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

Optimal. Leaf size=185 \[ \frac{3 a^2 b x \sqrt{a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2}}{2 \left (a^2 \sin (d+e x)+a b\right )}-\frac{a^2 b \sin (d+e x) \cos (d+e x) \sqrt{a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2}}{2 e \left (a^2 \sin (d+e x)+a b\right )}-\frac{\left (a^2+b^2\right ) \cos (d+e x) \sqrt{a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2}}{e (a \sin (d+e x)+b)} \]

[Out]

-(((a^2 + b^2)*Cos[d + e*x]*Sqrt[b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2])/(e*(b + a*Sin[d + e*x]))) + (
3*a^2*b*x*Sqrt[b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2])/(2*(a*b + a^2*Sin[d + e*x])) - (a^2*b*Cos[d + e
*x]*Sin[d + e*x]*Sqrt[b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2])/(2*e*(a*b + a^2*Sin[d + e*x]))

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Rubi [A]  time = 0.109408, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.049, Rules used = {3290, 2734} \[ \frac{3 a^2 b x \sqrt{a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2}}{2 \left (a^2 \sin (d+e x)+a b\right )}-\frac{a^2 b \sin (d+e x) \cos (d+e x) \sqrt{a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2}}{2 e \left (a^2 \sin (d+e x)+a b\right )}-\frac{\left (a^2+b^2\right ) \cos (d+e x) \sqrt{a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2}}{e (a \sin (d+e x)+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 + b^2)*Cos[d + e*x]*Sqrt[b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2])/(e*(b + a*Sin[d + e*x]))) + (
3*a^2*b*x*Sqrt[b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2])/(2*(a*b + a^2*Sin[d + e*x])) - (a^2*b*Cos[d + e
*x]*Sin[d + e*x]*Sqrt[b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2])/(2*e*(a*b + a^2*Sin[d + e*x]))

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 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)) \sqrt{b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx &=\frac{\sqrt{b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \int \left (2 a b+2 a^2 \sin (d+e x)\right ) (a+b \sin (d+e x)) \, dx}{2 a b+2 a^2 \sin (d+e x)}\\ &=-\frac{\left (a^2+b^2\right ) \cos (d+e x) \sqrt{b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}{e (b+a \sin (d+e x))}+\frac{3 a^2 b x \sqrt{b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}{2 \left (a b+a^2 \sin (d+e x)\right )}-\frac{a^2 b \cos (d+e x) \sin (d+e x) \sqrt{b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)}}{2 e \left (a b+a^2 \sin (d+e x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.193099, size = 70, normalized size = 0.38 \[ -\frac{\sqrt{(a \sin (d+e x)+b)^2} \left (4 \left (a^2+b^2\right ) \cos (d+e x)+a b (\sin (2 (d+e x))-6 (d+e x))\right )}{4 e (a \sin (d+e x)+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[(b + a*Sin[d + e*x])^2]*(4*(a^2 + b^2)*Cos[d + e*x] + a*b*(-6*(d + e*x) + Sin[2*(d + e*x)])))/(4*e*(b +
a*Sin[d + e*x]))

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Maple [A]  time = 0.209, size = 107, normalized size = 0.6 \begin{align*} -{\frac{\sin \left ( ex+d \right ) \cos \left ( ex+d \right ) ab+2\,{a}^{2}\cos \left ( ex+d \right ) +2\,\cos \left ( ex+d \right ){b}^{2}-3\, \left ( ex+d \right ) ab+2\,{a}^{2}+2\,{b}^{2}}{2\,e \left ( b+a\sin \left ( ex+d \right ) \right ) }\sqrt{-{a}^{2} \left ( \cos \left ( ex+d \right ) \right ) ^{2}+2\,ab\sin \left ( ex+d \right ) +{a}^{2}+{b}^{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)^(1/2),x)

[Out]

-1/2/e*(-a^2*cos(e*x+d)^2+2*a*b*sin(e*x+d)+a^2+b^2)^(1/2)*(sin(e*x+d)*cos(e*x+d)*a*b+2*a^2*cos(e*x+d)+2*cos(e*
x+d)*b^2-3*(e*x+d)*a*b+2*a^2+2*b^2)/(b+a*sin(e*x+d))

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Maxima [A]  time = 1.55904, size = 252, normalized size = 1.36 \begin{align*} \frac{2 \,{\left (b \arctan \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right ) - \frac{a}{\frac{\sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + 1}\right )} a +{\left (a \arctan \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right ) - \frac{2 \, b + \frac{a \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac{2 \, b \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} - \frac{a \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}}}{\frac{2 \, \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + \frac{\sin \left (e x + d\right )^{4}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{4}} + 1}\right )} b}{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)^(1/2),x, algorithm="maxima")

[Out]

(2*(b*arctan(sin(e*x + d)/(cos(e*x + d) + 1)) - a/(sin(e*x + d)^2/(cos(e*x + d) + 1)^2 + 1))*a + (a*arctan(sin
(e*x + d)/(cos(e*x + d) + 1)) - (2*b + a*sin(e*x + d)/(cos(e*x + d) + 1) + 2*b*sin(e*x + d)^2/(cos(e*x + d) +
1)^2 - a*sin(e*x + d)^3/(cos(e*x + d) + 1)^3)/(2*sin(e*x + d)^2/(cos(e*x + d) + 1)^2 + sin(e*x + d)^4/(cos(e*x
+ d) + 1)^4 + 1))*b)/e

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Fricas [A]  time = 1.81702, size = 108, normalized size = 0.58 \begin{align*} \frac{3 \, a b e x - a b \cos \left (e x + d\right ) \sin \left (e x + d\right ) - 2 \,{\left (a^{2} + b^{2}\right )} \cos \left (e x + d\right )}{2 \, 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)^(1/2),x, algorithm="fricas")

[Out]

1/2*(3*a*b*e*x - a*b*cos(e*x + d)*sin(e*x + d) - 2*(a^2 + b^2)*cos(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)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.22403, size = 132, normalized size = 0.71 \begin{align*} -a^{2} \cos \left (x e + d\right ) e^{\left (-1\right )} \mathrm{sgn}\left (a \sin \left (x e + d\right ) + b\right ) - b^{2} \cos \left (x e + d\right ) e^{\left (-1\right )} \mathrm{sgn}\left (a \sin \left (x e + d\right ) + b\right ) - \frac{1}{4} \, a b e^{\left (-1\right )} \mathrm{sgn}\left (a \sin \left (x e + d\right ) + b\right ) \sin \left (2 \, x e + 2 \, d\right ) + \frac{3}{2} \, a b x \mathrm{sgn}\left (a \sin \left (x e + d\right ) + b\right ) \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)^(1/2),x, algorithm="giac")

[Out]

-a^2*cos(x*e + d)*e^(-1)*sgn(a*sin(x*e + d) + b) - b^2*cos(x*e + d)*e^(-1)*sgn(a*sin(x*e + d) + b) - 1/4*a*b*e
^(-1)*sgn(a*sin(x*e + d) + b)*sin(2*x*e + 2*d) + 3/2*a*b*x*sgn(a*sin(x*e + d) + b)