3.248 \(\int \sec ^4(e+f x) (d \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=45 \[ \frac{2 (d \tan (e+f x))^{11/2}}{11 d^3 f}+\frac{2 (d \tan (e+f x))^{7/2}}{7 d f} \]

[Out]

(2*(d*Tan[e + f*x])^(7/2))/(7*d*f) + (2*(d*Tan[e + f*x])^(11/2))/(11*d^3*f)

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Rubi [A]  time = 0.0503122, 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} \[ \frac{2 (d \tan (e+f x))^{11/2}}{11 d^3 f}+\frac{2 (d \tan (e+f x))^{7/2}}{7 d f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^4*(d*Tan[e + f*x])^(5/2),x]

[Out]

(2*(d*Tan[e + f*x])^(7/2))/(7*d*f) + (2*(d*Tan[e + f*x])^(11/2))/(11*d^3*f)

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(e+f x) (d \tan (e+f x))^{5/2} \, dx &=\frac{\operatorname{Subst}\left (\int (d x)^{5/2} \left (1+x^2\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left ((d x)^{5/2}+\frac{(d x)^{9/2}}{d^2}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{2 (d \tan (e+f x))^{7/2}}{7 d f}+\frac{2 (d \tan (e+f x))^{11/2}}{11 d^3 f}\\ \end{align*}

Mathematica [A]  time = 0.262581, size = 42, normalized size = 0.93 \[ \frac{2 (2 \cos (2 (e+f x))+9) \sec ^2(e+f x) (d \tan (e+f x))^{7/2}}{77 d f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^4*(d*Tan[e + f*x])^(5/2),x]

[Out]

(2*(9 + 2*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(d*Tan[e + f*x])^(7/2))/(77*d*f)

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Maple [A]  time = 0.139, size = 50, normalized size = 1.1 \begin{align*}{\frac{ \left ( 8\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+14 \right ) \sin \left ( fx+e \right ) }{77\,f \left ( \cos \left ( fx+e \right ) \right ) ^{3}} \left ({\frac{d\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^4*(d*tan(f*x+e))^(5/2),x)

[Out]

2/77/f*(4*cos(f*x+e)^2+7)*(d*sin(f*x+e)/cos(f*x+e))^(5/2)*sin(f*x+e)/cos(f*x+e)^3

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Maxima [A]  time = 0.978024, size = 49, normalized size = 1.09 \begin{align*} \frac{2 \,{\left (7 \, \left (d \tan \left (f x + e\right )\right )^{\frac{11}{2}} + 11 \, \left (d \tan \left (f x + e\right )\right )^{\frac{7}{2}} d^{2}\right )}}{77 \, d^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

2/77*(7*(d*tan(f*x + e))^(11/2) + 11*(d*tan(f*x + e))^(7/2)*d^2)/(d^3*f)

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Fricas [A]  time = 1.92287, size = 171, normalized size = 3.8 \begin{align*} -\frac{2 \,{\left (4 \, d^{2} \cos \left (f x + e\right )^{4} + 3 \, d^{2} \cos \left (f x + e\right )^{2} - 7 \, d^{2}\right )} \sqrt{\frac{d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{77 \, f \cos \left (f x + e\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-2/77*(4*d^2*cos(f*x + e)^4 + 3*d^2*cos(f*x + e)^2 - 7*d^2)*sqrt(d*sin(f*x + e)/cos(f*x + e))*sin(f*x + e)/(f*
cos(f*x + e)^5)

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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(f*x+e)**4*(d*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.21743, size = 80, normalized size = 1.78 \begin{align*} \frac{2 \,{\left (7 \, \sqrt{d \tan \left (f x + e\right )} d^{5} \tan \left (f x + e\right )^{5} + 11 \, \sqrt{d \tan \left (f x + e\right )} d^{5} \tan \left (f x + e\right )^{3}\right )}}{77 \, d^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

2/77*(7*sqrt(d*tan(f*x + e))*d^5*tan(f*x + e)^5 + 11*sqrt(d*tan(f*x + e))*d^5*tan(f*x + e)^3)/(d^3*f)