^2)*EllipticF[(e
- Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(105*d*f*Sqrt[c + d*Sin[e + f*x]])

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Rubi [A]  time = 0.480587, antiderivative size = 298, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ -\frac{2 \left (56 a c d+15 b c^2+25 b d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{105 f}-\frac{2 \left (c^2-d^2\right ) \left (56 a c d+15 b c^2+25 b d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{105 d f \sqrt{c+d \sin (e+f x)}}+\frac{2 \left (161 a c^2 d+63 a d^3+15 b c^3+145 b c d^2\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{105 d f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 (7 a d+5 b c) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 f}-\frac{2 b \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(15*b*c^2 + 56*a*c*d + 25*b*d^2)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(105*f) - (2*(5*b*c + 7*a*d)*Cos[e
+ f*x]*(c + d*Sin[e + f*x])^(3/2))/(35*f) - (2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^(5/2))/(7*f) + (2*(15*b*c^
3 + 161*a*c^2*d + 145*b*c*d^2 + 63*a*d^3)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]
])/(105*d*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) - (2*(c^2 - d^2)*(15*b*c^2 + 56*a*c*d + 25*b*d^2)*EllipticF[(e
- Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(105*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))^{5/2} \, dx &=-\frac{2 b \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f}+\frac{2}{7} \int (c+d \sin (e+f x))^{3/2} \left (\frac{1}{2} (7 a c+5 b d)+\frac{1}{2} (5 b c+7 a d) \sin (e+f x)\right ) \, dx\\ &=-\frac{2 (5 b c+7 a d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 f}-\frac{2 b \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f}+\frac{4}{35} \int \sqrt{c+d \sin (e+f x)} \left (\frac{1}{4} \left (40 b c d+7 a \left (5 c^2+3 d^2\right )\right )+\frac{1}{4} \left (15 b c^2+56 a c d+25 b d^2\right ) \sin (e+f x)\right ) \, dx\\ &=-\frac{2 \left (15 b c^2+56 a c d+25 b d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{105 f}-\frac{2 (5 b c+7 a d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 f}-\frac{2 b \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f}+\frac{8}{105} \int \frac{\frac{1}{8} \left (105 a c^3+135 b c^2 d+119 a c d^2+25 b d^3\right )+\frac{1}{8} \left (15 b c^3+161 a c^2 d+145 b c d^2+63 a d^3\right ) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx\\ &=-\frac{2 \left (15 b c^2+56 a c d+25 b d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{105 f}-\frac{2 (5 b c+7 a d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 f}-\frac{2 b \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f}-\frac{\left (\left (c^2-d^2\right ) \left (15 b c^2+56 a c d+25 b d^2\right )\right ) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{105 d}+\frac{\left (15 b c^3+161 a c^2 d+145 b c d^2+63 a d^3\right ) \int \sqrt{c+d \sin (e+f x)} \, dx}{105 d}\\ &=-\frac{2 \left (15 b c^2+56 a c d+25 b d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{105 f}-\frac{2 (5 b c+7 a d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 f}-\frac{2 b \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f}+\frac{\left (\left (15 b c^3+161 a c^2 d+145 b c d^2+63 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}{105 d \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{\left (\left (c^2-d^2\right ) \left (15 b c^2+56 a c d+25 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}{105 d \sqrt{c+d \sin (e+f x)}}\\ &=-\frac{2 \left (15 b c^2+56 a c d+25 b d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{105 f}-\frac{2 (5 b c+7 a d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 f}-\frac{2 b \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f}+\frac{2 \left (15 b c^3+161 a c^2 d+145 b c d^2+63 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)}}{105 d f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 \left (c^2-d^2\right ) \left (15 b c^2+56 a c d+25 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}}}{105 d f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.07843, size = 275, normalized size = 0.92 \[ \frac{-d \cos (e+f x) (c+d \sin (e+f x)) \left (6 d (7 a d+15 b c) \sin (e+f x)+154 a c d+90 b c^2-15 b d^2 \cos (2 (e+f x))+65 b d^2\right )-2 d \left (7 a \left (15 c^3+17 c d^2\right )+5 b d \left (27 c^2+5 d^2\right )\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-2 \left (7 a d \left (23 c^2+9 d^2\right )+5 b \left (3 c^3+29 c d^2\right )\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \left ((c+d) E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-c F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )}{105 d f \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*(5*b*d*(27*c^2 + 5*d^2) + 7*a*(15*c^3 + 17*c*d^2))*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[
(c + d*Sin[e + f*x])/(c + d)] - 2*(7*a*d*(23*c^2 + 9*d^2) + 5*b*(3*c^3 + 29*c*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)])*Sqrt[(c + d*Sin[e + f*x])/
(c + d)] - d*Cos[e + f*x]*(c + d*Sin[e + f*x])*(90*b*c^2 + 154*a*c*d + 65*b*d^2 - 15*b*d^2*Cos[2*(e + f*x)] +
6*d*(15*b*c + 7*a*d)*Sin[e + f*x]))/(105*d*f*Sqrt[c + d*Sin[e + f*x]])

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Maple [B]  time = 1.23, size = 1839, 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))^(5/2),x)

[Out]

2/105*(-45*b*c^3*d^2-25*b*c*d^4-77*a*c^2*d^3-130*((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^3*d^2+145
*((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*d^4+56*a*c^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))*d^2-42*((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)*Ell
ipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*a*c^2*d^3-56*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+s
in(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^4+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))*b*c^4*d+120*((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^2+10*((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^3-120*((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^4-161*((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^4*d+105*a*c^4*
((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+98*((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^2*d^3
+63*((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*d^5+10*b*d^5*sin(f*x+e)^3-21*a*d^5*sin(f*x+e)^2-25*b*d^5
*sin(f*x+e)+15*b*d^5*sin(f*x+e)^5+21*a*d^5*sin(f*x+e)^4+60*b*c*d^4*sin(f*x+e)^4+98*a*c*d^4*sin(f*x+e)^3+90*b*c
^2*d^3*sin(f*x+e)^3+77*a*c^2*d^3*sin(f*x+e)^2+45*b*c^3*d^2*sin(f*x+e)^2-35*b*c*d^4*sin(f*x+e)^2-98*a*c*d^4*sin
(f*x+e)-90*b*c^2*d^3*sin(f*x+e)-25*((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^5-63*((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^5-15*((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)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*b*c^5)/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{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 ({\left (a c^{2} + 2 \, b c d + a d^{2} -{\left (2 \, b c d + a d^{2}\right )} \cos \left (f x + e\right )^{2} -{\left (b d^{2} \cos \left (f x + e\right )^{2} - b c^{2} - 2 \, a c d - b d^{2}\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))^(5/2),x, algorithm="fricas")

[Out]

integral((a*c^2 + 2*b*c*d + a*d^2 - (2*b*c*d + a*d^2)*cos(f*x + e)^2 - (b*d^2*cos(f*x + e)^2 - b*c^2 - 2*a*c*d
- b*d^2)*sin(f*x + e))*sqrt(d*sin(f*x + e) + c), 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(-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, algorithm="giac")

[Out]

Timed out