3.227 \(\int \sec ^4(e+f x) \sqrt {d \tan (e+f x)} \, dx\)

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

[Out]

2/3*(d*tan(f*x+e))^(3/2)/d/f+2/7*(d*tan(f*x+e))^(7/2)/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 = {2607, 14} \[ \frac {2 (d \tan (e+f x))^{7/2}}{7 d^3 f}+\frac {2 (d \tan (e+f x))^{3/2}}{3 d f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^4*Sqrt[d*Tan[e + f*x]],x]

[Out]

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

Rubi steps

\begin {align*} \int \sec ^4(e+f x) \sqrt {d \tan (e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \sqrt {d x} \left (1+x^2\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \left (\sqrt {d x}+\frac {(d x)^{5/2}}{d^2}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {2 (d \tan (e+f x))^{3/2}}{3 d f}+\frac {2 (d \tan (e+f x))^{7/2}}{7 d^3 f}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 34, normalized size = 0.76 \[ \frac {2 \left (3 \sec ^2(e+f x)+4\right ) (d \tan (e+f x))^{3/2}}{21 d f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^4*Sqrt[d*Tan[e + f*x]],x]

[Out]

(2*(4 + 3*Sec[e + f*x]^2)*(d*Tan[e + f*x])^(3/2))/(21*d*f)

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fricas [A]  time = 0.63, size = 49, normalized size = 1.09 \[ \frac {2 \, {\left (4 \, \cos \left (f x + e\right )^{2} + 3\right )} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{21 \, f \cos \left (f x + e\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/21*(4*cos(f*x + e)^2 + 3)*sqrt(d*sin(f*x + e)/cos(f*x + e))*sin(f*x + e)/(f*cos(f*x + e)^3)

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giac [A]  time = 0.62, size = 57, normalized size = 1.27 \[ \frac {2 \, {\left (3 \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )^{3} + 7 \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )\right )}}{21 \, d^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

2/21*(3*sqrt(d*tan(f*x + e))*d^3*tan(f*x + e)^3 + 7*sqrt(d*tan(f*x + e))*d^3*tan(f*x + e))/(d^3*f)

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maple [A]  time = 0.61, size = 50, normalized size = 1.11 \[ \frac {2 \left (4 \left (\cos ^{2}\left (f x +e \right )\right )+3\right ) \sqrt {\frac {d \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )}{21 f \cos \left (f x +e \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^4*(d*tan(f*x+e))^(1/2),x)

[Out]

2/21/f*(4*cos(f*x+e)^2+3)*(d*sin(f*x+e)/cos(f*x+e))^(1/2)*sin(f*x+e)/cos(f*x+e)^3

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maxima [A]  time = 0.33, size = 36, normalized size = 0.80 \[ \frac {2 \, {\left (3 \, \left (d \tan \left (f x + e\right )\right )^{\frac {7}{2}} + 7 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d^{2}\right )}}{21 \, d^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

2/21*(3*(d*tan(f*x + e))^(7/2) + 7*(d*tan(f*x + e))^(3/2)*d^2)/(d^3*f)

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mupad [B]  time = 6.09, size = 218, normalized size = 4.84 \[ -\frac {\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,8{}\mathrm {i}}{21\,f}-\frac {\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,8{}\mathrm {i}}{21\,f\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}+\frac {\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,24{}\mathrm {i}}{7\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,16{}\mathrm {i}}{7\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(e + f*x))^(1/2)/cos(e + f*x)^4,x)

[Out]

((-(d*(exp(e*2i + f*x*2i)*1i - 1i))/(exp(e*2i + f*x*2i) + 1))^(1/2)*24i)/(7*f*(exp(e*2i + f*x*2i) + 1)^2) - ((
-(d*(exp(e*2i + f*x*2i)*1i - 1i))/(exp(e*2i + f*x*2i) + 1))^(1/2)*8i)/(21*f*(exp(e*2i + f*x*2i) + 1)) - ((-(d*
(exp(e*2i + f*x*2i)*1i - 1i))/(exp(e*2i + f*x*2i) + 1))^(1/2)*8i)/(21*f) - ((-(d*(exp(e*2i + f*x*2i)*1i - 1i))
/(exp(e*2i + f*x*2i) + 1))^(1/2)*16i)/(7*f*(exp(e*2i + f*x*2i) + 1)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {d \tan {\left (e + f x \right )}} \sec ^{4}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**4*(d*tan(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(d*tan(e + f*x))*sec(e + f*x)**4, x)

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