3.259 \(\int \frac {\sec ^4(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=43 \[ \frac {2 (d \tan (a+b x))^{3/2}}{3 b d^3}-\frac {2}{b d \sqrt {d \tan (a+b x)}} \]

[Out]

-2/b/d/(d*tan(b*x+a))^(1/2)+2/3*(d*tan(b*x+a))^(3/2)/b/d^3

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Rubi [A]  time = 0.05, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2607, 14} \[ \frac {2 (d \tan (a+b x))^{3/2}}{3 b d^3}-\frac {2}{b d \sqrt {d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^4/(d*Tan[a + b*x])^(3/2),x]

[Out]

-2/(b*d*Sqrt[d*Tan[a + b*x]]) + (2*(d*Tan[a + b*x])^(3/2))/(3*b*d^3)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rubi steps

\begin {align*} \int \frac {\sec ^4(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1+x^2}{(d x)^{3/2}} \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{(d x)^{3/2}}+\frac {\sqrt {d x}}{d^2}\right ) \, dx,x,\tan (a+b x)\right )}{b}\\ &=-\frac {2}{b d \sqrt {d \tan (a+b x)}}+\frac {2 (d \tan (a+b x))^{3/2}}{3 b d^3}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 32, normalized size = 0.74 \[ \frac {2 \left (\tan ^2(a+b x)-3\right )}{3 b d \sqrt {d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^4/(d*Tan[a + b*x])^(3/2),x]

[Out]

(2*(-3 + Tan[a + b*x]^2))/(3*b*d*Sqrt[d*Tan[a + b*x]])

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fricas [A]  time = 0.53, size = 54, normalized size = 1.26 \[ -\frac {2 \, {\left (4 \, \cos \left (b x + a\right )^{2} - 1\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{3 \, b d^{2} \cos \left (b x + a\right ) \sin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4/(d*tan(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

-2/3*(4*cos(b*x + a)^2 - 1)*sqrt(d*sin(b*x + a)/cos(b*x + a))/(b*d^2*cos(b*x + a)*sin(b*x + a))

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giac [A]  time = 2.22, size = 44, normalized size = 1.02 \[ \frac {2 \, {\left (\frac {\sqrt {d \tan \left (b x + a\right )} \tan \left (b x + a\right )}{b d} - \frac {3}{\sqrt {d \tan \left (b x + a\right )} b}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4/(d*tan(b*x+a))^(3/2),x, algorithm="giac")

[Out]

2/3*(sqrt(d*tan(b*x + a))*tan(b*x + a)/(b*d) - 3/(sqrt(d*tan(b*x + a))*b))/d

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maple [A]  time = 0.60, size = 50, normalized size = 1.16 \[ -\frac {2 \left (4 \left (\cos ^{2}\left (b x +a \right )\right )-1\right ) \sin \left (b x +a \right )}{3 b \cos \left (b x +a \right )^{3} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^4/(d*tan(b*x+a))^(3/2),x)

[Out]

-2/3/b*(4*cos(b*x+a)^2-1)*sin(b*x+a)/cos(b*x+a)^3/(d*sin(b*x+a)/cos(b*x+a))^(3/2)

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maxima [A]  time = 0.72, size = 36, normalized size = 0.84 \[ -\frac {2 \, {\left (\frac {3}{\sqrt {d \tan \left (b x + a\right )}} - \frac {\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}}{d^{2}}\right )}}{3 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4/(d*tan(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

-2/3*(3/sqrt(d*tan(b*x + a)) - (d*tan(b*x + a))^(3/2)/d^2)/(b*d)

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mupad [B]  time = 2.97, size = 64, normalized size = 1.49 \[ -\frac {4\,\left (\sin \left (2\,a+2\,b\,x\right )+\sin \left (4\,a+4\,b\,x\right )\right )\,\sqrt {\frac {d\,\sin \left (2\,a+2\,b\,x\right )}{\cos \left (2\,a+2\,b\,x\right )+1}}}{3\,b\,d^2\,{\sin \left (2\,a+2\,b\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(a + b*x)^4*(d*tan(a + b*x))^(3/2)),x)

[Out]

-(4*(sin(2*a + 2*b*x) + sin(4*a + 4*b*x))*((d*sin(2*a + 2*b*x))/(cos(2*a + 2*b*x) + 1))^(1/2))/(3*b*d^2*sin(2*
a + 2*b*x)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{4}{\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**4/(d*tan(b*x+a))**(3/2),x)

[Out]

Integral(sec(a + b*x)**4/(d*tan(a + b*x))**(3/2), x)

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