\(\int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx\) [305]

Optimal result

Integrand size = 25, antiderivative size = 131 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx=\frac {4 b^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{21 d^4 f \sqrt {b \tan (e+f x)}}-\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}} \]

[Out]

-4/21*b^2*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticF(cos(1/2*e+1/4*Pi+1/2*f*x),2^
(1/2))*(d*sec(f*x+e))^(1/2)*sin(f*x+e)^(1/2)/d^4/f/(b*tan(f*x+e))^(1/2)-2/7*b*(b*tan(f*x+e))^(1/2)/f/(d*sec(f*
x+e))^(7/2)+2/21*b*(b*tan(f*x+e))^(1/2)/d^2/f/(d*sec(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2690, 2692, 2696, 2721, 2720} \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx=\frac {4 b^2 \sqrt {\sin (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),2\right ) \sqrt {d \sec (e+f x)}}{21 d^4 f \sqrt {b \tan (e+f x)}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}}-\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}} \]

[In]

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

[Out]

(4*b^2*EllipticF[(e - Pi/2 + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]]*Sqrt[Sin[e + f*x]])/(21*d^4*f*Sqrt[b*Tan[e + f*x]
]) - (2*b*Sqrt[b*Tan[e + f*x]])/(7*f*(d*Sec[e + f*x])^(7/2)) + (2*b*Sqrt[b*Tan[e + f*x]])/(21*d^2*f*(d*Sec[e +
f*x])^(3/2))

Rule 2690

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*m)), x] - Dist[b^2*((n - 1)/(a^2*m)), Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan
[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, 3/2]
)) && IntegersQ[2*m, 2*n]

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]

Rule 2696

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

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {b^2 \int \frac {1}{(d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx}{7 d^2} \\ & = -\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}}+\frac {\left (2 b^2\right ) \int \frac {\sqrt {d \sec (e+f x)}}{\sqrt {b \tan (e+f x)}} \, dx}{21 d^4} \\ & = -\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}}+\frac {\left (2 b^2 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \int \frac {1}{\sqrt {b \sin (e+f x)}} \, dx}{21 d^4 \sqrt {b \tan (e+f x)}} \\ & = -\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}}+\frac {\left (2 b^2 \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}\right ) \int \frac {1}{\sqrt {\sin (e+f x)}} \, dx}{21 d^4 \sqrt {b \tan (e+f x)}} \\ & = \frac {4 b^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{21 d^4 f \sqrt {b \tan (e+f x)}}-\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.93 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.62 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx=-\frac {b \left (1+3 \cos (2 (e+f x))-4 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{3/4}\right ) \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}} \]

[In]

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

[Out]

-1/21*(b*(1 + 3*Cos[2*(e + f*x)] - 4*Hypergeometric2F1[1/4, 3/4, 5/4, -Tan[e + f*x]^2]*(Sec[e + f*x]^2)^(3/4))
*Sqrt[b*Tan[e + f*x]])/(d^2*f*(d*Sec[e + f*x])^(3/2))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 3.12 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.23

method	result	size
default	\(\frac {\sin \left (f x +e \right ) \left (-2 i \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (f x +e \right )+3 \sqrt {2}\, \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-2 i \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {2}\, \cos \left (f x +e \right ) \sin \left (f x +e \right )\right ) \sqrt {b \tan \left (f x +e \right )}\, b \sqrt {2}}{21 f \left (\cos \left (f x +e \right )-1\right ) d^{3} \sqrt {d \sec \left (f x +e \right )}\, \left (\cos \left (f x +e \right )+1\right )}\)	\(292\)

[In]

int((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/21/f*sin(f*x+e)*(-2*I*(-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2)*(-I*(cot(f*x+e)-csc(f*x+e)+I))^(1/2)*(-I*(cot(f*x
+e)-csc(f*x+e)))^(1/2)*EllipticF((-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2),1/2*2^(1/2))*cos(f*x+e)+3*2^(1/2)*cos(f*
x+e)^3*sin(f*x+e)-2*I*(-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2)*(-I*(cot(f*x+e)-csc(f*x+e)+I))^(1/2)*(-I*(cot(f*x+e
)-csc(f*x+e)))^(1/2)*EllipticF((-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2),1/2*2^(1/2))-2^(1/2)*cos(f*x+e)*sin(f*x+e)
)*(b*tan(f*x+e))^(1/2)*b/(cos(f*x+e)-1)/d^3/(d*sec(f*x+e))^(1/2)/(cos(f*x+e)+1)*2^(1/2)

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.89 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx=\frac {2 \, {\left (\sqrt {-2 i \, b d} b {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + \sqrt {2 i \, b d} b {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - {\left (3 \, b \cos \left (f x + e\right )^{4} - b \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}\right )}}{21 \, d^{4} f} \]

[In]

integrate((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

2/21*(sqrt(-2*I*b*d)*b*weierstrassPInverse(4, 0, cos(f*x + e) + I*sin(f*x + e)) + sqrt(2*I*b*d)*b*weierstrassP
Inverse(4, 0, cos(f*x + e) - I*sin(f*x + e)) - (3*b*cos(f*x + e)^4 - b*cos(f*x + e)^2)*sqrt(b*sin(f*x + e)/cos
(f*x + e))*sqrt(d/cos(f*x + e)))/(d^4*f)

Sympy [F(-1)]

Timed out. \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate((b*tan(f*x+e))**(3/2)/(d*sec(f*x+e))**(7/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(7/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx=\int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{7/2}} \,d x \]

[In]

int((b*tan(e + f*x))^(3/2)/(d/cos(e + f*x))^(7/2),x)

[Out]

int((b*tan(e + f*x))^(3/2)/(d/cos(e + f*x))^(7/2), x)