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

Optimal. Leaf size=100 \[ \frac{2 a e^2 \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{3 d \sqrt{e \sin (c+d x)}}-\frac{2 a e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d}+\frac{2 b (e \sin (c+d x))^{5/2}}{5 d e} \]

[Out]

(2*a*e^2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*Sqrt[e*Sin[c + d*x]]) - (2*a*e*Cos[c + d*x]
*Sqrt[e*Sin[c + d*x]])/(3*d) + (2*b*(e*Sin[c + d*x])^(5/2))/(5*d*e)

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Rubi [A]  time = 0.0713243, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2669, 2635, 2642, 2641} \[ \frac{2 a e^2 \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{3 d \sqrt{e \sin (c+d x)}}-\frac{2 a e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d}+\frac{2 b (e \sin (c+d x))^{5/2}}{5 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*e^2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*Sqrt[e*Sin[c + d*x]]) - (2*a*e*Cos[c + d*x]
*Sqrt[e*Sin[c + d*x]])/(3*d) + (2*b*(e*Sin[c + d*x])^(5/2))/(5*d*e)

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2635

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int (a+b \cos (c+d x)) (e \sin (c+d x))^{3/2} \, dx &=\frac{2 b (e \sin (c+d x))^{5/2}}{5 d e}+a \int (e \sin (c+d x))^{3/2} \, dx\\ &=-\frac{2 a e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d}+\frac{2 b (e \sin (c+d x))^{5/2}}{5 d e}+\frac{1}{3} \left (a e^2\right ) \int \frac{1}{\sqrt{e \sin (c+d x)}} \, dx\\ &=-\frac{2 a e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d}+\frac{2 b (e \sin (c+d x))^{5/2}}{5 d e}+\frac{\left (a e^2 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx}{3 \sqrt{e \sin (c+d x)}}\\ &=\frac{2 a e^2 F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{3 d \sqrt{e \sin (c+d x)}}-\frac{2 a e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d}+\frac{2 b (e \sin (c+d x))^{5/2}}{5 d e}\\ \end{align*}

Mathematica [A]  time = 0.458702, size = 80, normalized size = 0.8 \[ \frac{2 (e \sin (c+d x))^{3/2} \left (\sqrt{\sin (c+d x)} \left (3 b \sin ^2(c+d x)-5 a \cos (c+d x)\right )-5 a F\left (\left .\frac{1}{4} (-2 c-2 d x+\pi )\right |2\right )\right )}{15 d \sin ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 1.426, size = 116, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({\frac{2\,b}{5\,e} \left ( e\sin \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{{e}^{2}a}{3\,\cos \left ( dx+c \right ) } \left ( \sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{1-\sin \left ( dx+c \right ) },{\frac{\sqrt{2}}{2}} \right ) -2\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+2\,\sin \left ( dx+c \right ) \right ){\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2/5/e*b*(e*sin(d*x+c))^(5/2)-1/3*e^2*a*((1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*Elliptic
F((1-sin(d*x+c))^(1/2),1/2*2^(1/2))-2*sin(d*x+c)^3+2*sin(d*x+c))/cos(d*x+c)/(e*sin(d*x+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 ) + a\right )} \left (e \sin \left (d x + c\right )\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))*(e*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*e*cos(d*x + c) + a*e)*sqrt(e*sin(d*x + c))*sin(d*x + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \cos \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\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))*(e*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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