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

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Rubi [A]  time = 0.11, antiderivative size = 185, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.19, 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]

-1/4*(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)])))/(e*(b
+ a*Sin[d + e*x]))

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fricas [A]  time = 0.87, size = 43, normalized size = 0.23 \[ \frac {3 \, a b e x - a b \cos \left (e x + d\right ) \sin \left (e x + d\right ) - 2 \, {\left (a^{2} + b^{2}\right )} \cos \left (e x + d\right )}{2 \, e} \]

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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giac [A]  time = 0.21, size = 98, normalized size = 0.53 \[ -a^{2} \cos \left (x e + d\right ) e^{\left (-1\right )} \mathrm {sgn}\left (a \sin \left (x e + d\right ) + b\right ) - b^{2} \cos \left (x e + d\right ) e^{\left (-1\right )} \mathrm {sgn}\left (a \sin \left (x e + d\right ) + b\right ) - \frac {1}{4} \, a b e^{\left (-1\right )} \mathrm {sgn}\left (a \sin \left (x e + d\right ) + b\right ) \sin \left (2 \, x e + 2 \, d\right ) + \frac {3}{2} \, a b x \mathrm {sgn}\left (a \sin \left (x e + d\right ) + b\right ) \]

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)

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maple [A]  time = 0.59, size = 107, normalized size = 0.58 \[ -\frac {\sqrt {-a^{2} \left (\cos ^{2}\left (e x +d \right )\right )+2 a b \sin \left (e x +d \right )+a^{2}+b^{2}}\, \left (\sin \left (e x +d \right ) \cos \left (e x +d \right ) a b +2 a^{2} \cos \left (e x +d \right )+2 \cos \left (e x +d \right ) b^{2}-3 \left (e x +d \right ) a b +2 a^{2}+2 b^{2}\right )}{2 e \left (b +a \sin \left (e x +d \right )\right )} \]

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 = 0.43, size = 187, normalized size = 1.01 \[ \frac {2 \, {\left (b \arctan \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right ) - \frac {a}{\frac {\sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + 1}\right )} a + {\left (a \arctan \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right ) - \frac {2 \, b + \frac {a \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {2 \, b \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} - \frac {a \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}}}{\frac {2 \, \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + \frac {\sin \left (e x + d\right )^{4}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{4}} + 1}\right )} b}{e} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (a+b\,\sin \left (d+e\,x\right )\right )\,\sqrt {a^2\,{\sin \left (d+e\,x\right )}^2+2\,a\,b\,\sin \left (d+e\,x\right )+b^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(d + e*x))*(b^2 + a^2*sin(d + e*x)^2 + 2*a*b*sin(d + e*x))^(1/2),x)

[Out]

int((a + b*sin(d + e*x))*(b^2 + a^2*sin(d + e*x)^2 + 2*a*b*sin(d + e*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (d + e x \right )}\right ) \sqrt {\left (a \sin {\left (d + e x \right )} + b\right )^{2}}\, dx \]

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]

Integral((a + b*sin(d + e*x))*sqrt((a*sin(d + e*x) + b)**2), x)

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