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

Optimal. Leaf size=132 \[ \frac {10 e^6 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 a d \sqrt {e \cos (c+d x)}}+\frac {10 e^5 \sin (c+d x) \sqrt {e \cos (c+d x)}}{21 a d}+\frac {2 e^3 \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 a d}+\frac {2 e (e \cos (c+d x))^{9/2}}{9 a d} \]

[Out]

2/9*e*(e*cos(d*x+c))^(9/2)/a/d+2/7*e^3*(e*cos(d*x+c))^(5/2)*sin(d*x+c)/a/d+10/21*e^6*(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)/a/d/(e*cos(d*x+c))^(1/2)+10/21*e
^5*sin(d*x+c)*(e*cos(d*x+c))^(1/2)/a/d

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Rubi [A]  time = 0.12, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2682, 2635, 2642, 2641} \[ \frac {10 e^5 \sin (c+d x) \sqrt {e \cos (c+d x)}}{21 a d}+\frac {2 e^3 \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 a d}+\frac {10 e^6 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 a d \sqrt {e \cos (c+d x)}}+\frac {2 e (e \cos (c+d x))^{9/2}}{9 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*(e*Cos[c + d*x])^(9/2))/(9*a*d) + (10*e^6*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*a*d*Sqrt[e*Co
s[c + d*x]]) + (10*e^5*Sqrt[e*Cos[c + d*x]]*Sin[c + d*x])/(21*a*d) + (2*e^3*(e*Cos[c + d*x])^(5/2)*Sin[c + d*x
])/(7*a*d)

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 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 {(e \cos (c+d x))^{11/2}}{a+a \sin (c+d x)} \, dx &=\frac {2 e (e \cos (c+d x))^{9/2}}{9 a d}+\frac {e^2 \int (e \cos (c+d x))^{7/2} \, dx}{a}\\ &=\frac {2 e (e \cos (c+d x))^{9/2}}{9 a d}+\frac {2 e^3 (e \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d}+\frac {\left (5 e^4\right ) \int (e \cos (c+d x))^{3/2} \, dx}{7 a}\\ &=\frac {2 e (e \cos (c+d x))^{9/2}}{9 a d}+\frac {10 e^5 \sqrt {e \cos (c+d x)} \sin (c+d x)}{21 a d}+\frac {2 e^3 (e \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d}+\frac {\left (5 e^6\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{21 a}\\ &=\frac {2 e (e \cos (c+d x))^{9/2}}{9 a d}+\frac {10 e^5 \sqrt {e \cos (c+d x)} \sin (c+d x)}{21 a d}+\frac {2 e^3 (e \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d}+\frac {\left (5 e^6 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 a \sqrt {e \cos (c+d x)}}\\ &=\frac {2 e (e \cos (c+d x))^{9/2}}{9 a d}+\frac {10 e^6 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 a d \sqrt {e \cos (c+d x)}}+\frac {10 e^5 \sqrt {e \cos (c+d x)} \sin (c+d x)}{21 a d}+\frac {2 e^3 (e \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d}\\ \end {align*}

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Mathematica [C]  time = 0.14, size = 66, normalized size = 0.50 \[ -\frac {8 \sqrt [4]{2} (e \cos (c+d x))^{13/2} \, _2F_1\left (-\frac {5}{4},\frac {13}{4};\frac {17}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{13 a d e (\sin (c+d x)+1)^{13/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*2^(1/4)*(e*Cos[c + d*x])^(13/2)*Hypergeometric2F1[-5/4, 13/4, 17/4, (1 - Sin[c + d*x])/2])/(13*a*d*e*(1 +
Sin[c + d*x])^(13/4))

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e \cos \left (d x + c\right )} e^{5} \cos \left (d x + c\right )^{5}}{a \sin \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(e*cos(d*x + c))*e^5*cos(d*x + c)^5/(a*sin(d*x + c) + a), x)

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giac [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))^(11/2)/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.89, size = 251, normalized size = 1.90 \[ -\frac {2 e^{6} \left (224 \left (\sin ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+144 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-560 \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-216 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+560 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+168 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-280 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}-48 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+70 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{63 a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/63/a/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^6*(224*sin(1/2*d*x+1/2*c)^11+144*cos(1/2*d*x+
1/2*c)*sin(1/2*d*x+1/2*c)^8-560*sin(1/2*d*x+1/2*c)^9-216*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+560*sin(1/2*d
*x+1/2*c)^7+168*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-280*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)-48*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+
1/2*c)+70*sin(1/2*d*x+1/2*c)^3-7*sin(1/2*d*x+1/2*c))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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