3.3.26 \(\int \sec ^6(e+f x) \sqrt {d \tan (e+f x)} \, dx\) [226]

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

[Out]

2/3*(d*tan(f*x+e))^(3/2)/d/f+4/7*(d*tan(f*x+e))^(7/2)/d^3/f+2/11*(d*tan(f*x+e))^(11/2)/d^5/f

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Rubi [A]
time = 0.04, antiderivative size = 67, 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, 276} \begin {gather*} \frac {2 (d \tan (e+f x))^{11/2}}{11 d^5 f}+\frac {4 (d \tan (e+f x))^{7/2}}{7 d^3 f}+\frac {2 (d \tan (e+f x))^{3/2}}{3 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^6*Sqrt[d*Tan[e + f*x]],x]

[Out]

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

Rule 276

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]

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 ^6(e+f x) \sqrt {d \tan (e+f x)} \, dx &=\frac {\text {Subst}\left (\int \sqrt {d x} \left (1+x^2\right )^2 \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\text {Subst}\left (\int \left (\sqrt {d x}+\frac {2 (d x)^{5/2}}{d^2}+\frac {(d x)^{9/2}}{d^4}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {2 (d \tan (e+f x))^{3/2}}{3 d f}+\frac {4 (d \tan (e+f x))^{7/2}}{7 d^3 f}+\frac {2 (d \tan (e+f x))^{11/2}}{11 d^5 f}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 52, normalized size = 0.78 \begin {gather*} \frac {2 (45+28 \cos (2 (e+f x))+4 \cos (4 (e+f x))) \sec ^4(e+f x) (d \tan (e+f x))^{3/2}}{231 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^6*Sqrt[d*Tan[e + f*x]],x]

[Out]

(2*(45 + 28*Cos[2*(e + f*x)] + 4*Cos[4*(e + f*x)])*Sec[e + f*x]^4*(d*Tan[e + f*x])^(3/2))/(231*d*f)

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Maple [A]
time = 3.82, size = 60, normalized size = 0.90

method	result	size
default	\(\frac {2 \left (32 \left (\cos ^{4}\left (f x +e \right )\right )+24 \left (\cos ^{2}\left (f x +e \right )\right )+21\right ) \sqrt {\frac {d \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )}{231 f \cos \left (f x +e \right )^{5}}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^6*(d*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/231/f*(32*cos(f*x+e)^4+24*cos(f*x+e)^2+21)*(d*sin(f*x+e)/cos(f*x+e))^(1/2)*sin(f*x+e)/cos(f*x+e)^5

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Maxima [A]
time = 0.28, size = 54, normalized size = 0.81 \begin {gather*} \frac {2 \, {\left (21 \, \left (d \tan \left (f x + e\right )\right )^{\frac {11}{2}} + 66 \, \left (d \tan \left (f x + e\right )\right )^{\frac {7}{2}} d^{2} + 77 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d^{4}\right )}}{231 \, d^{5} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

2/231*(21*(d*tan(f*x + e))^(11/2) + 66*(d*tan(f*x + e))^(7/2)*d^2 + 77*(d*tan(f*x + e))^(3/2)*d^4)/(d^5*f)

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Fricas [A]
time = 0.39, size = 65, normalized size = 0.97 \begin {gather*} \frac {2 \, {\left (32 \, \cos \left (f x + e\right )^{4} + 24 \, \cos \left (f x + e\right )^{2} + 21\right )} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{231 \, 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)^6*(d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/231*(32*cos(f*x + e)^4 + 24*cos(f*x + e)^2 + 21)*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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {d \tan {\left (e + f x \right )}} \sec ^{6}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**6*(d*tan(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(d*tan(e + f*x))*sec(e + f*x)**6, x)

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Giac [A]
time = 0.49, size = 82, normalized size = 1.22 \begin {gather*} \frac {2 \, {\left (21 \, \sqrt {d \tan \left (f x + e\right )} d^{5} \tan \left (f x + e\right )^{5} + 66 \, \sqrt {d \tan \left (f x + e\right )} d^{5} \tan \left (f x + e\right )^{3} + 77 \, \sqrt {d \tan \left (f x + e\right )} d^{5} \tan \left (f x + e\right )\right )}}{231 \, d^{5} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

2/231*(21*sqrt(d*tan(f*x + e))*d^5*tan(f*x + e)^5 + 66*sqrt(d*tan(f*x + e))*d^5*tan(f*x + e)^3 + 77*sqrt(d*tan
(f*x + e))*d^5*tan(f*x + e))/(d^5*f)

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Mupad [B]
time = 7.24, size = 334, normalized size = 4.99 \begin {gather*} -\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}}\,64{}\mathrm {i}}{231\,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}}\,64{}\mathrm {i}}{231\,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}}\,32{}\mathrm {i}}{77\,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}}\,768{}\mathrm {i}}{77\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3}-\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}}\,160{}\mathrm {i}}{11\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^4}+\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}}\,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))^(1/2)/cos(e + f*x)^6,x)

[Out]

((-(d*(exp(e*2i + f*x*2i)*1i - 1i))/(exp(e*2i + f*x*2i) + 1))^(1/2)*768i)/(77*f*(exp(e*2i + f*x*2i) + 1)^3) -
((-(d*(exp(e*2i + f*x*2i)*1i - 1i))/(exp(e*2i + f*x*2i) + 1))^(1/2)*64i)/(231*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)*32i)/(77*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)*64i)/(231*f) - ((-(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*(exp(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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