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

Optimal. Leaf size=106 \[ \frac {4 a^5 (e \cos (c+d x))^{7/2}}{d e^5 (a-a \sin (c+d x))^2}+\frac {14 a^3 (e \cos (c+d x))^{3/2}}{3 d e^3}-\frac {14 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{d e^2 \sqrt {\cos (c+d x)}} \]

[Out]

14/3*a^3*(e*cos(d*x+c))^(3/2)/d/e^3+4*a^5*(e*cos(d*x+c))^(7/2)/d/e^5/(a-a*sin(d*x+c))^2-14*a^3*(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/e^2/cos(d*x+c)^(
1/2)

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Rubi [A]  time = 0.20, antiderivative size = 106, 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 = {2670, 2680, 2682, 2640, 2639} \[ \frac {14 a^3 (e \cos (c+d x))^{3/2}}{3 d e^3}+\frac {4 a^5 (e \cos (c+d x))^{7/2}}{d e^5 (a-a \sin (c+d x))^2}-\frac {14 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{d e^2 \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(14*a^3*(e*Cos[c + d*x])^(3/2))/(3*d*e^3) - (14*a^3*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(d*e^2*Sqr
t[Cos[c + d*x]]) + (4*a^5*(e*Cos[c + d*x])^(7/2))/(d*e^5*(a - a*Sin[c + d*x])^2)

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]

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 2670

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

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2682

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e
+ f*x])^(p - 1))/(b*f*(p - 1)), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{3/2}} \, dx &=\frac {a^6 \int \frac {(e \cos (c+d x))^{9/2}}{(a-a \sin (c+d x))^3} \, dx}{e^6}\\ &=\frac {4 a^5 (e \cos (c+d x))^{7/2}}{d e^5 (a-a \sin (c+d x))^2}-\frac {\left (7 a^4\right ) \int \frac {(e \cos (c+d x))^{5/2}}{a-a \sin (c+d x)} \, dx}{e^4}\\ &=\frac {14 a^3 (e \cos (c+d x))^{3/2}}{3 d e^3}+\frac {4 a^5 (e \cos (c+d x))^{7/2}}{d e^5 (a-a \sin (c+d x))^2}-\frac {\left (7 a^3\right ) \int \sqrt {e \cos (c+d x)} \, dx}{e^2}\\ &=\frac {14 a^3 (e \cos (c+d x))^{3/2}}{3 d e^3}+\frac {4 a^5 (e \cos (c+d x))^{7/2}}{d e^5 (a-a \sin (c+d x))^2}-\frac {\left (7 a^3 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{e^2 \sqrt {\cos (c+d x)}}\\ &=\frac {14 a^3 (e \cos (c+d x))^{3/2}}{3 d e^3}-\frac {14 a^3 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d e^2 \sqrt {\cos (c+d x)}}+\frac {4 a^5 (e \cos (c+d x))^{7/2}}{d e^5 (a-a \sin (c+d x))^2}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 64, normalized size = 0.60 \[ \frac {8\ 2^{3/4} a^3 \sqrt [4]{\sin (c+d x)+1} \, _2F_1\left (-\frac {7}{4},-\frac {1}{4};\frac {3}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{d e \sqrt {e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*2^(3/4)*a^3*Hypergeometric2F1[-7/4, -1/4, 3/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(1/4))/(d*e*Sqrt[e*
Cos[c + d*x]])

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{e^{2} \cos \left (d x + c\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(3*a^3*cos(d*x + c)^2 - 4*a^3 + (a^3*cos(d*x + c)^2 - 4*a^3)*sin(d*x + c))*sqrt(e*cos(d*x + c))/(e^2
*cos(d*x + c)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.18, size = 146, normalized size = 1.38 \[ -\frac {2 \left (-4 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+21 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}-24 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-13 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{3 e \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/e/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/sin(1/2*d*x+1/2*c)*(-4*sin(1/2*d*x+1/2*c)^5+21*EllipticE(cos(1/2*d*
x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)-24*sin(1/2*d*x+1/2*c)^2*cos(1/
2*d*x+1/2*c)+4*sin(1/2*d*x+1/2*c)^3-13*sin(1/2*d*x+1/2*c))*a^3/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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