3.732 \(\int (a+b \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)} \, dx\)

Optimal. Leaf size=254 \[ \frac{2 \left (3 d^2 \left (5 a^2+3 b^2\right )-2 b c (b c-5 a d)\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{4 b \left (c^2-d^2\right ) (b c-5 a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d^2 f \sqrt{c+d \sin (e+f x)}}+\frac{4 b (b c-5 a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f}-\frac{2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f} \]

[Out]

(4*b*(b*c - 5*a*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(15*d*f) - (2*b^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^
(3/2))/(5*d*f) + (2*(3*(5*a^2 + 3*b^2)*d^2 - 2*b*c*(b*c - 5*a*d))*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]
*Sqrt[c + d*Sin[e + f*x]])/(15*d^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (4*b*(b*c - 5*a*d)*(c^2 - d^2)*Elli
pticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(15*d^2*f*Sqrt[c + d*Sin[e + f*x]
])

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Rubi [A]  time = 0.411492, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2791, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \left (3 d^2 \left (5 a^2+3 b^2\right )-2 b c (b c-5 a d)\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{4 b \left (c^2-d^2\right ) (b c-5 a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d^2 f \sqrt{c+d \sin (e+f x)}}+\frac{4 b (b c-5 a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f}-\frac{2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*(b*c - 5*a*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(15*d*f) - (2*b^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^
(3/2))/(5*d*f) + (2*(3*(5*a^2 + 3*b^2)*d^2 - 2*b*c*(b*c - 5*a*d))*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]
*Sqrt[c + d*Sin[e + f*x]])/(15*d^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (4*b*(b*c - 5*a*d)*(c^2 - d^2)*Elli
pticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(15*d^2*f*Sqrt[c + d*Sin[e + f*x]
])

Rule 2791

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[
(d^2*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[b*(d^2*(m + 1) + c^2*(m + 2)) - d*(a*d - 2*b*c*(m + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

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))^2 \sqrt{c+d \sin (e+f x)} \, dx &=-\frac{2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac{2 \int \sqrt{c+d \sin (e+f x)} \left (\frac{1}{2} \left (5 a^2+3 b^2\right ) d-b (b c-5 a d) \sin (e+f x)\right ) \, dx}{5 d}\\ &=\frac{4 b (b c-5 a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f}-\frac{2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac{4 \int \frac{\frac{1}{4} d \left (15 a^2 c+7 b^2 c+10 a b d\right )+\frac{1}{4} \left (3 \left (5 a^2+3 b^2\right ) d^2-2 b c (b c-5 a d)\right ) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx}{15 d}\\ &=\frac{4 b (b c-5 a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f}-\frac{2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac{\left (2 b (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^2}+\frac{1}{15} \left (15 a^2+9 b^2-\frac{2 b c (b c-5 a d)}{d^2}\right ) \int \sqrt{c+d \sin (e+f x)} \, dx\\ &=\frac{4 b (b c-5 a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f}-\frac{2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac{\left (\left (15 a^2+9 b^2-\frac{2 b c (b c-5 a d)}{d^2}\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 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left (2 b (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^2 \sqrt{c+d \sin (e+f x)}}\\ &=\frac{4 b (b c-5 a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f}-\frac{2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac{2 \left (15 a^2+9 b^2-\frac{2 b c (b c-5 a d)}{d^2}\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 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{4 b (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^2 f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.895404, size = 214, normalized size = 0.84 \[ \frac{2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \left (\left (-15 a^2 d^2-10 a b c d+b^2 \left (2 c^2-9 d^2\right )\right ) \left ((c+d) E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-c F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )-d^2 \left (15 a^2 c+10 a b d+7 b^2 c\right ) F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )-2 b d \cos (e+f x) (c+d \sin (e+f x)) (10 a d+b c+3 b d \sin (e+f x))}{15 d^2 f \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-(d^2*(15*a^2*c + 7*b^2*c + 10*a*b*d)*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]) + (-10*a*b*c*d - 15
*a^2*d^2 + b^2*(2*c^2 - 9*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)] - 2*b*d*Cos[e + f*x]*(c + d*Sin[e + f*x])*
(b*c + 10*a*d + 3*b*d*Sin[e + f*x]))/(15*d^2*f*Sqrt[c + d*Sin[e + f*x]])

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Maple [B]  time = 3.353, size = 1100, normalized size = 4.3 \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))^2*(c+d*sin(f*x+e))^(1/2),x)

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*(b^2*d*(-2/5/d*sin(f*x+e)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+8/15
*c/d^2*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+4/15*c/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)*EllipticF(((c+d*sin(f*
x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+2*(3/5+8/15*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)*((-c/d-1)*Ellipti
cE(((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))))+(2*a*b*d+b^2*c)*(-2/3/d*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+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)*E
llipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))-4/3*c/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))))+2*(a^2*d+2*a*b*c)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-si
n(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)))+2*a^2*c*(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)))/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 )}^{2} \sqrt{d \sin \left (f x + e\right ) + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^2*(c+d*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}\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))^2*(c+d*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(-(b^2*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2)*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 )^{2} \sqrt{c + d \sin{\left (e + f x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))**2*(c+d*sin(f*x+e))**(1/2),x)

[Out]

Integral((a + b*sin(e + f*x))**2*sqrt(c + d*sin(e + f*x)), 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))^2*(c+d*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out