3.3.6 \(\int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2 \, dx\) [206]

Optimal. Leaf size=137 \[ -\frac {18 a^2 (e \cos (c+d x))^{5/2}}{35 d e}+\frac {6 a^2 e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{7 d \sqrt {e \cos (c+d x)}}+\frac {6 a^2 e \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 d}-\frac {2 (e \cos (c+d x))^{5/2} \left (a^2+a^2 \sin (c+d x)\right )}{7 d e} \]

[Out]

-18/35*a^2*(e*cos(d*x+c))^(5/2)/d/e-2/7*(e*cos(d*x+c))^(5/2)*(a^2+a^2*sin(d*x+c))/d/e+6/7*a^2*e^2*(cos(1/2*d*x
+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/(e*cos(d*x+c))^(1
/2)+6/7*a^2*e*sin(d*x+c)*(e*cos(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.09, antiderivative size = 137, 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 = {2757, 2748, 2715, 2721, 2720} \begin {gather*} \frac {6 a^2 e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{7 d \sqrt {e \cos (c+d x)}}-\frac {18 a^2 (e \cos (c+d x))^{5/2}}{35 d e}-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{5/2}}{7 d e}+\frac {6 a^2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-18*a^2*(e*Cos[c + d*x])^(5/2))/(35*d*e) + (6*a^2*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(7*d*Sqrt
[e*Cos[c + d*x]]) + (6*a^2*e*Sqrt[e*Cos[c + d*x]]*Sin[c + d*x])/(7*d) - (2*(e*Cos[c + d*x])^(5/2)*(a^2 + a^2*S
in[c + d*x]))/(7*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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 2757

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[a*((2*m + p - 1)/(m + p)), Int
[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2,
0] && GtQ[m, 0] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

Rubi steps

\begin {align*} \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2 \, dx &=-\frac {2 (e \cos (c+d x))^{5/2} \left (a^2+a^2 \sin (c+d x)\right )}{7 d e}+\frac {1}{7} (9 a) \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x)) \, dx\\ &=-\frac {18 a^2 (e \cos (c+d x))^{5/2}}{35 d e}-\frac {2 (e \cos (c+d x))^{5/2} \left (a^2+a^2 \sin (c+d x)\right )}{7 d e}+\frac {1}{7} \left (9 a^2\right ) \int (e \cos (c+d x))^{3/2} \, dx\\ &=-\frac {18 a^2 (e \cos (c+d x))^{5/2}}{35 d e}+\frac {6 a^2 e \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 d}-\frac {2 (e \cos (c+d x))^{5/2} \left (a^2+a^2 \sin (c+d x)\right )}{7 d e}+\frac {1}{7} \left (3 a^2 e^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx\\ &=-\frac {18 a^2 (e \cos (c+d x))^{5/2}}{35 d e}+\frac {6 a^2 e \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 d}-\frac {2 (e \cos (c+d x))^{5/2} \left (a^2+a^2 \sin (c+d x)\right )}{7 d e}+\frac {\left (3 a^2 e^2 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{7 \sqrt {e \cos (c+d x)}}\\ &=-\frac {18 a^2 (e \cos (c+d x))^{5/2}}{35 d e}+\frac {6 a^2 e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{7 d \sqrt {e \cos (c+d x)}}+\frac {6 a^2 e \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 d}-\frac {2 (e \cos (c+d x))^{5/2} \left (a^2+a^2 \sin (c+d x)\right )}{7 d e}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.05, size = 66, normalized size = 0.48 \begin {gather*} -\frac {16 \sqrt [4]{2} a^2 (e \cos (c+d x))^{5/2} \, _2F_1\left (-\frac {9}{4},\frac {5}{4};\frac {9}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{5 d e (1+\sin (c+d x))^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*2^(1/4)*a^2*(e*Cos[c + d*x])^(5/2)*Hypergeometric2F1[-9/4, 5/4, 9/4, (1 - Sin[c + d*x])/2])/(5*d*e*(1 + S
in[c + d*x])^(5/4))

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Maple [A]
time = 1.74, size = 203, normalized size = 1.48

method	result	size
default	\(-\frac {2 a^{2} e^{2} \left (-80 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+120 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-112 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+168 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-20 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-84 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+14 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(203\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[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^2*e^2*(-80*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^8+120*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-112*sin(1/2*d*x+1/2*c)^7+168*sin(1/2*d*x+1/2*c)^5+15*(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))-20*sin(1/2*d*x+1/2*
c)^2*cos(1/2*d*x+1/2*c)-84*sin(1/2*d*x+1/2*c)^3+14*sin(1/2*d*x+1/2*c))/d

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 118, normalized size = 0.86 \begin {gather*} \frac {-15 i \, \sqrt {2} a^{2} e^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 i \, \sqrt {2} a^{2} e^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (14 \, a^{2} \cos \left (d x + c\right )^{2} e^{\frac {3}{2}} + 5 \, {\left (a^{2} \cos \left (d x + c\right )^{2} e^{\frac {3}{2}} - 3 \, a^{2} e^{\frac {3}{2}}\right )} \sin \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}}{35 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(-15*I*sqrt(2)*a^2*e^(3/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 15*I*sqrt(2)*a^2*e
^(3/2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*(14*a^2*cos(d*x + c)^2*e^(3/2) + 5*(a^2*c
os(d*x + c)^2*e^(3/2) - 3*a^2*e^(3/2))*sin(d*x + c))*sqrt(cos(d*x + c)))/d

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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))**(3/2)*(a+a*sin(d*x+c))**2,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3064 deep

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}\,{\left (a+a\,\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))^(3/2)*(a + a*sin(c + d*x))^2,x)

[Out]

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

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