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

Optimal. Leaf size=149 \[ \frac{2 e^2 \left (9 a^2+2 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{15 d \sqrt{\cos (c+d x)}}+\frac{2 e \left (9 a^2+2 b^2\right ) \sin (c+d x) (e \cos (c+d x))^{3/2}}{45 d}-\frac{22 a b (e \cos (c+d x))^{7/2}}{63 d e}-\frac{2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{9 d e} \]

[Out]

(-22*a*b*(e*Cos[c + d*x])^(7/2))/(63*d*e) + (2*(9*a^2 + 2*b^2)*e^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2,
2])/(15*d*Sqrt[Cos[c + d*x]]) + (2*(9*a^2 + 2*b^2)*e*(e*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(45*d) - (2*b*(e*Co
s[c + d*x])^(7/2)*(a + b*Sin[c + d*x]))/(9*d*e)

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Rubi [A]  time = 0.166829, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2692, 2669, 2635, 2640, 2639} \[ \frac{2 e^2 \left (9 a^2+2 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{15 d \sqrt{\cos (c+d x)}}+\frac{2 e \left (9 a^2+2 b^2\right ) \sin (c+d x) (e \cos (c+d x))^{3/2}}{45 d}-\frac{22 a b (e \cos (c+d x))^{7/2}}{63 d e}-\frac{2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{9 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-22*a*b*(e*Cos[c + d*x])^(7/2))/(63*d*e) + (2*(9*a^2 + 2*b^2)*e^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2,
2])/(15*d*Sqrt[Cos[c + d*x]]) + (2*(9*a^2 + 2*b^2)*e*(e*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(45*d) - (2*b*(e*Co
s[c + d*x])^(7/2)*(a + b*Sin[c + d*x]))/(9*d*e)

Rule 2692

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

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+b \sin (c+d x))^2 \, dx &=-\frac{2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{9 d e}+\frac{2}{9} \int (e \cos (c+d x))^{5/2} \left (\frac{9 a^2}{2}+b^2+\frac{11}{2} a b \sin (c+d x)\right ) \, dx\\ &=-\frac{22 a b (e \cos (c+d x))^{7/2}}{63 d e}-\frac{2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{9 d e}+\frac{1}{9} \left (9 a^2+2 b^2\right ) \int (e \cos (c+d x))^{5/2} \, dx\\ &=-\frac{22 a b (e \cos (c+d x))^{7/2}}{63 d e}+\frac{2 \left (9 a^2+2 b^2\right ) e (e \cos (c+d x))^{3/2} \sin (c+d x)}{45 d}-\frac{2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{9 d e}+\frac{1}{15} \left (\left (9 a^2+2 b^2\right ) e^2\right ) \int \sqrt{e \cos (c+d x)} \, dx\\ &=-\frac{22 a b (e \cos (c+d x))^{7/2}}{63 d e}+\frac{2 \left (9 a^2+2 b^2\right ) e (e \cos (c+d x))^{3/2} \sin (c+d x)}{45 d}-\frac{2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{9 d e}+\frac{\left (\left (9 a^2+2 b^2\right ) e^2 \sqrt{e \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{15 \sqrt{\cos (c+d x)}}\\ &=-\frac{22 a b (e \cos (c+d x))^{7/2}}{63 d e}+\frac{2 \left (9 a^2+2 b^2\right ) e^2 \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d \sqrt{\cos (c+d x)}}+\frac{2 \left (9 a^2+2 b^2\right ) e (e \cos (c+d x))^{3/2} \sin (c+d x)}{45 d}-\frac{2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{9 d e}\\ \end{align*}

Mathematica [A]  time = 0.88886, size = 113, normalized size = 0.76 \[ \frac{(e \cos (c+d x))^{5/2} \left (84 \left (9 a^2+2 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\cos ^{\frac{3}{2}}(c+d x) \left (21 \left (12 a^2+b^2\right ) \sin (c+d x)-5 b (36 a+7 b \sin (3 (c+d x)))-180 a b \cos (2 (c+d x))\right )\right )}{630 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Cos[c + d*x])^(5/2)*(84*(9*a^2 + 2*b^2)*EllipticE[(c + d*x)/2, 2] + Cos[c + d*x]^(3/2)*(-180*a*b*Cos[2*(c
+ d*x)] + 21*(12*a^2 + b^2)*Sin[c + d*x] - 5*b*(36*a + 7*b*Sin[3*(c + d*x)]))))/(630*d*Cos[c + d*x]^(5/2))

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Maple [B]  time = 1.271, size = 408, normalized size = 2.7 \begin{align*}{\frac{2\,{e}^{3}}{315\,d} \left ( -1120\,{b}^{2} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{10}\cos \left ( 1/2\,dx+c/2 \right ) -1440\,ab \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{9}+2240\,{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+504\,{a}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+2880\,ab \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}-1568\,{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-504\,{a}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-2160\,ab \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+448\,{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+189\,\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 ){a}^{2}+42\,\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 ){b}^{2}+126\,{a}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+720\,ab \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-42\,{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-90\,\sin \left ( 1/2\,dx+c/2 \right ) ab \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+b*sin(d*x+c))^2,x)

[Out]

2/315/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^3*(-1120*b^2*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+
1/2*c)-1440*a*b*sin(1/2*d*x+1/2*c)^9+2240*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+504*a^2*cos(1/2*d*x+1/2*
c)*sin(1/2*d*x+1/2*c)^6+2880*a*b*sin(1/2*d*x+1/2*c)^7-1568*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-504*a^2
*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-2160*a*b*sin(1/2*d*x+1/2*c)^5+448*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+
1/2*c)^4+189*(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
))*a^2+42*(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))*
b^2+126*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+720*a*b*sin(1/2*d*x+1/2*c)^3-42*b^2*cos(1/2*d*x+1/2*c)*sin
(1/2*d*x+1/2*c)^2-90*sin(1/2*d*x+1/2*c)*a*b)/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 (b \sin \left (d x + c\right ) + a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (b^{2} e^{2} \cos \left (d x + c\right )^{4} - 2 \, a b e^{2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) -{\left (a^{2} + b^{2}\right )} 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+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

integral(-(b^2*e^2*cos(d*x + c)^4 - 2*a*b*e^2*cos(d*x + c)^2*sin(d*x + c) - (a^2 + b^2)*e^2*cos(d*x + c)^2)*sq
rt(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+b*sin(d*x+c))**2,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 (b \sin \left (d x + c\right ) + a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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