3.263 \(\int \frac {\sec ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=138 \[ \frac {24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac {12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac {24 \cos (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {d \tan (a+b x)}}{5 b d^2 \sqrt {\sin (2 a+2 b x)}}-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}} \]

[Out]

-2*sec(b*x+a)^3/b/d/(d*tan(b*x+a))^(1/2)+24/5*cos(b*x+a)*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*Ellipti
cE(cos(a+1/4*Pi+b*x),2^(1/2))*(d*tan(b*x+a))^(1/2)/b/d^2/sin(2*b*x+2*a)^(1/2)+24/5*cos(b*x+a)*(d*tan(b*x+a))^(
3/2)/b/d^3+12/5*sec(b*x+a)*(d*tan(b*x+a))^(3/2)/b/d^3

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Rubi [A]  time = 0.17, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2608, 2613, 2615, 2572, 2639} \[ \frac {24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac {12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac {24 \cos (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {d \tan (a+b x)}}{5 b d^2 \sqrt {\sin (2 a+2 b x)}}-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sec[a + b*x]^3)/(b*d*Sqrt[d*Tan[a + b*x]]) - (24*Cos[a + b*x]*EllipticE[a - Pi/4 + b*x, 2]*Sqrt[d*Tan[a +
b*x]])/(5*b*d^2*Sqrt[Sin[2*a + 2*b*x]]) + (24*Cos[a + b*x]*(d*Tan[a + b*x])^(3/2))/(5*b*d^3) + (12*Sec[a + b*x
]*(d*Tan[a + b*x])^(3/2))/(5*b*d^3)

Rule 2572

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(Sqrt[a*Sin[e +
f*x]]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[2*e + 2*f*x]], Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
x]

Rule 2608

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a^2*(a*Sec[
e + f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 1))/(b*f*(n + 1)), x] - Dist[(a^2*(m - 2))/(b^2*(n + 1)), Int[(a*Sec[e
+ f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && LtQ[n, -1] && (GtQ[m, 1] || (Eq
Q[m, 1] && EqQ[n, -3/2])) && IntegersQ[2*m, 2*n]

Rule 2613

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a^2*(a*Sec[
e + f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(a^2*(m - 2))/(m + n - 1), Int[(a*Sec
[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[
n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 2615

Int[Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]]/sec[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[(Sqrt[Cos[e + f*x]]*Sqrt[b*
Tan[e + f*x]])/Sqrt[Sin[e + f*x]], Int[Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]], x], x] /; FreeQ[{b, e, f}, 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 {\sec ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx &=-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}+\frac {6 \int \sec ^3(a+b x) \sqrt {d \tan (a+b x)} \, dx}{d^2}\\ &=-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}+\frac {12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac {12 \int \sec (a+b x) \sqrt {d \tan (a+b x)} \, dx}{5 d^2}\\ &=-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}+\frac {24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac {12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac {24 \int \cos (a+b x) \sqrt {d \tan (a+b x)} \, dx}{5 d^2}\\ &=-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}+\frac {24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac {12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac {\left (24 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}\right ) \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)} \, dx}{5 d^2 \sqrt {\sin (a+b x)}}\\ &=-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}+\frac {24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac {12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac {\left (24 \cos (a+b x) \sqrt {d \tan (a+b x)}\right ) \int \sqrt {\sin (2 a+2 b x)} \, dx}{5 d^2 \sqrt {\sin (2 a+2 b x)}}\\ &=-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}-\frac {24 \cos (a+b x) E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {d \tan (a+b x)}}{5 b d^2 \sqrt {\sin (2 a+2 b x)}}+\frac {24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac {12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}\\ \end {align*}

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Mathematica [C]  time = 0.90, size = 104, normalized size = 0.75 \[ \frac {2 \csc (a+b x) \sqrt {d \tan (a+b x)} \left (\sqrt {\sec ^2(a+b x)} \left (12 \sin ^2(a+b x)+\tan ^2(a+b x)-5\right )-8 \tan ^2(a+b x) \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\tan ^2(a+b x)\right )\right )}{5 b d^2 \sqrt {\sec ^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Csc[a + b*x]*Sqrt[d*Tan[a + b*x]]*(-8*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Tan[a + b*x]^2 + Sq
rt[Sec[a + b*x]^2]*(-5 + 12*Sin[a + b*x]^2 + Tan[a + b*x]^2)))/(5*b*d^2*Sqrt[Sec[a + b*x]^2])

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \tan \left (b x + a\right )} \sec \left (b x + a\right )^{5}}{d^{2} \tan \left (b x + a\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*tan(b*x + a))*sec(b*x + a)^5/(d^2*tan(b*x + a)^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (b x + a\right )^{5}}{\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5/(d*tan(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(b*x + a)^5/(d*tan(b*x + a))^(3/2), x)

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maple [B]  time = 0.66, size = 537, normalized size = 3.89 \[ -\frac {\left (-24 \left (\cos ^{3}\left (b x +a \right )\right ) \EllipticE \left (\sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}+12 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticF \left (\sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )-24 \left (\cos ^{2}\left (b x +a \right )\right ) \EllipticE \left (\sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}+12 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticF \left (\sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )+12 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}-6 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}-\sqrt {2}\right ) \sin \left (b x +a \right ) \sqrt {2}}{5 b \cos \left (b x +a \right )^{4} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^5/(d*tan(b*x+a))^(3/2),x)

[Out]

-1/5/b*(-24*cos(b*x+a)^3*EllipticE((-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*(-(-sin(b*x+a)-
1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/
2)+12*cos(b*x+a)^3*(-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2
)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticF((-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-24*
cos(b*x+a)^2*EllipticE((-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*(-(-sin(b*x+a)-1+cos(b*x+a)
)/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)+12*cos(b*
x+a)^2*(-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b
*x+a))/sin(b*x+a))^(1/2)*EllipticF((-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+12*cos(b*x+a)^3
*2^(1/2)-6*cos(b*x+a)^2*2^(1/2)-2^(1/2))*sin(b*x+a)/cos(b*x+a)^4/(d*sin(b*x+a)/cos(b*x+a))^(3/2)*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (b x + a\right )^{5}}{\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5/(d*tan(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(b*x + a)^5/(d*tan(b*x + a))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (a+b\,x\right )}^5\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(a + b*x)^5*(d*tan(a + b*x))^(3/2)),x)

[Out]

int(1/(cos(a + b*x)^5*(d*tan(a + b*x))^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{5}{\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**5/(d*tan(b*x+a))**(3/2),x)

[Out]

Integral(sec(a + b*x)**5/(d*tan(a + b*x))**(3/2), x)

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