3.197 \(\int (e \cos (c+d x))^{5/2} (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=95 \[ \frac{6 a e^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d \sqrt{\cos (c+d x)}}-\frac{2 a (e \cos (c+d x))^{7/2}}{7 d e}+\frac{2 a e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d} \]

[Out]

(-2*a*(e*Cos[c + d*x])^(7/2))/(7*d*e) + (6*a*e^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos
[c + d*x]]) + (2*a*e*(e*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.0671417, antiderivative size = 95, 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, 2640, 2639} \[ \frac{6 a e^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d \sqrt{\cos (c+d x)}}-\frac{2 a (e \cos (c+d x))^{7/2}}{7 d e}+\frac{2 a e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(5/2)*(a + a*Sin[c + d*x]),x]

[Out]

(-2*a*(e*Cos[c + d*x])^(7/2))/(7*d*e) + (6*a*e^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos
[c + d*x]]) + (2*a*e*(e*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d)

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 2640

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

Rule 2639

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

Rubi steps

\begin{align*} \int (e \cos (c+d x))^{5/2} (a+a \sin (c+d x)) \, dx &=-\frac{2 a (e \cos (c+d x))^{7/2}}{7 d e}+a \int (e \cos (c+d x))^{5/2} \, dx\\ &=-\frac{2 a (e \cos (c+d x))^{7/2}}{7 d e}+\frac{2 a e (e \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac{1}{5} \left (3 a e^2\right ) \int \sqrt{e \cos (c+d x)} \, dx\\ &=-\frac{2 a (e \cos (c+d x))^{7/2}}{7 d e}+\frac{2 a e (e \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac{\left (3 a e^2 \sqrt{e \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 \sqrt{\cos (c+d x)}}\\ &=-\frac{2 a (e \cos (c+d x))^{7/2}}{7 d e}+\frac{6 a e^2 \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a e (e \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [C]  time = 2.60873, size = 264, normalized size = 2.78 \[ \frac{a e^3 \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \left (168 (\cos (d x)-i \sin (d x)) \sqrt{i \sin (2 (c+d x))+\cos (2 (c+d x))+1} \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )+56 (\cos (d x)+i \sin (d x)) \sqrt{i \sin (2 (c+d x))+\cos (2 (c+d x))+1} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )+20 \sin (c+2 d x)-20 \sin (3 c+2 d x)+5 \sin (3 c+4 d x)-5 \sin (5 c+4 d x)-182 \cos (2 c+d x)+14 \cos (2 c+3 d x)-14 \cos (4 c+3 d x)-30 \sin (c)-154 \cos (d x)\right )}{560 d \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(5/2)*(a + a*Sin[c + d*x]),x]

[Out]

(a*e^3*Csc[c/2]*Sec[c/2]*(-154*Cos[d*x] - 182*Cos[2*c + d*x] + 14*Cos[2*c + 3*d*x] - 14*Cos[4*c + 3*d*x] - 30*
Sin[c] + 168*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*(Cos[d*x] - I*Sin[d*x])
*Sqrt[1 + Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]] + 56*Hypergeometric2F1[1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c]
+ I*Sin[c])^2)]*(Cos[d*x] + I*Sin[d*x])*Sqrt[1 + Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]] + 20*Sin[c + 2*d*x] -
20*Sin[3*c + 2*d*x] + 5*Sin[3*c + 4*d*x] - 5*Sin[5*c + 4*d*x]))/(560*d*Sqrt[e*Cos[c + d*x]])

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Maple [A]  time = 0.411, size = 214, normalized size = 2.3 \begin{align*}{\frac{2\,a{e}^{3}}{35\,d} \left ( -80\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{9}+56\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +160\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}-56\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) -120\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+21\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +14\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) +40\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-5\,\sin \left ( 1/2\,dx+c/2 \right ) \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(5/2)*(a+a*sin(d*x+c)),x)

[Out]

2/35/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*a*e^3*(-80*sin(1/2*d*x+1/2*c)^9+56*sin(1/2*d*x+1/2
*c)^6*cos(1/2*d*x+1/2*c)+160*sin(1/2*d*x+1/2*c)^7-56*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-120*sin(1/2*d*x+1
/2*c)^5+21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))
+14*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+40*sin(1/2*d*x+1/2*c)^3-5*sin(1/2*d*x+1/2*c))/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{\frac{5}{2}}{\left (a \sin \left (d x + c\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/2)*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(5/2)*(a*sin(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/2)*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

integral((a*e^2*cos(d*x + c)^2*sin(d*x + c) + a*e^2*cos(d*x + c)^2)*sqrt(e*cos(d*x + c)), 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((e*cos(d*x+c))**(5/2)*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{\frac{5}{2}}{\left (a \sin \left (d x + c\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/2)*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(5/2)*(a*sin(d*x + c) + a), x)