3.574 \(\int (a+b \cos (c+d x) \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=212 \[ -\frac{\left (4 a^2-b^2\right ) \sqrt{\frac{2 a+b \sin (2 c+2 d x)}{2 a+b}} \text{EllipticF}\left (c+d x-\frac{\pi }{4},\frac{2 b}{2 a+b}\right )}{6 \sqrt{2} d \sqrt{2 a+b \sin (2 c+2 d x)}}-\frac{b \cos (2 c+2 d x) \sqrt{2 a+b \sin (2 c+2 d x)}}{6 \sqrt{2} d}+\frac{2 \sqrt{2} a \sqrt{2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac{\pi }{4}|\frac{2 b}{2 a+b}\right )}{3 d \sqrt{\frac{2 a+b \sin (2 c+2 d x)}{2 a+b}}} \]

[Out]

-(b*Cos[2*c + 2*d*x]*Sqrt[2*a + b*Sin[2*c + 2*d*x]])/(6*Sqrt[2]*d) + (2*Sqrt[2]*a*EllipticE[c - Pi/4 + d*x, (2
*b)/(2*a + b)]*Sqrt[2*a + b*Sin[2*c + 2*d*x]])/(3*d*Sqrt[(2*a + b*Sin[2*c + 2*d*x])/(2*a + b)]) - ((4*a^2 - b^
2)*EllipticF[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a + b*Sin[2*c + 2*d*x])/(2*a + b)])/(6*Sqrt[2]*d*Sqrt[2*
a + b*Sin[2*c + 2*d*x]])

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Rubi [A]  time = 0.219818, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {2666, 2656, 2752, 2663, 2661, 2655, 2653} \[ -\frac{\left (4 a^2-b^2\right ) \sqrt{\frac{2 a+b \sin (2 c+2 d x)}{2 a+b}} F\left (c+d x-\frac{\pi }{4}|\frac{2 b}{2 a+b}\right )}{6 \sqrt{2} d \sqrt{2 a+b \sin (2 c+2 d x)}}-\frac{b \cos (2 c+2 d x) \sqrt{2 a+b \sin (2 c+2 d x)}}{6 \sqrt{2} d}+\frac{2 \sqrt{2} a \sqrt{2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac{\pi }{4}|\frac{2 b}{2 a+b}\right )}{3 d \sqrt{\frac{2 a+b \sin (2 c+2 d x)}{2 a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Cos[2*c + 2*d*x]*Sqrt[2*a + b*Sin[2*c + 2*d*x]])/(6*Sqrt[2]*d) + (2*Sqrt[2]*a*EllipticE[c - Pi/4 + d*x, (2
*b)/(2*a + b)]*Sqrt[2*a + b*Sin[2*c + 2*d*x]])/(3*d*Sqrt[(2*a + b*Sin[2*c + 2*d*x])/(2*a + b)]) - ((4*a^2 - b^
2)*EllipticF[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a + b*Sin[2*c + 2*d*x])/(2*a + b)])/(6*Sqrt[2]*d*Sqrt[2*
a + b*Sin[2*c + 2*d*x]])

Rule 2666

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[(a + (b*Sin[2*c + 2*
d*x])/2)^n, x] /; FreeQ[{a, b, c, d, n}, x]

Rule 2656

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n -
1))/(d*n), x] + Dist[1/n, Int[(a + b*Sin[c + d*x])^(n - 2)*Simp[a^2*n + b^2*(n - 1) + a*b*(2*n - 1)*Sin[c + d*
x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

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 \cos (c+d x) \sin (c+d x))^{3/2} \, dx &=\int \left (a+\frac{1}{2} b \sin (2 c+2 d x)\right )^{3/2} \, dx\\ &=-\frac{b \cos (2 c+2 d x) \sqrt{2 a+b \sin (2 c+2 d x)}}{6 \sqrt{2} d}+\frac{2}{3} \int \frac{\frac{1}{8} \left (12 a^2+b^2\right )+a b \sin (2 c+2 d x)}{\sqrt{a+\frac{1}{2} b \sin (2 c+2 d x)}} \, dx\\ &=-\frac{b \cos (2 c+2 d x) \sqrt{2 a+b \sin (2 c+2 d x)}}{6 \sqrt{2} d}+\frac{1}{3} (4 a) \int \sqrt{a+\frac{1}{2} b \sin (2 c+2 d x)} \, dx+\frac{1}{12} \left (-4 a^2+b^2\right ) \int \frac{1}{\sqrt{a+\frac{1}{2} b \sin (2 c+2 d x)}} \, dx\\ &=-\frac{b \cos (2 c+2 d x) \sqrt{2 a+b \sin (2 c+2 d x)}}{6 \sqrt{2} d}+\frac{\left (4 a \sqrt{a+\frac{1}{2} b \sin (2 c+2 d x)}\right ) \int \sqrt{\frac{a}{a+\frac{b}{2}}+\frac{b \sin (2 c+2 d x)}{2 \left (a+\frac{b}{2}\right )}} \, dx}{3 \sqrt{\frac{a+\frac{1}{2} b \sin (2 c+2 d x)}{a+\frac{b}{2}}}}+\frac{\left (\left (-4 a^2+b^2\right ) \sqrt{\frac{a+\frac{1}{2} b \sin (2 c+2 d x)}{a+\frac{b}{2}}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+\frac{b}{2}}+\frac{b \sin (2 c+2 d x)}{2 \left (a+\frac{b}{2}\right )}}} \, dx}{12 \sqrt{a+\frac{1}{2} b \sin (2 c+2 d x)}}\\ &=-\frac{b \cos (2 c+2 d x) \sqrt{2 a+b \sin (2 c+2 d x)}}{6 \sqrt{2} d}+\frac{2 \sqrt{2} a E\left (c-\frac{\pi }{4}+d x|\frac{2 b}{2 a+b}\right ) \sqrt{2 a+b \sin (2 c+2 d x)}}{3 d \sqrt{\frac{2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\frac{\left (4 a^2-b^2\right ) F\left (c-\frac{\pi }{4}+d x|\frac{2 b}{2 a+b}\right ) \sqrt{\frac{2 a+b \sin (2 c+2 d x)}{2 a+b}}}{6 \sqrt{2} d \sqrt{2 a+b \sin (2 c+2 d x)}}\\ \end{align*}

Mathematica [A]  time = 1.50225, size = 167, normalized size = 0.79 \[ \frac{-\left (4 a^2-b^2\right ) \sqrt{\frac{2 a+b \sin (2 (c+d x))}{2 a+b}} \text{EllipticF}\left (c+d x-\frac{\pi }{4},\frac{2 b}{2 a+b}\right )-b \cos (2 (c+d x)) (2 a+b \sin (2 (c+d x)))+8 a (2 a+b) \sqrt{\frac{2 a+b \sin (2 (c+d x))}{2 a+b}} E\left (c+d x-\frac{\pi }{4}|\frac{2 b}{2 a+b}\right )}{6 d \sqrt{4 a+2 b \sin (2 (c+d x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b*Cos[2*(c + d*x)]*(2*a + b*Sin[2*(c + d*x)])) + 8*a*(2*a + b)*EllipticE[c - Pi/4 + d*x, (2*b)/(2*a + b)]*S
qrt[(2*a + b*Sin[2*(c + d*x)])/(2*a + b)] - (4*a^2 - b^2)*EllipticF[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a
+ b*Sin[2*(c + d*x)])/(2*a + b)])/(6*d*Sqrt[4*a + 2*b*Sin[2*(c + d*x)]])

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Maple [B]  time = 2.825, size = 844, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(24*a^3*((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2)*(-(sin(2*d*x+2*c)-1)*b/(2*a+b))^(1/2)*(-(1+sin(2*d*x+2*c))*
b/(2*a-b))^(1/2)*EllipticF(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))+4*((2*a+b*sin(2*d*x
+2*c))/(2*a-b))^(1/2)*EllipticF(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*(-(sin(2*d*x+2
*c)-1)*b/(2*a+b))^(1/2)*(-(1+sin(2*d*x+2*c))*b/(2*a-b))^(1/2)*a^2*b-6*((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2)*(
-(sin(2*d*x+2*c)-1)*b/(2*a+b))^(1/2)*(-(1+sin(2*d*x+2*c))*b/(2*a-b))^(1/2)*EllipticF(((2*a+b*sin(2*d*x+2*c))/(
2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*b^2*a-((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2)*(-(sin(2*d*x+2*c)-1)*b/(2*
a+b))^(1/2)*(-(1+sin(2*d*x+2*c))*b/(2*a-b))^(1/2)*EllipticF(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2
*a+b))^(1/2))*b^3-32*((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2)*EllipticE(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),(
(2*a-b)/(2*a+b))^(1/2))*(-(sin(2*d*x+2*c)-1)*b/(2*a+b))^(1/2)*(-(1+sin(2*d*x+2*c))*b/(2*a-b))^(1/2)*a^3+8*((2*
a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2)*EllipticE(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*(
-(sin(2*d*x+2*c)-1)*b/(2*a+b))^(1/2)*(-(1+sin(2*d*x+2*c))*b/(2*a-b))^(1/2)*a*b^2+b^3*sin(2*d*x+2*c)^3+2*a*b^2*
sin(2*d*x+2*c)^2-sin(2*d*x+2*c)*b^3-2*a*b^2)/b/cos(2*d*x+2*c)/(4*a+2*b*sin(2*d*x+2*c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*cos(d*x + c)*sin(d*x + c) + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*cos(d*x + c)*sin(d*x + c) + a)^(3/2), 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*cos(d*x+c)*sin(d*x+c))**(3/2),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*cos(d*x+c)*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out