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

Optimal. Leaf size=45 \[ \frac{2 (d \tan (a+b x))^{9/2}}{9 b d^3}+\frac{2 (d \tan (a+b x))^{5/2}}{5 b d} \]

[Out]

(2*(d*Tan[a + b*x])^(5/2))/(5*b*d) + (2*(d*Tan[a + b*x])^(9/2))/(9*b*d^3)

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Rubi [A]  time = 0.0526398, antiderivative size = 45, 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} \[ \frac{2 (d \tan (a+b x))^{9/2}}{9 b d^3}+\frac{2 (d \tan (a+b x))^{5/2}}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d*Tan[a + b*x])^(5/2))/(5*b*d) + (2*(d*Tan[a + b*x])^(9/2))/(9*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 \sec ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int (d x)^{3/2} \left (1+x^2\right ) \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left ((d x)^{3/2}+\frac{(d x)^{7/2}}{d^2}\right ) \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac{2 (d \tan (a+b x))^{5/2}}{5 b d}+\frac{2 (d \tan (a+b x))^{9/2}}{9 b d^3}\\ \end{align*}

Mathematica [A]  time = 0.133345, size = 42, normalized size = 0.93 \[ \frac{2 d \left (5 \sec ^4(a+b x)-\sec ^2(a+b x)-4\right ) \sqrt{d \tan (a+b x)}}{45 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(-4 - Sec[a + b*x]^2 + 5*Sec[a + b*x]^4)*Sqrt[d*Tan[a + b*x]])/(45*b)

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Maple [A]  time = 0.24, size = 50, normalized size = 1.1 \begin{align*}{\frac{ \left ( 8\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}+10 \right ) \sin \left ( bx+a \right ) }{45\,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]

2/45/b*(4*cos(b*x+a)^2+5)*(d*sin(b*x+a)/cos(b*x+a))^(3/2)*sin(b*x+a)/cos(b*x+a)^3

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Maxima [A]  time = 0.941261, size = 49, normalized size = 1.09 \begin{align*} \frac{2 \,{\left (5 \, \left (d \tan \left (b x + a\right )\right )^{\frac{9}{2}} + 9 \, \left (d \tan \left (b x + a\right )\right )^{\frac{5}{2}} d^{2}\right )}}{45 \, 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="maxima")

[Out]

2/45*(5*(d*tan(b*x + a))^(9/2) + 9*(d*tan(b*x + a))^(5/2)*d^2)/(b*d^3)

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Fricas [A]  time = 1.89385, size = 143, normalized size = 3.18 \begin{align*} -\frac{2 \,{\left (4 \, d \cos \left (b x + a\right )^{4} + d \cos \left (b x + a\right )^{2} - 5 \, d\right )} \sqrt{\frac{d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{45 \, b \cos \left (b x + a\right )^{4}} \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]

-2/45*(4*d*cos(b*x + a)^4 + d*cos(b*x + a)^2 - 5*d)*sqrt(d*sin(b*x + a)/cos(b*x + a))/(b*cos(b*x + a)^4)

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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)**4*(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 )^{4}\,{d x} \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]

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