3.197 \(\int \frac{1}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=207 \[ \frac{11 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{2 a^3 d}+\frac{119 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{119 \sin (c+d x)}{30 d \cos ^{\frac{3}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}+\frac{11 \sin (c+d x)}{2 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{119 \sin (c+d x)}{10 a^3 d \sqrt{\cos (c+d x)}}-\frac{2 \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^2}-\frac{\sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^3} \]

[Out]

(119*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + (11*EllipticF[(c + d*x)/2, 2])/(2*a^3*d) + (11*Sin[c + d*x])/(2*a
^3*d*Cos[c + d*x]^(3/2)) - (119*Sin[c + d*x])/(10*a^3*d*Sqrt[Cos[c + d*x]]) - Sin[c + d*x]/(5*d*Cos[c + d*x]^(
3/2)*(a + a*Cos[c + d*x])^3) - (2*Sin[c + d*x])/(3*a*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^2) - (119*Sin[c
+ d*x])/(30*d*Cos[c + d*x]^(3/2)*(a^3 + a^3*Cos[c + d*x]))

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Rubi [A]  time = 0.359069, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2766, 2978, 2748, 2636, 2641, 2639} \[ \frac{11 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{2 a^3 d}+\frac{119 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{119 \sin (c+d x)}{30 d \cos ^{\frac{3}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}+\frac{11 \sin (c+d x)}{2 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{119 \sin (c+d x)}{10 a^3 d \sqrt{\cos (c+d x)}}-\frac{2 \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^2}-\frac{\sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^3),x]

[Out]

(119*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + (11*EllipticF[(c + d*x)/2, 2])/(2*a^3*d) + (11*Sin[c + d*x])/(2*a
^3*d*Cos[c + d*x]^(3/2)) - (119*Sin[c + d*x])/(10*a^3*d*Sqrt[Cos[c + d*x]]) - Sin[c + d*x]/(5*d*Cos[c + d*x]^(
3/2)*(a + a*Cos[c + d*x])^3) - (2*Sin[c + d*x])/(3*a*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^2) - (119*Sin[c
+ d*x])/(30*d*Cos[c + d*x]^(3/2)*(a^3 + a^3*Cos[c + d*x]))

Rule 2766

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

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (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))^3} \, dx &=-\frac{\sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3}+\frac{\int \frac{\frac{13 a}{2}-\frac{7}{2} a \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{\sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac{2 \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2}+\frac{\int \frac{\frac{69 a^2}{2}-25 a^2 \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx}{15 a^4}\\ &=-\frac{\sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac{2 \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac{119 \sin (c+d x)}{30 d \cos ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )}+\frac{\int \frac{\frac{495 a^3}{4}-\frac{357}{4} a^3 \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{15 a^6}\\ &=-\frac{\sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac{2 \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac{119 \sin (c+d x)}{30 d \cos ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )}-\frac{119 \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{20 a^3}+\frac{33 \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{4 a^3}\\ &=\frac{11 \sin (c+d x)}{2 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{119 \sin (c+d x)}{10 a^3 d \sqrt{\cos (c+d x)}}-\frac{\sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac{2 \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac{119 \sin (c+d x)}{30 d \cos ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )}+\frac{11 \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{4 a^3}+\frac{119 \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}\\ &=\frac{119 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{11 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{2 a^3 d}+\frac{11 \sin (c+d x)}{2 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{119 \sin (c+d x)}{10 a^3 d \sqrt{\cos (c+d x)}}-\frac{\sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac{2 \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac{119 \sin (c+d x)}{30 d \cos ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 2.45878, size = 394, normalized size = 1.9 \[ \frac{\cos ^6\left (\frac{1}{2} (c+d x)\right ) \left (-\frac{\csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \left (5134 \cos \left (\frac{1}{2} (c-d x)\right )+4148 \cos \left (\frac{1}{2} (3 c+d x)\right )+4664 \cos \left (\frac{1}{2} (c+3 d x)\right )+2476 \cos \left (\frac{1}{2} (5 c+3 d x)\right )+3340 \cos \left (\frac{1}{2} (3 c+5 d x)\right )+944 \cos \left (\frac{1}{2} (7 c+5 d x)\right )+1620 \cos \left (\frac{1}{2} (5 c+7 d x)\right )+165 \cos \left (\frac{1}{2} (9 c+7 d x)\right )+357 \cos \left (\frac{1}{2} (7 c+9 d x)\right )\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right )}{96 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 i \sqrt{2} e^{-i (c+d x)} \left (119 \left (-1+e^{2 i c}\right ) \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )-55 \left (-1+e^{2 i c}\right ) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+119 \left (1+e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) d \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}\right )}{5 a^3 (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^3),x]

[Out]

(Cos[(c + d*x)/2]^6*(((4*I)*Sqrt[2]*(119*(1 + E^((2*I)*(c + d*x))) + 119*(-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))] - 55*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sq
rt[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))]) - ((5134*Cos[(c - d*x)/2] + 4148*Cos[(3*c + d*x)
/2] + 4664*Cos[(c + 3*d*x)/2] + 2476*Cos[(5*c + 3*d*x)/2] + 3340*Cos[(3*c + 5*d*x)/2] + 944*Cos[(7*c + 5*d*x)/
2] + 1620*Cos[(5*c + 7*d*x)/2] + 165*Cos[(9*c + 7*d*x)/2] + 357*Cos[(7*c + 9*d*x)/2])*Csc[c/2]*Sec[c/2]*Sec[(c
+ d*x)/2]^5)/(96*d*Cos[c + d*x]^(3/2))))/(5*a^3*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 4.527, size = 453, normalized size = 2.2 \begin{align*} \text{result too large to display} \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)^3,x)

[Out]

-1/4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^3*(118/5*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x
+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)+238/5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*s
in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1
/2*c),2^(1/2)))-4/3*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c
)^2-1/2)^2-128/5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1
/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+48*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*
cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)+1/5*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/c
os(1/2*d*x+1/2*c)^5+32/15*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)^3)/sin(1/2*d
*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [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))^3,x, algorithm="maxima")

[Out]

Timed out

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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^{3} \cos \left (d x + c\right )^{6} + 3 \, a^{3} \cos \left (d x + c\right )^{5} + 3 \, a^{3} \cos \left (d x + c\right )^{4} + a^{3} \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))^3,x, algorithm="fricas")

[Out]

integral(sqrt(cos(d*x + c))/(a^3*cos(d*x + c)^6 + 3*a^3*cos(d*x + c)^5 + 3*a^3*cos(d*x + c)^4 + a^3*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))**3,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 )}^{3} \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))^3,x, algorithm="giac")

[Out]

integrate(1/((a*cos(d*x + c) + a)^3*cos(d*x + c)^(5/2)), x)