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

Optimal. Leaf size=247 \[ -\frac{21 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{2 a^3 d}-\frac{63 \sin (c+d x)}{10 d \sec ^{\frac{3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}+\frac{77 \sin (c+d x)}{10 a^3 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{21 \sin (c+d x)}{2 a^3 d \sqrt{\sec (c+d x)}}+\frac{231 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{4 \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^2}-\frac{\sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^3} \]

[Out]

(231*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) - (21*Sqrt[Cos[c + d*x]]*Elli
pticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(2*a^3*d) + (77*Sin[c + d*x])/(10*a^3*d*Sec[c + d*x]^(3/2)) - (21*Si
n[c + d*x])/(2*a^3*d*Sqrt[Sec[c + d*x]]) - Sin[c + d*x]/(5*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^3) - (4*S
in[c + d*x])/(5*a*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2) - (63*Sin[c + d*x])/(10*d*Sec[c + d*x]^(3/2)*(a
^3 + a^3*Sec[c + d*x]))

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Rubi [A]  time = 0.373765, antiderivative size = 247, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3817, 4020, 3787, 3769, 3771, 2639, 2641} \[ -\frac{63 \sin (c+d x)}{10 d \sec ^{\frac{3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}+\frac{77 \sin (c+d x)}{10 a^3 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{21 \sin (c+d x)}{2 a^3 d \sqrt{\sec (c+d x)}}-\frac{21 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{2 a^3 d}+\frac{231 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{4 \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^2}-\frac{\sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(231*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) - (21*Sqrt[Cos[c + d*x]]*Elli
pticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(2*a^3*d) + (77*Sin[c + d*x])/(10*a^3*d*Sec[c + d*x]^(3/2)) - (21*Si
n[c + d*x])/(2*a^3*d*Sqrt[Sec[c + d*x]]) - Sin[c + d*x]/(5*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^3) - (4*S
in[c + d*x])/(5*a*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2) - (63*Sin[c + d*x])/(10*d*Sec[c + d*x]^(3/2)*(a
^3 + a^3*Sec[c + d*x]))

Rule 3817

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[
e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*(2*m + 1)), x] + Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc
[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a*(2*m + n + 1) - b*(m + n + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d
, e, f, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m])

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3769

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

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

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 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]

Rubi steps

\begin{align*} \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx &=-\frac{\sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac{\int \frac{-\frac{15 a}{2}+\frac{9}{2} a \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{\sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac{4 \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{\int \frac{-\frac{105 a^2}{2}+42 a^2 \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{15 a^4}\\ &=-\frac{\sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac{4 \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{63 \sin (c+d x)}{10 d \sec ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right )}-\frac{\int \frac{-\frac{1155 a^3}{4}+\frac{945}{4} a^3 \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{15 a^6}\\ &=-\frac{\sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac{4 \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{63 \sin (c+d x)}{10 d \sec ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right )}-\frac{63 \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{4 a^3}+\frac{77 \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{4 a^3}\\ &=\frac{77 \sin (c+d x)}{10 a^3 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{21 \sin (c+d x)}{2 a^3 d \sqrt{\sec (c+d x)}}-\frac{\sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac{4 \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{63 \sin (c+d x)}{10 d \sec ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right )}-\frac{21 \int \sqrt{\sec (c+d x)} \, dx}{4 a^3}+\frac{231 \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{20 a^3}\\ &=\frac{77 \sin (c+d x)}{10 a^3 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{21 \sin (c+d x)}{2 a^3 d \sqrt{\sec (c+d x)}}-\frac{\sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac{4 \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{63 \sin (c+d x)}{10 d \sec ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right )}-\frac{\left (21 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{4 a^3}+\frac{\left (231 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}\\ &=\frac{231 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{10 a^3 d}-\frac{21 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{2 a^3 d}+\frac{77 \sin (c+d x)}{10 a^3 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{21 \sin (c+d x)}{2 a^3 d \sqrt{\sec (c+d x)}}-\frac{\sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac{4 \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{63 \sin (c+d x)}{10 d \sec ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 2.66608, size = 297, normalized size = 1.2 \[ -\frac{e^{-i d x} \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{7}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (77 i e^{-\frac{3}{2} i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \left (1+e^{i (c+d x)}\right )^5 \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+3360 \cos ^5\left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+125 \sin \left (\frac{1}{2} (c+d x)\right )+359 \sin \left (\frac{3}{2} (c+d x)\right )+350 \sin \left (\frac{5}{2} (c+d x)\right )+138 \sin \left (\frac{7}{2} (c+d x)\right )+5 \sin \left (\frac{9}{2} (c+d x)\right )-\sin \left (\frac{11}{2} (c+d x)\right )-3465 i \cos \left (\frac{1}{2} (c+d x)\right )-2541 i \cos \left (\frac{3}{2} (c+d x)\right )-1155 i \cos \left (\frac{5}{2} (c+d x)\right )-231 i \cos \left (\frac{7}{2} (c+d x)\right )\right )}{40 a^3 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[(c + d*x)/2]*Sec[c + d*x]^(7/2)*(Cos[d*x] + I*Sin[d*x])*((-3465*I)*Cos[(c + d*x)/2] - (2541*I)*Cos[(3*(c
+ d*x))/2] - (1155*I)*Cos[(5*(c + d*x))/2] - (231*I)*Cos[(7*(c + d*x))/2] + 3360*Cos[(c + d*x)/2]^5*Sqrt[Cos[
c + d*x]]*EllipticF[(c + d*x)/2, 2] + ((77*I)*(1 + E^(I*(c + d*x)))^5*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeome
tric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(((3*I)/2)*(c + d*x)) + 125*Sin[(c + d*x)/2] + 359*Sin[(3*(c +
d*x))/2] + 350*Sin[(5*(c + d*x))/2] + 138*Sin[(7*(c + d*x))/2] + 5*Sin[(9*(c + d*x))/2] - Sin[(11*(c + d*x))/
2]))/(40*a^3*d*E^(I*d*x)*(1 + Sec[c + d*x])^3)

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Maple [A]  time = 1.65, size = 296, normalized size = 1.2 \begin{align*} -{\frac{1}{20\,{a}^{3}d}\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}} \left ( 64\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{12}-288\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{10}-76\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}-210\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}-462\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +530\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-248\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+19\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ){\frac{1}{\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 ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-5} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \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/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x)

[Out]

-1/20/a^3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(64*cos(1/2*d*x+1/2*c)^12-288*cos(1/2*d*x+1/
2*c)^10-76*cos(1/2*d*x+1/2*c)^8-210*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(c
os(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5-462*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(
1/2)*cos(1/2*d*x+1/2*c)^5*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+530*cos(1/2*d*x+1/2*c)^6-248*cos(1/2*d*x+1/2*c
)^4+19*cos(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)/cos(1/2*d*x+1/2*c)^5/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/sec(d*x+c)^(5/2)/(a+a*sec(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{\sec \left (d x + c\right )}}{a^{3} \sec \left (d x + c\right )^{6} + 3 \, a^{3} \sec \left (d x + c\right )^{5} + 3 \, a^{3} \sec \left (d x + c\right )^{4} + a^{3} \sec \left (d x + c\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

integral(sqrt(sec(d*x + c))/(a^3*sec(d*x + c)^6 + 3*a^3*sec(d*x + c)^5 + 3*a^3*sec(d*x + c)^4 + a^3*sec(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/sec(d*x+c)**(5/2)/(a+a*sec(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 \sec \left (d x + c\right ) + a\right )}^{3} \sec \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/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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