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

Optimal. Leaf size=80 \[ \frac {2 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b}-\frac {d^2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{3 b \sqrt {d \tan (a+b x)}} \]

[Out]

1/3*d^2*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))*sec(b*x+a)*sin(2*b*
x+2*a)^(1/2)/b/(d*tan(b*x+a))^(1/2)+2/3*d*sec(b*x+a)*(d*tan(b*x+a))^(1/2)/b

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Rubi [A]  time = 0.08, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2611, 2614, 2573, 2641} \[ \frac {2 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b}-\frac {d^2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{3 b \sqrt {d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 (a+b x) (d \tan (a+b x))^{3/2} \, dx &=\frac {2 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b}-\frac {1}{3} d^2 \int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\\ &=\frac {2 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b}-\frac {\left (d^2 \sqrt {\sin (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}} \, dx}{3 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=\frac {2 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b}-\frac {\left (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}{3 \sqrt {d \tan (a+b x)}}\\ &=-\frac {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)}}{3 b \sqrt {d \tan (a+b x)}}+\frac {2 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b}\\ \end {align*}

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Mathematica [C]  time = 0.31, size = 69, normalized size = 0.86 \[ \frac {2 d \cos (a+b x) \sqrt {d \tan (a+b x)} \left (\sec ^2(a+b x)-\sqrt {\sec ^2(a+b x)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\tan ^2(a+b x)\right )\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*Cos[a + b*x]*(Sec[a + b*x]^2 - Hypergeometric2F1[1/4, 1/2, 5/4, -Tan[a + b*x]^2]*Sqrt[Sec[a + b*x]^2])*Sq
rt[d*Tan[a + b*x]])/(3*b)

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d \tan \left (b x + a\right )} d \sec \left (b x + a\right ) \tan \left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*tan(b*x + a))*d*sec(b*x + a)*tan(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.42, size = 188, normalized size = 2.35 \[ \frac {\left (-1+\cos \left (b x +a \right )\right ) \left (\sin \left (b x +a \right ) \cos \left (b x +a \right ) \EllipticF \left (\sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}+\cos \left (b x +a \right ) \sqrt {2}-\sqrt {2}\right ) \left (\cos \left (b x +a \right )+1\right )^{2} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \sqrt {2}}{3 b \sin \left (b x +a \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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