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

Optimal. Leaf size=197 \[ \frac{2 a e^2 \left (7 a^2+6 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d \sqrt{e \cos (c+d x)}}-\frac{2 b \left (89 a^2+28 b^2\right ) (e \cos (c+d x))^{5/2}}{315 d e}+\frac{2 a e \left (7 a^2+6 b^2\right ) \sin (c+d x) \sqrt{e \cos (c+d x)}}{21 d}-\frac{2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^2}{9 d e}-\frac{26 a b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{63 d e} \]

[Out]

(-2*b*(89*a^2 + 28*b^2)*(e*Cos[c + d*x])^(5/2))/(315*d*e) + (2*a*(7*a^2 + 6*b^2)*e^2*Sqrt[Cos[c + d*x]]*Ellipt
icF[(c + d*x)/2, 2])/(21*d*Sqrt[e*Cos[c + d*x]]) + (2*a*(7*a^2 + 6*b^2)*e*Sqrt[e*Cos[c + d*x]]*Sin[c + d*x])/(
21*d) - (26*a*b*(e*Cos[c + d*x])^(5/2)*(a + b*Sin[c + d*x]))/(63*d*e) - (2*b*(e*Cos[c + d*x])^(5/2)*(a + b*Sin
[c + d*x])^2)/(9*d*e)

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Rubi [A]  time = 0.288608, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2692, 2862, 2669, 2635, 2642, 2641} \[ \frac{2 a e^2 \left (7 a^2+6 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d \sqrt{e \cos (c+d x)}}-\frac{2 b \left (89 a^2+28 b^2\right ) (e \cos (c+d x))^{5/2}}{315 d e}+\frac{2 a e \left (7 a^2+6 b^2\right ) \sin (c+d x) \sqrt{e \cos (c+d x)}}{21 d}-\frac{2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^2}{9 d e}-\frac{26 a b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{63 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*(89*a^2 + 28*b^2)*(e*Cos[c + d*x])^(5/2))/(315*d*e) + (2*a*(7*a^2 + 6*b^2)*e^2*Sqrt[Cos[c + d*x]]*Ellipt
icF[(c + d*x)/2, 2])/(21*d*Sqrt[e*Cos[c + d*x]]) + (2*a*(7*a^2 + 6*b^2)*e*Sqrt[e*Cos[c + d*x]]*Sin[c + d*x])/(
21*d) - (26*a*b*(e*Cos[c + d*x])^(5/2)*(a + b*Sin[c + d*x]))/(63*d*e) - (2*b*(e*Cos[c + d*x])^(5/2)*(a + b*Sin
[c + d*x])^2)/(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 2862

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[1/(m + p + 1), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(m + p + 1) + b*d*m + (a*d*
m + b*c*(m + p + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && Gt
Q[m, 0] &&  !LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x])

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 (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^3 \, dx &=-\frac{2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^2}{9 d e}+\frac{2}{9} \int (e \cos (c+d x))^{3/2} (a+b \sin (c+d x)) \left (\frac{9 a^2}{2}+2 b^2+\frac{13}{2} a b \sin (c+d x)\right ) \, dx\\ &=-\frac{26 a b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{63 d e}-\frac{2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^2}{9 d e}+\frac{4}{63} \int (e \cos (c+d x))^{3/2} \left (\frac{9}{4} a \left (7 a^2+6 b^2\right )+\frac{1}{4} b \left (89 a^2+28 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac{2 b \left (89 a^2+28 b^2\right ) (e \cos (c+d x))^{5/2}}{315 d e}-\frac{26 a b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{63 d e}-\frac{2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^2}{9 d e}+\frac{1}{7} \left (a \left (7 a^2+6 b^2\right )\right ) \int (e \cos (c+d x))^{3/2} \, dx\\ &=-\frac{2 b \left (89 a^2+28 b^2\right ) (e \cos (c+d x))^{5/2}}{315 d e}+\frac{2 a \left (7 a^2+6 b^2\right ) e \sqrt{e \cos (c+d x)} \sin (c+d x)}{21 d}-\frac{26 a b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{63 d e}-\frac{2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^2}{9 d e}+\frac{1}{21} \left (a \left (7 a^2+6 b^2\right ) e^2\right ) \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx\\ &=-\frac{2 b \left (89 a^2+28 b^2\right ) (e \cos (c+d x))^{5/2}}{315 d e}+\frac{2 a \left (7 a^2+6 b^2\right ) e \sqrt{e \cos (c+d x)} \sin (c+d x)}{21 d}-\frac{26 a b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{63 d e}-\frac{2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^2}{9 d e}+\frac{\left (a \left (7 a^2+6 b^2\right ) e^2 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{21 \sqrt{e \cos (c+d x)}}\\ &=-\frac{2 b \left (89 a^2+28 b^2\right ) (e \cos (c+d x))^{5/2}}{315 d e}+\frac{2 a \left (7 a^2+6 b^2\right ) e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d \sqrt{e \cos (c+d x)}}+\frac{2 a \left (7 a^2+6 b^2\right ) e \sqrt{e \cos (c+d x)} \sin (c+d x)}{21 d}-\frac{26 a b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{63 d e}-\frac{2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^2}{9 d e}\\ \end{align*}

Mathematica [A]  time = 1.35052, size = 153, normalized size = 0.78 \[ \frac{(e \cos (c+d x))^{3/2} \left (80 \left (7 a^3+6 a b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{2}{3} \sqrt{\cos (c+d x)} \left (-28 \left (27 a^2 b+4 b^3\right ) \cos (2 (c+d x))-756 a^2 b+840 a^3 \sin (c+d x)+450 a b^2 \sin (c+d x)-270 a b^2 \sin (3 (c+d x))+35 b^3 \cos (4 (c+d x))-147 b^3\right )\right )}{840 d \cos ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Cos[c + d*x])^(3/2)*(80*(7*a^3 + 6*a*b^2)*EllipticF[(c + d*x)/2, 2] + (2*Sqrt[Cos[c + d*x]]*(-756*a^2*b -
147*b^3 - 28*(27*a^2*b + 4*b^3)*Cos[2*(c + d*x)] + 35*b^3*Cos[4*(c + d*x)] + 840*a^3*Sin[c + d*x] + 450*a*b^2*
Sin[c + d*x] - 270*a*b^2*Sin[3*(c + d*x)]))/3))/(840*d*Cos[c + d*x]^(3/2))

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Maple [B]  time = 1.996, size = 450, normalized size = 2.3 \begin{align*} -{\frac{2\,{e}^{2}}{315\,d} \left ( 1120\,{b}^{3} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{11}-2160\,a{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}-2800\,{b}^{3} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{9}-1512\,{a}^{2}b \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}+3240\,a{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+2296\,{b}^{3} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}+420\,{a}^{3}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+2268\,{a}^{2}b \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}-1260\,a{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-644\,{b}^{3} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+105\,\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 EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){a}^{3}+90\,\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 EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) a{b}^{2}-210\,{a}^{3}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1134\,{a}^{2}b \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+90\,a{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-28\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}{b}^{3}+189\,{a}^{2}b\sin \left ( 1/2\,dx+c/2 \right ) +28\,{b}^{3}\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))^(3/2)*(a+b*sin(d*x+c))^3,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^2*(1120*b^3*sin(1/2*d*x+1/2*c)^11-2160*a*b^2*c
os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-2800*b^3*sin(1/2*d*x+1/2*c)^9-1512*a^2*b*sin(1/2*d*x+1/2*c)^7+3240*a*b^
2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+2296*b^3*sin(1/2*d*x+1/2*c)^7+420*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^4+2268*a^2*b*sin(1/2*d*x+1/2*c)^5-1260*a*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-644*b^3*sin(1/2*d
*x+1/2*c)^5+105*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(
1/2))*a^3+90*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2
))*a*b^2-210*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-1134*a^2*b*sin(1/2*d*x+1/2*c)^3+90*a*b^2*cos(1/2*d*x+
1/2*c)*sin(1/2*d*x+1/2*c)^2-28*sin(1/2*d*x+1/2*c)^3*b^3+189*a^2*b*sin(1/2*d*x+1/2*c)+28*b^3*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{3}{2}}{\left (b \sin \left (d x + c\right ) + a\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integral(-(3*a*b^2*e*cos(d*x + c)^3 - (a^3 + 3*a*b^2)*e*cos(d*x + c) + (b^3*e*cos(d*x + c)^3 - (3*a^2*b + b^3)
*e*cos(d*x + c))*sin(d*x + c))*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))**(3/2)*(a+b*sin(d*x+c))**3,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{3}{2}}{\left (b \sin \left (d x + c\right ) + a\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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