3.222 \(\int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx\)

Optimal. Leaf size=127 \[ \frac {4 a^5 \sqrt {e \cos (c+d x)}}{7 d e^5 (a-a \sin (c+d x))^2}-\frac {2 a^3 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d e^4 \sqrt {e \cos (c+d x)}}-\frac {2 a^6 \sqrt {e \cos (c+d x)}}{21 d e^5 \left (a^3-a^3 \sin (c+d x)\right )} \]

[Out]

-2/21*a^3*(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^4/(e*cos(d*x+c))^(1/2)+4/7*a^5*(e*cos(d*x+c))^(1/2)/d/e^5/(a-a*sin(d*x+c))^2-2/21*a^6*(e*cos(d*x+c))^(1
/2)/d/e^5/(a^3-a^3*sin(d*x+c))

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Rubi [A]  time = 0.20, antiderivative size = 127, 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, 2683, 2642, 2641} \[ -\frac {2 a^6 \sqrt {e \cos (c+d x)}}{21 d e^5 \left (a^3-a^3 \sin (c+d x)\right )}+\frac {4 a^5 \sqrt {e \cos (c+d x)}}{7 d e^5 (a-a \sin (c+d x))^2}-\frac {2 a^3 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d e^4 \sqrt {e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^3/(e*Cos[c + d*x])^(9/2),x]

[Out]

(-2*a^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*e^4*Sqrt[e*Cos[c + d*x]]) + (4*a^5*Sqrt[e*Cos[c +
d*x]])/(7*d*e^5*(a - a*Sin[c + d*x])^2) - (2*a^6*Sqrt[e*Cos[c + d*x]])/(21*d*e^5*(a^3 - a^3*Sin[c + d*x]))

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 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 2683

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

Rubi steps

\begin {align*} \int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx &=\frac {a^6 \int \frac {(e \cos (c+d x))^{3/2}}{(a-a \sin (c+d x))^3} \, dx}{e^6}\\ &=\frac {4 a^5 \sqrt {e \cos (c+d x)}}{7 d e^5 (a-a \sin (c+d x))^2}-\frac {a^4 \int \frac {1}{\sqrt {e \cos (c+d x)} (a-a \sin (c+d x))} \, dx}{7 e^4}\\ &=\frac {4 a^5 \sqrt {e \cos (c+d x)}}{7 d e^5 (a-a \sin (c+d x))^2}-\frac {2 a^4 \sqrt {e \cos (c+d x)}}{21 d e^5 (a-a \sin (c+d x))}-\frac {a^3 \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{21 e^4}\\ &=\frac {4 a^5 \sqrt {e \cos (c+d x)}}{7 d e^5 (a-a \sin (c+d x))^2}-\frac {2 a^4 \sqrt {e \cos (c+d x)}}{21 d e^5 (a-a \sin (c+d x))}-\frac {\left (a^3 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 e^4 \sqrt {e \cos (c+d x)}}\\ &=-\frac {2 a^3 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d e^4 \sqrt {e \cos (c+d x)}}+\frac {4 a^5 \sqrt {e \cos (c+d x)}}{7 d e^5 (a-a \sin (c+d x))^2}-\frac {2 a^4 \sqrt {e \cos (c+d x)}}{21 d e^5 (a-a \sin (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 66, normalized size = 0.52 \[ \frac {4 \sqrt [4]{2} a^3 (\sin (c+d x)+1)^{7/4} \, _2F_1\left (-\frac {7}{4},-\frac {1}{4};-\frac {3}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{7 d e (e \cos (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^3/(e*Cos[c + d*x])^(9/2),x]

[Out]

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

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fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{e^{5} \cos \left (d x + c\right )^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(9/2),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^5
*cos(d*x + c)^5), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}{\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(9/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(9/2), x)

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maple [B]  time = 2.00, size = 401, normalized size = 3.16 \[ \frac {2 \left (8 \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 ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \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 ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+6 \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}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+28 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\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}-22 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-28 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{21 \left (8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \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}\, e^{4} 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))^(9/2),x)

[Out]

2/21/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*
d*x+1/2*c)^2*e+e)^(1/2)/e^4*(8*(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))*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^6-12*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4+8*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+6*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)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2
-8*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+28*sin(1/2*d*x+1/2*c)^5-(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)-22*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-28*sin(1/2
*d*x+1/2*c)^3-5*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 \[ \int \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}{\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(9/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(9/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^3/(e*cos(c + d*x))^(9/2),x)

[Out]

int((a + a*sin(c + d*x))^3/(e*cos(c + d*x))^(9/2), 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))**(9/2),x)

[Out]

Timed out

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