3.180 \(\int \frac{1}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx\)

Optimal. Leaf size=124 \[ \frac{5 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}-\frac{\sin (c+d x)}{d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)}+\frac{5 \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x)}-\frac{3 \sin (c+d x)}{a d \sqrt{\cos (c+d x)}} \]

[Out]

(3*EllipticE[(c + d*x)/2, 2])/(a*d) + (5*EllipticF[(c + d*x)/2, 2])/(3*a*d) + (5*Sin[c + d*x])/(3*a*d*Cos[c +
d*x]^(3/2)) - (3*Sin[c + d*x])/(a*d*Sqrt[Cos[c + d*x]]) - Sin[c + d*x]/(d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*
x]))

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Rubi [A]  time = 0.112024, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2768, 2748, 2636, 2641, 2639} \[ \frac{5 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}-\frac{\sin (c+d x)}{d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)}+\frac{5 \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x)}-\frac{3 \sin (c+d x)}{a d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])),x]

[Out]

(3*EllipticE[(c + d*x)/2, 2])/(a*d) + (5*EllipticF[(c + d*x)/2, 2])/(3*a*d) + (5*Sin[c + d*x])/(3*a*d*Cos[c +
d*x]^(3/2)) - (3*Sin[c + d*x])/(a*d*Sqrt[Cos[c + d*x]]) - Sin[c + d*x]/(d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*
x]))

Rule 2768

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b
^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(b*c - a*d)*(a + b*Sin[e + f*x])), x] + Dist[d/(a*(b*c - a*
d)), Int[(c + d*Sin[e + f*x])^n*(a*n - b*(n + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 0] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[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 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]

Rubi steps

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

Mathematica [C]  time = 3.57571, size = 332, normalized size = 2.68 \[ \frac{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \left (-\frac{\csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \left (10 \cos \left (\frac{1}{2} (c-d x)\right )+8 \cos \left (\frac{1}{2} (3 c+d x)\right )+4 \cos \left (\frac{1}{2} (c+3 d x)\right )+5 \cos \left (\frac{1}{2} (5 c+3 d x)\right )+9 \cos \left (\frac{1}{2} (3 c+5 d x)\right )\right ) \sec \left (\frac{1}{2} (c+d x)\right )}{4 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 i \sqrt{2} e^{-i (c+d x)} \left (9 \left (-1+e^{2 i c}\right ) \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )-5 \left (-1+e^{2 i c}\right ) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+9 \left (1+e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) d \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}\right )}{3 a (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])),x]

[Out]

(Cos[(c + d*x)/2]^2*(((2*I)*Sqrt[2]*(9*(1 + E^((2*I)*(c + d*x))) + 9*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c +
d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] - 5*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1
+ E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]))/(d*E^(I*(c + d*x))*(-1 + E^((2
*I)*c))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]) - ((10*Cos[(c - d*x)/2] + 8*Cos[(3*c + d*x)/2] + 4*Co
s[(c + 3*d*x)/2] + 5*Cos[(5*c + 3*d*x)/2] + 9*Cos[(3*c + 5*d*x)/2])*Csc[c/2]*Sec[c/2]*Sec[(c + d*x)/2])/(4*d*C
os[c + d*x]^(3/2))))/(3*a*(1 + Cos[c + d*x]))

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Maple [B]  time = 4.016, size = 413, normalized size = 3.3 \begin{align*}{\frac{1}{3\,da}\sqrt{- \left ( -2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 10\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-18\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-36\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-5\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\cos \left ( 1/2\,dx+c/2 \right ) +9\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\cos \left ( 1/2\,dx+c/2 \right ) +44\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-11\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3} \left ( 4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(d*x+c)^(5/2)/(a+cos(d*x+c)*a),x)

[Out]

1/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a/cos(1/2*d*x+1/2*c)/sin(1/2*d*x+1/2*c)^3/(4*sin
(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(10*(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)*cos(1/2*d*x+1/2*c)*
sin(1/2*d*x+1/2*c)^2-18*(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)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-36*sin(1/2*d*x+1/2*c)^6-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*cos(1/2*d*x+1/2*c)+9*(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)*cos(1/2*d*x+1/2*c)+44*sin(1/2
*d*x+1/2*c)^4-11*sin(1/2*d*x+1/2*c)^2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c)),x, algorithm="maxima")

[Out]

integrate(1/((a*cos(d*x + c) + a)*cos(d*x + c)^(5/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{\cos \left (d x + c\right )}}{a \cos \left (d x + c\right )^{4} + a \cos \left (d x + c\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)**(5/2)/(a+a*cos(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c)),x, algorithm="giac")

[Out]

integrate(1/((a*cos(d*x + c) + a)*cos(d*x + c)^(5/2)), x)