3.504 \(\int (a+b \sin (d+e x)) (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x))^{3/2} \, dx\)

Optimal. Leaf size=331 \[ \frac{5 a^4 b x \left (3 a^2+4 b^2\right ) \left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}{8 \left (a^2 \sin (d+e x)+a b\right )^3}-\frac{a^4 b \left (29 a^2+6 b^2\right ) \sin (d+e x) \cos (d+e x) \left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}{24 e \left (a^2 \sin (d+e x)+a b\right )^3}-\frac{b \cos (d+e x) \left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}{4 e}-\frac{\left (4 a^2+3 b^2\right ) \cos (d+e x) \left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}{12 e (a \sin (d+e x)+b)}-\frac{\left (28 a^2 b^2+4 a^4+3 b^4\right ) \cos (d+e x) \left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}{6 e (a \sin (d+e x)+b)^3} \]

[Out]

-(b*Cos[d + e*x]*(b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2)^(3/2))/(4*e) - ((4*a^4 + 28*a^2*b^2 + 3*b^4)*
Cos[d + e*x]*(b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2)^(3/2))/(6*e*(b + a*Sin[d + e*x])^3) - ((4*a^2 + 3
*b^2)*Cos[d + e*x]*(b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2)^(3/2))/(12*e*(b + a*Sin[d + e*x])) + (5*a^4
*b*(3*a^2 + 4*b^2)*x*(b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2)^(3/2))/(8*(a*b + a^2*Sin[d + e*x])^3) - (
a^4*b*(29*a^2 + 6*b^2)*Cos[d + e*x]*Sin[d + e*x]*(b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2)^(3/2))/(24*e*
(a*b + a^2*Sin[d + e*x])^3)

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Rubi [A]  time = 0.322718, antiderivative size = 331, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.073, Rules used = {3290, 2753, 2734} \[ \frac{5 a^4 b x \left (3 a^2+4 b^2\right ) \left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}{8 \left (a^2 \sin (d+e x)+a b\right )^3}-\frac{a^4 b \left (29 a^2+6 b^2\right ) \sin (d+e x) \cos (d+e x) \left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}{24 e \left (a^2 \sin (d+e x)+a b\right )^3}-\frac{b \cos (d+e x) \left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}{4 e}-\frac{\left (4 a^2+3 b^2\right ) \cos (d+e x) \left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}{12 e (a \sin (d+e x)+b)}-\frac{\left (28 a^2 b^2+4 a^4+3 b^4\right ) \cos (d+e x) \left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}{6 e (a \sin (d+e x)+b)^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[d + e*x])*(b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2)^(3/2),x]

[Out]

-(b*Cos[d + e*x]*(b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2)^(3/2))/(4*e) - ((4*a^4 + 28*a^2*b^2 + 3*b^4)*
Cos[d + e*x]*(b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2)^(3/2))/(6*e*(b + a*Sin[d + e*x])^3) - ((4*a^2 + 3
*b^2)*Cos[d + e*x]*(b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2)^(3/2))/(12*e*(b + a*Sin[d + e*x])) + (5*a^4
*b*(3*a^2 + 4*b^2)*x*(b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2)^(3/2))/(8*(a*b + a^2*Sin[d + e*x])^3) - (
a^4*b*(29*a^2 + 6*b^2)*Cos[d + e*x]*Sin[d + e*x]*(b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2)^(3/2))/(24*e*
(a*b + a^2*Sin[d + e*x])^3)

Rule 3290

Int[((A_) + (B_.)*sin[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*sin[(d_.) + (e_.)*(x_)] + (c_.)*sin[(d_.) + (e_.)*(x_
)]^2)^(n_), x_Symbol] :> Dist[(a + b*Sin[d + e*x] + c*Sin[d + e*x]^2)^n/(b + 2*c*Sin[d + e*x])^(2*n), Int[(A +
B*Sin[d + e*x])*(b + 2*c*Sin[d + e*x])^(2*n), x], x] /; FreeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 - 4*a*c, 0
] &&  !IntegerQ[n]

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 2734

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

Rubi steps

\begin{align*} \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \, dx &=\frac{\left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \int \left (2 a b+2 a^2 \sin (d+e x)\right )^3 (a+b \sin (d+e x)) \, dx}{\left (2 a b+2 a^2 \sin (d+e x)\right )^3}\\ &=-\frac{b \cos (d+e x) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{4 e}+\frac{\left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \int \left (2 a b+2 a^2 \sin (d+e x)\right )^2 \left (14 a^2 b+2 a \left (4 a^2+3 b^2\right ) \sin (d+e x)\right ) \, dx}{4 \left (2 a b+2 a^2 \sin (d+e x)\right )^3}\\ &=-\frac{b \cos (d+e x) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{4 e}-\frac{\left (4 a^2+3 b^2\right ) \cos (d+e x) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{12 e (b+a \sin (d+e x))}+\frac{\left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \int \left (2 a b+2 a^2 \sin (d+e x)\right ) \left (4 a^3 \left (8 a^2+27 b^2\right )+4 a^2 b \left (29 a^2+6 b^2\right ) \sin (d+e x)\right ) \, dx}{12 \left (2 a b+2 a^2 \sin (d+e x)\right )^3}\\ &=-\frac{b \cos (d+e x) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{4 e}-\frac{\left (4 a^4+28 a^2 b^2+3 b^4\right ) \cos (d+e x) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{6 e (b+a \sin (d+e x))^3}-\frac{\left (4 a^2+3 b^2\right ) \cos (d+e x) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{12 e (b+a \sin (d+e x))}+\frac{5 a^4 b \left (3 a^2+4 b^2\right ) x \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{8 \left (a b+a^2 \sin (d+e x)\right )^3}-\frac{a^4 b \left (29 a^2+6 b^2\right ) \cos (d+e x) \sin (d+e x) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{24 e \left (a b+a^2 \sin (d+e x)\right )^3}\\ \end{align*}

Mathematica [A]  time = 0.894213, size = 140, normalized size = 0.42 \[ \frac{\sqrt{(a \sin (d+e x)+b)^2} \left (3 a b \left (20 \left (3 a^2+4 b^2\right ) (d+e x)-8 \left (4 a^2+3 b^2\right ) \sin (2 (d+e x))+a^2 \sin (4 (d+e x))\right )-24 \left (21 a^2 b^2+3 a^4+4 b^4\right ) \cos (d+e x)+8 a \left (a^3+3 a b^2\right ) \cos (3 (d+e x))\right )}{96 e (a \sin (d+e x)+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[d + e*x])*(b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2)^(3/2),x]

[Out]

(Sqrt[(b + a*Sin[d + e*x])^2]*(-24*(3*a^4 + 21*a^2*b^2 + 4*b^4)*Cos[d + e*x] + 8*a*(a^3 + 3*a*b^2)*Cos[3*(d +
e*x)] + 3*a*b*(20*(3*a^2 + 4*b^2)*(d + e*x) - 8*(4*a^2 + 3*b^2)*Sin[2*(d + e*x)] + a^2*Sin[4*(d + e*x)])))/(96
*e*(b + a*Sin[d + e*x]))

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Maple [A]  time = 0.306, size = 269, normalized size = 0.8 \begin{align*} -{\frac{6\, \left ( \cos \left ( ex+d \right ) \right ) ^{3}\sin \left ( ex+d \right ){a}^{3}b+8\,{a}^{4} \left ( \cos \left ( ex+d \right ) \right ) ^{3}+24\,{a}^{2}{b}^{2} \left ( \cos \left ( ex+d \right ) \right ) ^{3}-51\,\sin \left ( ex+d \right ) \cos \left ( ex+d \right ){a}^{3}b-36\,\cos \left ( ex+d \right ) \sin \left ( ex+d \right ) a{b}^{3}-24\,{a}^{4}\cos \left ( ex+d \right ) -144\,{a}^{2}{b}^{2}\cos \left ( ex+d \right ) -24\,\cos \left ( ex+d \right ){b}^{4}+45\, \left ( ex+d \right ){a}^{3}b+60\, \left ( ex+d \right ) a{b}^{3}-16\,{a}^{4}-120\,{a}^{2}{b}^{2}-24\,{b}^{4}}{24\,e \left ( \left ( \cos \left ( ex+d \right ) \right ) ^{2}\sin \left ( ex+d \right ){a}^{3}+3\, \left ( \cos \left ( ex+d \right ) \right ) ^{2}{a}^{2}b-{a}^{3}\sin \left ( ex+d \right ) -3\,\sin \left ( ex+d \right ) a{b}^{2}-3\,{a}^{2}b-{b}^{3} \right ) } \left ( -{a}^{2} \left ( \cos \left ( ex+d \right ) \right ) ^{2}+2\,ab\sin \left ( ex+d \right ) +{a}^{2}+{b}^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(e*x+d))*(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(3/2),x)

[Out]

-1/24/e*(-a^2*cos(e*x+d)^2+2*a*b*sin(e*x+d)+a^2+b^2)^(3/2)*(6*cos(e*x+d)^3*sin(e*x+d)*a^3*b+8*a^4*cos(e*x+d)^3
+24*a^2*b^2*cos(e*x+d)^3-51*sin(e*x+d)*cos(e*x+d)*a^3*b-36*cos(e*x+d)*sin(e*x+d)*a*b^3-24*a^4*cos(e*x+d)-144*a
^2*b^2*cos(e*x+d)-24*cos(e*x+d)*b^4+45*(e*x+d)*a^3*b+60*(e*x+d)*a*b^3-16*a^4-120*a^2*b^2-24*b^4)/(cos(e*x+d)^2
*sin(e*x+d)*a^3+3*cos(e*x+d)^2*a^2*b-a^3*sin(e*x+d)-3*sin(e*x+d)*a*b^2-3*a^2*b-b^3)

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Maxima [A]  time = 1.58749, size = 751, normalized size = 2.27 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(e*x+d))*(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(3/2),x, algorithm="maxima")

[Out]

1/12*(4*(3*(3*a^2*b + 2*b^3)*arctan(sin(e*x + d)/(cos(e*x + d) + 1)) - (4*a^3 + 18*a*b^2 + 9*a^2*b*sin(e*x + d
)/(cos(e*x + d) + 1) + 18*a*b^2*sin(e*x + d)^4/(cos(e*x + d) + 1)^4 - 9*a^2*b*sin(e*x + d)^5/(cos(e*x + d) + 1
)^5 + 12*(a^3 + 3*a*b^2)*sin(e*x + d)^2/(cos(e*x + d) + 1)^2)/(3*sin(e*x + d)^2/(cos(e*x + d) + 1)^2 + 3*sin(e
*x + d)^4/(cos(e*x + d) + 1)^4 + sin(e*x + d)^6/(cos(e*x + d) + 1)^6 + 1))*a + 3*(3*(a^3 + 4*a*b^2)*arctan(sin
(e*x + d)/(cos(e*x + d) + 1)) - (16*a^2*b + 8*b^3 + 8*b^3*sin(e*x + d)^6/(cos(e*x + d) + 1)^6 + 3*(a^3 + 4*a*b
^2)*sin(e*x + d)/(cos(e*x + d) + 1) + 8*(8*a^2*b + 3*b^3)*sin(e*x + d)^2/(cos(e*x + d) + 1)^2 + (11*a^3 + 12*a
*b^2)*sin(e*x + d)^3/(cos(e*x + d) + 1)^3 + 24*(2*a^2*b + b^3)*sin(e*x + d)^4/(cos(e*x + d) + 1)^4 - (11*a^3 +
12*a*b^2)*sin(e*x + d)^5/(cos(e*x + d) + 1)^5 - 3*(a^3 + 4*a*b^2)*sin(e*x + d)^7/(cos(e*x + d) + 1)^7)/(4*sin
(e*x + d)^2/(cos(e*x + d) + 1)^2 + 6*sin(e*x + d)^4/(cos(e*x + d) + 1)^4 + 4*sin(e*x + d)^6/(cos(e*x + d) + 1)
^6 + sin(e*x + d)^8/(cos(e*x + d) + 1)^8 + 1))*b)/e

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Fricas [A]  time = 1.83632, size = 263, normalized size = 0.79 \begin{align*} \frac{8 \,{\left (a^{4} + 3 \, a^{2} b^{2}\right )} \cos \left (e x + d\right )^{3} + 15 \,{\left (3 \, a^{3} b + 4 \, a b^{3}\right )} e x - 24 \,{\left (a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (e x + d\right ) + 3 \,{\left (2 \, a^{3} b \cos \left (e x + d\right )^{3} -{\left (17 \, a^{3} b + 12 \, a b^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{24 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(e*x+d))*(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(3/2),x, algorithm="fricas")

[Out]

1/24*(8*(a^4 + 3*a^2*b^2)*cos(e*x + d)^3 + 15*(3*a^3*b + 4*a*b^3)*e*x - 24*(a^4 + 6*a^2*b^2 + b^4)*cos(e*x + d
) + 3*(2*a^3*b*cos(e*x + d)^3 - (17*a^3*b + 12*a*b^3)*cos(e*x + d))*sin(e*x + d))/e

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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(e*x+d))*(b**2+2*a*b*sin(e*x+d)+a**2*sin(e*x+d)**2)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.33431, size = 323, normalized size = 0.98 \begin{align*} \frac{1}{32} \, a^{3} b e^{\left (-1\right )} \mathrm{sgn}\left (a \sin \left (x e + d\right ) + b\right ) \sin \left (4 \, x e + 4 \, d\right ) + \frac{1}{12} \,{\left (a^{4} \mathrm{sgn}\left (a \sin \left (x e + d\right ) + b\right ) + 3 \, a^{2} b^{2} \mathrm{sgn}\left (a \sin \left (x e + d\right ) + b\right )\right )} \cos \left (3 \, x e + 3 \, d\right ) e^{\left (-1\right )} - \frac{1}{4} \,{\left (3 \, a^{4} \mathrm{sgn}\left (a \sin \left (x e + d\right ) + b\right ) + 21 \, a^{2} b^{2} \mathrm{sgn}\left (a \sin \left (x e + d\right ) + b\right ) + 4 \, b^{4} \mathrm{sgn}\left (a \sin \left (x e + d\right ) + b\right )\right )} \cos \left (x e + d\right ) e^{\left (-1\right )} - \frac{1}{4} \,{\left (4 \, a^{3} b \mathrm{sgn}\left (a \sin \left (x e + d\right ) + b\right ) + 3 \, a b^{3} \mathrm{sgn}\left (a \sin \left (x e + d\right ) + b\right )\right )} e^{\left (-1\right )} \sin \left (2 \, x e + 2 \, d\right ) + \frac{5}{8} \,{\left (3 \, a^{3} b \mathrm{sgn}\left (a \sin \left (x e + d\right ) + b\right ) + 4 \, a b^{3} \mathrm{sgn}\left (a \sin \left (x e + d\right ) + b\right )\right )} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(e*x+d))*(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(3/2),x, algorithm="giac")

[Out]

1/32*a^3*b*e^(-1)*sgn(a*sin(x*e + d) + b)*sin(4*x*e + 4*d) + 1/12*(a^4*sgn(a*sin(x*e + d) + b) + 3*a^2*b^2*sgn
(a*sin(x*e + d) + b))*cos(3*x*e + 3*d)*e^(-1) - 1/4*(3*a^4*sgn(a*sin(x*e + d) + b) + 21*a^2*b^2*sgn(a*sin(x*e
+ d) + b) + 4*b^4*sgn(a*sin(x*e + d) + b))*cos(x*e + d)*e^(-1) - 1/4*(4*a^3*b*sgn(a*sin(x*e + d) + b) + 3*a*b^
3*sgn(a*sin(x*e + d) + b))*e^(-1)*sin(2*x*e + 2*d) + 5/8*(3*a^3*b*sgn(a*sin(x*e + d) + b) + 4*a*b^3*sgn(a*sin(
x*e + d) + b))*x