3.3.97 \(\int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{7/2}} \, dx\) [297]

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

[Out]

2/7*(b*tan(f*x+e))^(3/2)/b/f/(d*sec(f*x+e))^(7/2)+8/21*(b*tan(f*x+e))^(3/2)/b/d^2/f/(d*sec(f*x+e))^(3/2)

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2692, 2685} \begin {gather*} \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}} \end {gather*}

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 2685

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]

Rule 2692

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*Ta
n[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, -2^(-1)])) && Integ
ersQ[2*m, 2*n]

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 = 0.18, size = 53, normalized size = 0.74 \begin {gather*} \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)}} \end {gather*}

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]])

________________________________________________________________________________________

Maple [A]
time = 0.54, size = 62, normalized size = 0.86

method	result	size
default	\(\frac {2 \left (3 \left (\cos ^{2}\left (f x +e \right )\right )+4\right ) \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )}{21 f \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {7}{2}} \cos \left (f x +e \right )^{3}}\)	\(62\)

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,method=_RETURNVERBOSE)

[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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [A]
time = 0.40, size = 69, normalized size = 0.96 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6191 deep

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [B]
time = 3.35, size = 69, normalized size = 0.96 \begin {gather*} \frac {\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}\,\left (22\,\sin \left (2\,e+2\,f\,x\right )+3\,\sin \left (4\,e+4\,f\,x\right )\right )\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}}{84\,d^4\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d/cos(e + f*x))^(1/2)*(22*sin(2*e + 2*f*x) + 3*sin(4*e + 4*f*x))*((b*sin(2*e + 2*f*x))/(cos(2*e + 2*f*x) + 1
))^(1/2))/(84*d^4*f)

________________________________________________________________________________________