3.247 \(\int \sec ^6(e+f x) (d \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=67 \[ \frac{2 (d \tan (e+f x))^{15/2}}{15 d^5 f}+\frac{4 (d \tan (e+f x))^{11/2}}{11 d^3 f}+\frac{2 (d \tan (e+f x))^{7/2}}{7 d f} \]

[Out]

(2*(d*Tan[e + f*x])^(7/2))/(7*d*f) + (4*(d*Tan[e + f*x])^(11/2))/(11*d^3*f) + (2*(d*Tan[e + f*x])^(15/2))/(15*
d^5*f)

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Rubi [A]  time = 0.0584089, antiderivative size = 67, 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, 270} \[ \frac{2 (d \tan (e+f x))^{15/2}}{15 d^5 f}+\frac{4 (d \tan (e+f x))^{11/2}}{11 d^3 f}+\frac{2 (d \tan (e+f x))^{7/2}}{7 d f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^6*(d*Tan[e + f*x])^(5/2),x]

[Out]

(2*(d*Tan[e + f*x])^(7/2))/(7*d*f) + (4*(d*Tan[e + f*x])^(11/2))/(11*d^3*f) + (2*(d*Tan[e + f*x])^(15/2))/(15*
d^5*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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \sec ^6(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 )^2 \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left ((d x)^{5/2}+\frac{2 (d x)^{9/2}}{d^2}+\frac{(d x)^{13/2}}{d^4}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{2 (d \tan (e+f x))^{7/2}}{7 d f}+\frac{4 (d \tan (e+f x))^{11/2}}{11 d^3 f}+\frac{2 (d \tan (e+f x))^{15/2}}{15 d^5 f}\\ \end{align*}

Mathematica [A]  time = 0.430815, size = 52, normalized size = 0.78 \[ \frac{2 (44 \cos (2 (e+f x))+4 \cos (4 (e+f x))+117) \sec ^4(e+f x) (d \tan (e+f x))^{7/2}}{1155 d f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^6*(d*Tan[e + f*x])^(5/2),x]

[Out]

(2*(117 + 44*Cos[2*(e + f*x)] + 4*Cos[4*(e + f*x)])*Sec[e + f*x]^4*(d*Tan[e + f*x])^(7/2))/(1155*d*f)

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Maple [A]  time = 0.181, size = 60, normalized size = 0.9 \begin{align*}{\frac{ \left ( 64\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+112\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+154 \right ) \sin \left ( fx+e \right ) }{1155\,f \left ( \cos \left ( fx+e \right ) \right ) ^{5}} \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)^6*(d*tan(f*x+e))^(5/2),x)

[Out]

2/1155/f*(32*cos(f*x+e)^4+56*cos(f*x+e)^2+77)*(d*sin(f*x+e)/cos(f*x+e))^(5/2)*sin(f*x+e)/cos(f*x+e)^5

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Maxima [A]  time = 0.947682, size = 69, normalized size = 1.03 \begin{align*} \frac{2 \,{\left (77 \, \left (d \tan \left (f x + e\right )\right )^{\frac{15}{2}} + 210 \, \left (d \tan \left (f x + e\right )\right )^{\frac{11}{2}} d^{2} + 165 \, \left (d \tan \left (f x + e\right )\right )^{\frac{7}{2}} d^{4}\right )}}{1155 \, d^{5} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

2/1155*(77*(d*tan(f*x + e))^(15/2) + 210*(d*tan(f*x + e))^(11/2)*d^2 + 165*(d*tan(f*x + e))^(7/2)*d^4)/(d^5*f)

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Fricas [A]  time = 2.35071, size = 211, normalized size = 3.15 \begin{align*} -\frac{2 \,{\left (32 \, d^{2} \cos \left (f x + e\right )^{6} + 24 \, d^{2} \cos \left (f x + e\right )^{4} + 21 \, d^{2} \cos \left (f x + e\right )^{2} - 77 \, d^{2}\right )} \sqrt{\frac{d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{1155 \, f \cos \left (f x + e\right )^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-2/1155*(32*d^2*cos(f*x + e)^6 + 24*d^2*cos(f*x + e)^4 + 21*d^2*cos(f*x + e)^2 - 77*d^2)*sqrt(d*sin(f*x + e)/c
os(f*x + e))*sin(f*x + e)/(f*cos(f*x + e)^7)

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

[Out]

Timed out

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Giac [A]  time = 1.32158, size = 113, normalized size = 1.69 \begin{align*} \frac{2 \,{\left (77 \, \sqrt{d \tan \left (f x + e\right )} d^{7} \tan \left (f x + e\right )^{7} + 210 \, \sqrt{d \tan \left (f x + e\right )} d^{7} \tan \left (f x + e\right )^{5} + 165 \, \sqrt{d \tan \left (f x + e\right )} d^{7} \tan \left (f x + e\right )^{3}\right )}}{1155 \, d^{5} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

2/1155*(77*sqrt(d*tan(f*x + e))*d^7*tan(f*x + e)^7 + 210*sqrt(d*tan(f*x + e))*d^7*tan(f*x + e)^5 + 165*sqrt(d*
tan(f*x + e))*d^7*tan(f*x + e)^3)/(d^5*f)