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

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Rubi [A]  time = 0.0522945, antiderivative size = 43, normalized size of antiderivative = 1., 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 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])

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]]

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*}

Mathematica [A]  time = 0.0958876, 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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Maple [A]  time = 0.138, size = 50, normalized size = 1.2 \begin{align*} -{\frac{ \left ( 8\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-2 \right ) \sin \left ( bx+a \right ) }{3\,b \left ( \cos \left ( bx+a \right ) \right ) ^{3}} \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)^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 = 0.945544, size = 49, normalized size = 1.14 \begin{align*} -\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} \end{align*}

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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Fricas [A]  time = 1.70958, size = 131, normalized size = 3.05 \begin{align*} -\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 )} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{4}{\left (a + b x \right )}}{\left (d \tan{\left (a + b x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

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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Giac [A]  time = 1.34156, size = 57, normalized size = 1.33 \begin{align*} \frac{2 \,{\left (\sqrt{d \tan \left (b x + a\right )} d \tan \left (b x + a\right ) - \frac{3 \, d^{2}}{\sqrt{d \tan \left (b x + a\right )}}\right )}}{3 \, b d^{3}} \end{align*}

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