3.223 \(\int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{11/2}} \, dx\)

Optimal. Leaf size=165 \[ \frac {2 a^6 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}+\frac {2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}-\frac {2 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{15 d e^6 \sqrt {\cos (c+d x)}}+\frac {2 a^6 (e \cos (c+d x))^{3/2}}{15 d e^7 \left (a^3-a^3 \sin (c+d x)\right )} \]

[Out]

2/9*a^6*(e*cos(d*x+c))^(3/2)/d/e^7/(a-a*sin(d*x+c))^3+2/15*a^5*(e*cos(d*x+c))^(3/2)/d/e^7/(a-a*sin(d*x+c))^2+2
/15*a^6*(e*cos(d*x+c))^(3/2)/d/e^7/(a^3-a^3*sin(d*x+c))-2/15*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^6/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2670, 2681, 2683, 2640, 2639} \[ \frac {2 a^6 (e \cos (c+d x))^{3/2}}{15 d e^7 \left (a^3-a^3 \sin (c+d x)\right )}+\frac {2 a^6 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}+\frac {2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}-\frac {2 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{15 d e^6 \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^3/(e*Cos[c + d*x])^(11/2),x]

[Out]

(-2*a^3*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*d*e^6*Sqrt[Cos[c + d*x]]) + (2*a^6*(e*Cos[c + d*x]
)^(3/2))/(9*d*e^7*(a - a*Sin[c + d*x])^3) + (2*a^5*(e*Cos[c + d*x])^(3/2))/(15*d*e^7*(a - a*Sin[c + d*x])^2) +
(2*a^6*(e*Cos[c + d*x])^(3/2))/(15*d*e^7*(a^3 - a^3*Sin[c + d*x]))

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 2681

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)/(a*f*g*(2*m + p + 1)), x] + Dist[(m + p + 1)/(a*(2*m + p + 1)),
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] && LtQ[m, -1] && NeQ[2*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))^{11/2}} \, dx &=\frac {a^6 \int \frac {\sqrt {e \cos (c+d x)}}{(a-a \sin (c+d x))^3} \, dx}{e^6}\\ &=\frac {2 a^6 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}+\frac {a^5 \int \frac {\sqrt {e \cos (c+d x)}}{(a-a \sin (c+d x))^2} \, dx}{3 e^6}\\ &=\frac {2 a^6 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}+\frac {2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}+\frac {a^4 \int \frac {\sqrt {e \cos (c+d x)}}{a-a \sin (c+d x)} \, dx}{15 e^6}\\ &=\frac {2 a^6 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}+\frac {2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}+\frac {2 a^4 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))}-\frac {a^3 \int \sqrt {e \cos (c+d x)} \, dx}{15 e^6}\\ &=\frac {2 a^6 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}+\frac {2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}+\frac {2 a^4 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))}-\frac {\left (a^3 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{15 e^6 \sqrt {\cos (c+d x)}}\\ &=-\frac {2 a^3 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d e^6 \sqrt {\cos (c+d x)}}+\frac {2 a^6 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}+\frac {2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}+\frac {2 a^4 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 0.13, size = 66, normalized size = 0.40 \[ \frac {2\ 2^{3/4} a^3 (\sin (c+d x)+1)^{9/4} \, _2F_1\left (-\frac {9}{4},\frac {1}{4};-\frac {5}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{9 d e (e \cos (c+d x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^3/(e*Cos[c + d*x])^(11/2),x]

[Out]

(2*2^(3/4)*a^3*Hypergeometric2F1[-9/4, 1/4, -5/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(9/4))/(9*d*e*(e*Co
s[c + d*x])^(9/2))

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{e^{6} \cos \left (d x + c\right )^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(11/2),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^6
*cos(d*x + c)^6), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}{\left (e \cos \left (d x + c\right )\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(11/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(11/2), x)

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maple [B]  time = 2.68, size = 514, normalized size = 3.12 \[ -\frac {2 \left (48 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \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}\, \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-96 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-96 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \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}\, \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+192 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+72 \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}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-152 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-24 \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}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+56 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+36 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \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}-48 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-36 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-11 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{45 \left (16 \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-32 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \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^{5} 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))^(11/2),x)

[Out]

-2/45/(16*sin(1/2*d*x+1/2*c)^8-32*sin(1/2*d*x+1/2*c)^6+24*sin(1/2*d*x+1/2*c)^4-8*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^5*(48*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(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)^8-96*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*
c)-96*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(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)^6+192*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+72*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-152*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*
x+1/2*c)-24*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)
)*sin(1/2*d*x+1/2*c)^2+56*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+36*sin(1/2*d*x+1/2*c)^5+3*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)-48*sin(1/2*d*x+1/2*c)^2*cos(
1/2*d*x+1/2*c)-36*sin(1/2*d*x+1/2*c)^3-11*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 \[ \int \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}{\left (e \cos \left (d x + c\right )\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(11/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(11/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{11/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^3/(e*cos(c + d*x))^(11/2),x)

[Out]

int((a + a*sin(c + d*x))^3/(e*cos(c + d*x))^(11/2), 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))**(11/2),x)

[Out]

Timed out

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