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

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

[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 = 0.12, 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 = {2678, 2669, 2635, 2642, 2641} \[ \frac {6 a^2 e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{7 d \sqrt {e \cos (c+d x)}}-\frac {18 a^2 (e \cos (c+d x))^{5/2}}{35 d e}-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{5/2}}{7 d e}+\frac {6 a^2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{7 d} \]

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 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 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]

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 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 2678

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]  time = 0.08, size = 66, normalized size = 0.48 \[ -\frac {16 \sqrt [4]{2} a^2 (e \cos (c+d x))^{5/2} \, _2F_1\left (-\frac {9}{4},\frac {5}{4};\frac {9}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{5 d e (\sin (c+d x)+1)^{5/4}} \]

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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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a^{2} e \cos \left (d x + c\right )^{3} - 2 \, a^{2} e \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} e \cos \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}, x\right ) \]

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]

integral(-(a^2*e*cos(d*x + c)^3 - 2*a^2*e*cos(d*x + c)*sin(d*x + c) - 2*a^2*e*cos(d*x + c))*sqrt(e*cos(d*x + c
)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}^{2}\,{d x} \]

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

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maple [A]  time = 0.75, size = 203, normalized size = 1.48 \[ -\frac {2 a^{2} e^{2} \left (-80 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+120 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-112 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+168 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}-20 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-84 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+14 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d} \]

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)

[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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(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 \[ \int \left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}^{2}\,{d x} \]

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]

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

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

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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