3.556 \(\int (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=197 \[ \frac{2 a e^2 \left (3 a^2+2 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d \sqrt{\cos (c+d x)}}-\frac{2 b \left (43 a^2+12 b^2\right ) (e \cos (c+d x))^{7/2}}{231 d e}+\frac{2 a e \left (3 a^2+2 b^2\right ) \sin (c+d x) (e \cos (c+d x))^{3/2}}{15 d}-\frac{2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2}{11 d e}-\frac{10 a b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{33 d e} \]

[Out]

(-2*b*(43*a^2 + 12*b^2)*(e*Cos[c + d*x])^(7/2))/(231*d*e) + (2*a*(3*a^2 + 2*b^2)*e^2*Sqrt[e*Cos[c + d*x]]*Elli
pticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]) + (2*a*(3*a^2 + 2*b^2)*e*(e*Cos[c + d*x])^(3/2)*Sin[c + d*x])/
(15*d) - (10*a*b*(e*Cos[c + d*x])^(7/2)*(a + b*Sin[c + d*x]))/(33*d*e) - (2*b*(e*Cos[c + d*x])^(7/2)*(a + b*Si
n[c + d*x])^2)/(11*d*e)

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Rubi [A]  time = 0.286019, 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, 2640, 2639} \[ \frac{2 a e^2 \left (3 a^2+2 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d \sqrt{\cos (c+d x)}}-\frac{2 b \left (43 a^2+12 b^2\right ) (e \cos (c+d x))^{7/2}}{231 d e}+\frac{2 a e \left (3 a^2+2 b^2\right ) \sin (c+d x) (e \cos (c+d x))^{3/2}}{15 d}-\frac{2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2}{11 d e}-\frac{10 a b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{33 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*(43*a^2 + 12*b^2)*(e*Cos[c + d*x])^(7/2))/(231*d*e) + (2*a*(3*a^2 + 2*b^2)*e^2*Sqrt[e*Cos[c + d*x]]*Elli
pticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]) + (2*a*(3*a^2 + 2*b^2)*e*(e*Cos[c + d*x])^(3/2)*Sin[c + d*x])/
(15*d) - (10*a*b*(e*Cos[c + d*x])^(7/2)*(a + b*Sin[c + d*x]))/(33*d*e) - (2*b*(e*Cos[c + d*x])^(7/2)*(a + b*Si
n[c + d*x])^2)/(11*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 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))^3 \, dx &=-\frac{2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2}{11 d e}+\frac{2}{11} \int (e \cos (c+d x))^{5/2} (a+b \sin (c+d x)) \left (\frac{11 a^2}{2}+2 b^2+\frac{15}{2} a b \sin (c+d x)\right ) \, dx\\ &=-\frac{10 a b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{33 d e}-\frac{2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2}{11 d e}+\frac{4}{99} \int (e \cos (c+d x))^{5/2} \left (\frac{33}{4} a \left (3 a^2+2 b^2\right )+\frac{3}{4} b \left (43 a^2+12 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac{2 b \left (43 a^2+12 b^2\right ) (e \cos (c+d x))^{7/2}}{231 d e}-\frac{10 a b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{33 d e}-\frac{2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2}{11 d e}+\frac{1}{3} \left (a \left (3 a^2+2 b^2\right )\right ) \int (e \cos (c+d x))^{5/2} \, dx\\ &=-\frac{2 b \left (43 a^2+12 b^2\right ) (e \cos (c+d x))^{7/2}}{231 d e}+\frac{2 a \left (3 a^2+2 b^2\right ) e (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 d}-\frac{10 a b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{33 d e}-\frac{2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2}{11 d e}+\frac{1}{5} \left (a \left (3 a^2+2 b^2\right ) e^2\right ) \int \sqrt{e \cos (c+d x)} \, dx\\ &=-\frac{2 b \left (43 a^2+12 b^2\right ) (e \cos (c+d x))^{7/2}}{231 d e}+\frac{2 a \left (3 a^2+2 b^2\right ) e (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 d}-\frac{10 a b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{33 d e}-\frac{2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2}{11 d e}+\frac{\left (a \left (3 a^2+2 b^2\right ) e^2 \sqrt{e \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 \sqrt{\cos (c+d x)}}\\ &=-\frac{2 b \left (43 a^2+12 b^2\right ) (e \cos (c+d x))^{7/2}}{231 d e}+\frac{2 a \left (3 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 )}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a \left (3 a^2+2 b^2\right ) e (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 d}-\frac{10 a b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{33 d e}-\frac{2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2}{11 d e}\\ \end{align*}

Mathematica [A]  time = 1.38534, size = 150, normalized size = 0.76 \[ \frac{(e \cos (c+d x))^{5/2} \left (1848 \left (3 a^3+2 a b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\cos ^{\frac{3}{2}}(c+d x) \left (-60 \left (33 a^2 b+4 b^3\right ) \cos (2 (c+d x))-1980 a^2 b+1848 a^3 \sin (c+d x)+462 a b^2 \sin (c+d x)-770 a b^2 \sin (3 (c+d x))+105 b^3 \cos (4 (c+d x))-345 b^3\right )\right )}{4620 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Cos[c + d*x])^(5/2)*(1848*(3*a^3 + 2*a*b^2)*EllipticE[(c + d*x)/2, 2] + Cos[c + d*x]^(3/2)*(-1980*a^2*b -
345*b^3 - 60*(33*a^2*b + 4*b^3)*Cos[2*(c + d*x)] + 105*b^3*Cos[4*(c + d*x)] + 1848*a^3*Sin[c + d*x] + 462*a*b^
2*Sin[c + d*x] - 770*a*b^2*Sin[3*(c + d*x)])))/(4620*d*Cos[c + d*x]^(5/2))

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Maple [B]  time = 2.238, size = 534, normalized size = 2.7 \begin{align*} \text{result too large to display} \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))^3,x)

[Out]

2/1155/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^3*(6720*b^3*sin(1/2*d*x+1/2*c)^13-12320*a*b^2*
sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)-20160*b^3*sin(1/2*d*x+1/2*c)^11-7920*a^2*b*sin(1/2*d*x+1/2*c)^9+24640
*a*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+22560*b^3*sin(1/2*d*x+1/2*c)^9+1848*a^3*cos(1/2*d*x+1/2*c)*sin(
1/2*d*x+1/2*c)^6+15840*a^2*b*sin(1/2*d*x+1/2*c)^7-17248*a*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-11520*b^
3*sin(1/2*d*x+1/2*c)^7-1848*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-11880*a^2*b*sin(1/2*d*x+1/2*c)^5+4928*
a*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+2340*b^3*sin(1/2*d*x+1/2*c)^5+693*(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^3+462*(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*b^2+462*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d
*x+1/2*c)^2+3960*a^2*b*sin(1/2*d*x+1/2*c)^3-462*a*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+60*sin(1/2*d*x+1
/2*c)^3*b^3-495*a^2*b*sin(1/2*d*x+1/2*c)-60*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{5}{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))^(5/2)*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(5/2)*(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^{2} \cos \left (d x + c\right )^{4} -{\left (a^{3} + 3 \, a b^{2}\right )} e^{2} \cos \left (d x + c\right )^{2} +{\left (b^{3} e^{2} \cos \left (d x + c\right )^{4} -{\left (3 \, a^{2} b + b^{3}\right )} e^{2} \cos \left (d x + c\right )^{2}\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))^(5/2)*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

integral(-(3*a*b^2*e^2*cos(d*x + c)^4 - (a^3 + 3*a*b^2)*e^2*cos(d*x + c)^2 + (b^3*e^2*cos(d*x + c)^4 - (3*a^2*
b + b^3)*e^2*cos(d*x + c)^2)*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))**(5/2)*(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{5}{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))^(5/2)*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(5/2)*(b*sin(d*x + c) + a)^3, x)