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

Optimal. Leaf size=136 \[ -\frac{4 d^2 \sqrt{\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{77 b \sqrt{d \tan (a+b x)}}+\frac{2 d \sec ^5(a+b x) \sqrt{d \tan (a+b x)}}{11 b}-\frac{2 d \sec ^3(a+b x) \sqrt{d \tan (a+b x)}}{77 b}-\frac{4 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{77 b} \]

[Out]

(-4*d^2*EllipticF[a - Pi/4 + b*x, 2]*Sec[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(77*b*Sqrt[d*Tan[a + b*x]]) - (4*d*S
ec[a + b*x]*Sqrt[d*Tan[a + b*x]])/(77*b) - (2*d*Sec[a + b*x]^3*Sqrt[d*Tan[a + b*x]])/(77*b) + (2*d*Sec[a + b*x
]^5*Sqrt[d*Tan[a + b*x]])/(11*b)

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Rubi [A]  time = 0.183455, antiderivative size = 136, normalized size of antiderivative = 1., 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 = {2611, 2613, 2614, 2573, 2641} \[ -\frac{4 d^2 \sqrt{\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{77 b \sqrt{d \tan (a+b x)}}+\frac{2 d \sec ^5(a+b x) \sqrt{d \tan (a+b x)}}{11 b}-\frac{2 d \sec ^3(a+b x) \sqrt{d \tan (a+b x)}}{77 b}-\frac{4 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{77 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*d^2*EllipticF[a - Pi/4 + b*x, 2]*Sec[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(77*b*Sqrt[d*Tan[a + b*x]]) - (4*d*S
ec[a + b*x]*Sqrt[d*Tan[a + b*x]])/(77*b) - (2*d*Sec[a + b*x]^3*Sqrt[d*Tan[a + b*x]])/(77*b) + (2*d*Sec[a + b*x
]^5*Sqrt[d*Tan[a + b*x]])/(11*b)

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[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 2614

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x
]

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, 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 \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx &=\frac{2 d \sec ^5(a+b x) \sqrt{d \tan (a+b x)}}{11 b}-\frac{1}{11} d^2 \int \frac{\sec ^5(a+b x)}{\sqrt{d \tan (a+b x)}} \, dx\\ &=-\frac{2 d \sec ^3(a+b x) \sqrt{d \tan (a+b x)}}{77 b}+\frac{2 d \sec ^5(a+b x) \sqrt{d \tan (a+b x)}}{11 b}-\frac{1}{77} \left (6 d^2\right ) \int \frac{\sec ^3(a+b x)}{\sqrt{d \tan (a+b x)}} \, dx\\ &=-\frac{4 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{77 b}-\frac{2 d \sec ^3(a+b x) \sqrt{d \tan (a+b x)}}{77 b}+\frac{2 d \sec ^5(a+b x) \sqrt{d \tan (a+b x)}}{11 b}-\frac{1}{77} \left (4 d^2\right ) \int \frac{\sec (a+b x)}{\sqrt{d \tan (a+b x)}} \, dx\\ &=-\frac{4 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{77 b}-\frac{2 d \sec ^3(a+b x) \sqrt{d \tan (a+b x)}}{77 b}+\frac{2 d \sec ^5(a+b x) \sqrt{d \tan (a+b x)}}{11 b}-\frac{\left (4 d^2 \sqrt{\sin (a+b x)}\right ) \int \frac{1}{\sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)}} \, dx}{77 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}}\\ &=-\frac{4 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{77 b}-\frac{2 d \sec ^3(a+b x) \sqrt{d \tan (a+b x)}}{77 b}+\frac{2 d \sec ^5(a+b x) \sqrt{d \tan (a+b x)}}{11 b}-\frac{\left (4 d^2 \sec (a+b x) \sqrt{\sin (2 a+2 b x)}\right ) \int \frac{1}{\sqrt{\sin (2 a+2 b x)}} \, dx}{77 \sqrt{d \tan (a+b x)}}\\ &=-\frac{4 d^2 F\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sec (a+b x) \sqrt{\sin (2 a+2 b x)}}{77 b \sqrt{d \tan (a+b x)}}-\frac{4 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{77 b}-\frac{2 d \sec ^3(a+b x) \sqrt{d \tan (a+b x)}}{77 b}+\frac{2 d \sec ^5(a+b x) \sqrt{d \tan (a+b x)}}{11 b}\\ \end{align*}

Mathematica [C]  time = 0.799281, size = 90, normalized size = 0.66 \[ -\frac{d \sec ^5(a+b x) \sqrt{d \tan (a+b x)} \left (16 \cos ^6(a+b x) \sqrt{\sec ^2(a+b x)} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\tan ^2(a+b x)\right )+6 \cos (2 (a+b x))+\cos (4 (a+b x))-23\right )}{154 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*Sec[a + b*x]^5*(-23 + 6*Cos[2*(a + b*x)] + Cos[4*(a + b*x)] + 16*Cos[a + b*x]^6*Hypergeometric2F1[1/4, 1/2
, 5/4, -Tan[a + b*x]^2]*Sqrt[Sec[a + b*x]^2])*Sqrt[d*Tan[a + b*x]])/(154*b)

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Maple [A]  time = 0.135, size = 251, normalized size = 1.9 \begin{align*}{\frac{\sqrt{2} \left ( \cos \left ( bx+a \right ) -1 \right ) \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}{77\,b \left ( \cos \left ( bx+a \right ) \right ) ^{4} \left ( \sin \left ( bx+a \right ) \right ) ^{5}} \left ( 4\,\sin \left ( bx+a \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{\cos \left ( bx+a \right ) -1+\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}} \left ( \cos \left ( bx+a \right ) \right ) ^{5}{\it EllipticF} \left ( \sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},1/2\,\sqrt{2} \right ) -2\,\sqrt{2} \left ( \cos \left ( bx+a \right ) \right ) ^{5}+2\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}\sqrt{2}- \left ( \cos \left ( bx+a \right ) \right ) ^{3}\sqrt{2}+ \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sqrt{2}+7\,\cos \left ( bx+a \right ) \sqrt{2}-7\,\sqrt{2} \right ) \left ({\frac{d\sin \left ( bx+a \right ) }{\cos \left ( bx+a \right ) }} \right ) ^{{\frac{3}{2}}}} \end{align*}

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]

1/77/b*2^(1/2)*(cos(b*x+a)-1)*(4*sin(b*x+a)*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b
*x+a))^(1/2)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)*cos(b*x+a)^5*EllipticF((-(cos(b*x+a)-1-sin(b*x+a))/
sin(b*x+a))^(1/2),1/2*2^(1/2))-2*2^(1/2)*cos(b*x+a)^5+2*cos(b*x+a)^4*2^(1/2)-cos(b*x+a)^3*2^(1/2)+cos(b*x+a)^2
*2^(1/2)+7*cos(b*x+a)*2^(1/2)-7*2^(1/2))*(cos(b*x+a)+1)^2*(d*sin(b*x+a)/cos(b*x+a))^(3/2)/cos(b*x+a)^4/sin(b*x
+a)^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}} \sec \left (b x + a\right )^{5}\,{d x} \end{align*}

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{d \tan \left (b x + a\right )} d \sec \left (b x + a\right )^{5} \tan \left (b x + a\right ), x\right ) \end{align*}

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}} \sec \left (b x + a\right )^{5}\,{d x} \end{align*}

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