3.3.48 \(\int \sec ^4(e+f x) (d \tan (e+f x))^{5/2} \, dx\) [248]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 45, 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 = {2687, 14} \begin {gather*} \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} \end {gather*}

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 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 2687

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 \sec ^4(e+f x) (d \tan (e+f x))^{5/2} \, dx &=\frac {\text {Subst}\left (\int (d x)^{5/2} \left (1+x^2\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\text {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*}

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Mathematica [A]
time = 0.30, size = 42, normalized size = 0.93 \begin {gather*} \frac {2 (9+2 \cos (2 (e+f x))) \sec ^2(e+f x) (d \tan (e+f x))^{7/2}}{77 d f} \end {gather*}

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 = 0.32, size = 50, normalized size = 1.11

method	result	size
default	\(\frac {2 \left (4 \left (\cos ^{2}\left (f x +e \right )\right )+7\right ) \left (\frac {d \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \sin \left (f x +e \right )}{77 f \cos \left (f x +e \right )^{3}}\)	\(50\)

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,method=_RETURNVERBOSE)

[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 = 0.28, size = 38, normalized size = 0.84 \begin {gather*} \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 {gather*}

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 = 0.46, size = 75, normalized size = 1.67 \begin {gather*} -\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 {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 3878 deep

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Giac [A]
time = 0.52, size = 59, normalized size = 1.31 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 7.22, size = 352, normalized size = 7.82 \begin {gather*} \frac {d^2\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,8{}\mathrm {i}}{77\,f}+\frac {d^2\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,8{}\mathrm {i}}{77\,f\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}-\frac {d^2\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,296{}\mathrm {i}}{77\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {d^2\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,944{}\mathrm {i}}{77\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {d^2\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,160{}\mathrm {i}}{11\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^4}+\frac {d^2\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{11\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d^2*(-(d*(exp(e*2i + f*x*2i)*1i - 1i))/(exp(e*2i + f*x*2i) + 1))^(1/2)*8i)/(77*f) + (d^2*(-(d*(exp(e*2i + f*x
*2i)*1i - 1i))/(exp(e*2i + f*x*2i) + 1))^(1/2)*8i)/(77*f*(exp(e*2i + f*x*2i) + 1)) - (d^2*(-(d*(exp(e*2i + f*x
*2i)*1i - 1i))/(exp(e*2i + f*x*2i) + 1))^(1/2)*296i)/(77*f*(exp(e*2i + f*x*2i) + 1)^2) + (d^2*(-(d*(exp(e*2i +
f*x*2i)*1i - 1i))/(exp(e*2i + f*x*2i) + 1))^(1/2)*944i)/(77*f*(exp(e*2i + f*x*2i) + 1)^3) - (d^2*(-(d*(exp(e*
2i + f*x*2i)*1i - 1i))/(exp(e*2i + f*x*2i) + 1))^(1/2)*160i)/(11*f*(exp(e*2i + f*x*2i) + 1)^4) + (d^2*(-(d*(ex
p(e*2i + f*x*2i)*1i - 1i))/(exp(e*2i + f*x*2i) + 1))^(1/2)*64i)/(11*f*(exp(e*2i + f*x*2i) + 1)^5)

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