3.297 \(\int \frac{\sqrt{b \tan (e+f x)}}{(d \sec (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=72 \[ \frac{8 (b \tan (e+f x))^{3/2}}{21 b d^2 f (d \sec (e+f x))^{3/2}}+\frac{2 (b \tan (e+f x))^{3/2}}{7 b f (d \sec (e+f x))^{7/2}} \]

[Out]

(2*(b*Tan[e + f*x])^(3/2))/(7*b*f*(d*Sec[e + f*x])^(7/2)) + (8*(b*Tan[e + f*x])^(3/2))/(21*b*d^2*f*(d*Sec[e +
f*x])^(3/2))

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Rubi [A]  time = 0.103374, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {2612, 2605} \[ \frac{8 (b \tan (e+f x))^{3/2}}{21 b d^2 f (d \sec (e+f x))^{3/2}}+\frac{2 (b \tan (e+f x))^{3/2}}{7 b f (d \sec (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Tan[e + f*x]]/(d*Sec[e + f*x])^(7/2),x]

[Out]

(2*(b*Tan[e + f*x])^(3/2))/(7*b*f*(d*Sec[e + f*x])^(7/2)) + (8*(b*Tan[e + f*x])^(3/2))/(21*b*d^2*f*(d*Sec[e +
f*x])^(3/2))

Rule 2612

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1))/(b*f*m), x] + Dist[(m + n + 1)/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan[
e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, -2^(-1)])) && Integer
sQ[2*m, 2*n]

Rule 2605

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> -Simp[((a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n + 1))/(b*f*m), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 1, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt{b \tan (e+f x)}}{(d \sec (e+f x))^{7/2}} \, dx &=\frac{2 (b \tan (e+f x))^{3/2}}{7 b f (d \sec (e+f x))^{7/2}}+\frac{4 \int \frac{\sqrt{b \tan (e+f x)}}{(d \sec (e+f x))^{3/2}} \, dx}{7 d^2}\\ &=\frac{2 (b \tan (e+f x))^{3/2}}{7 b f (d \sec (e+f x))^{7/2}}+\frac{8 (b \tan (e+f x))^{3/2}}{21 b d^2 f (d \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.169996, size = 53, normalized size = 0.74 \[ \frac{(19 \sin (e+f x)+3 \sin (3 (e+f x))) \sqrt{b \tan (e+f x)}}{42 d^3 f \sqrt{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Tan[e + f*x]]/(d*Sec[e + f*x])^(7/2),x]

[Out]

((19*Sin[e + f*x] + 3*Sin[3*(e + f*x)])*Sqrt[b*Tan[e + f*x]])/(42*d^3*f*Sqrt[d*Sec[e + f*x]])

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Maple [A]  time = 0.191, size = 62, normalized size = 0.9 \begin{align*}{\frac{ \left ( 6\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+8 \right ) \sin \left ( fx+e \right ) }{21\,f \left ( \cos \left ( fx+e \right ) \right ) ^{3}}\sqrt{{\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}} \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(7/2),x)

[Out]

2/21/f*(3*cos(f*x+e)^2+4)*(b*sin(f*x+e)/cos(f*x+e))^(1/2)*sin(f*x+e)/(d/cos(f*x+e))^(7/2)/cos(f*x+e)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b \tan \left (f x + e\right )}}{\left (d \sec \left (f x + e\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*tan(f*x + e))/(d*sec(f*x + e))^(7/2), x)

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Fricas [A]  time = 1.7139, size = 159, normalized size = 2.21 \begin{align*} \frac{2 \,{\left (3 \, \cos \left (f x + e\right )^{3} + 4 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt{\frac{d}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{21 \, d^{4} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

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

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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((b*tan(f*x+e))**(1/2)/(d*sec(f*x+e))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b \tan \left (f x + e\right )}}{\left (d \sec \left (f x + e\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*tan(f*x + e))/(d*sec(f*x + e))^(7/2), x)