3.6.48 \(\int (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^2 \, dx\) [548]

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

[Out]

-22/63*a*b*(e*cos(d*x+c))^(7/2)/d/e+2/45*(9*a^2+2*b^2)*e*(e*cos(d*x+c))^(3/2)*sin(d*x+c)/d-2/9*b*(e*cos(d*x+c)
)^(7/2)*(a+b*sin(d*x+c))/d/e+2/15*(9*a^2+2*b^2)*e^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(
sin(1/2*d*x+1/2*c),2^(1/2))*(e*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2771, 2748, 2715, 2721, 2719} \begin {gather*} \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} \end {gather*}

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 2715

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] && Integ
erQ[2*n]

Rule 2719

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

Rule 2721

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

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[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 2771

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])

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 = 0.92, size = 113, normalized size = 0.76 \begin {gather*} \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 (-180 a b \cos (2 (c+d x))+21 \left (12 a^2+b^2\right ) \sin (c+d x)-5 b (36 a+7 b \sin (3 (c+d x)))\right )\right )}{630 d \cos ^{\frac {5}{2}}(c+d x)} \end {gather*}

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))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(407\) vs. \(2(157)=314\).
time = 5.08, size = 408, normalized size = 2.74

method	result	size
default	\(\frac {2 e^{3} \left (-1120 b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1440 a b \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2240 b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+504 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2880 a b \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1568 b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-504 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2160 a b \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+448 b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+189 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}+42 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}+126 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+720 a b \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-42 b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-90 a b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{315 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(408\)

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,method=_RETURNVERBOSE)

[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*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^10-1440*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*a*b*sin(1/2*d*x+1/2*c))/d

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 154, normalized size = 1.03 \begin {gather*} \frac {21 i \, \sqrt {2} {\left (9 \, a^{2} + 2 \, b^{2}\right )} e^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} {\left (9 \, a^{2} + 2 \, b^{2}\right )} e^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (90 \, a b \cos \left (d x + c\right )^{3} e^{\frac {5}{2}} + 7 \, {\left (5 \, b^{2} \cos \left (d x + c\right )^{3} e^{\frac {5}{2}} - {\left (9 \, a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right ) e^{\frac {5}{2}}\right )} \sin \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}}{315 \, d} \end {gather*}

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]

1/315*(21*I*sqrt(2)*(9*a^2 + 2*b^2)*e^(5/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I
*sin(d*x + c))) - 21*I*sqrt(2)*(9*a^2 + 2*b^2)*e^(5/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d
*x + c) - I*sin(d*x + c))) - 2*(90*a*b*cos(d*x + c)^3*e^(5/2) + 7*(5*b^2*cos(d*x + c)^3*e^(5/2) - (9*a^2 + 2*b
^2)*cos(d*x + c)*e^(5/2))*sin(d*x + c))*sqrt(cos(d*x + c)))/d

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 7319 deep

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________