\(\int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{3/2}} \, dx\) [590]

Optimal result

Integrand size = 25, antiderivative size = 139 \[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{3/2}} \, dx=\frac {2 a b}{3 f (d \sec (e+f x))^{3/2}}+\frac {2 \left (a^2+2 b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{3 d^2 f}+\frac {2 \left (a^2+2 b^2\right ) \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}-\frac {2 b (a+b \tan (e+f x))}{f (d \sec (e+f x))^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 139, 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 = {3589, 3567, 3854, 3856, 2720} \[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{3/2}} \, dx=\frac {2 \left (a^2+2 b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{3 d^2 f}+\frac {2 \left (a^2+2 b^2\right ) \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}+\frac {2 a b}{3 f (d \sec (e+f x))^{3/2}}-\frac {2 b (a+b \tan (e+f x))}{f (d \sec (e+f x))^{3/2}} \]

[In]

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

[Out]

(2*a*b)/(3*f*(d*Sec[e + f*x])^(3/2)) + (2*(a^2 + 2*b^2)*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[d*Se
c[e + f*x]])/(3*d^2*f) + (2*(a^2 + 2*b^2)*Sin[e + f*x])/(3*d*f*Sqrt[d*Sec[e + f*x]]) - (2*b*(a + b*Tan[e + f*x
]))/(f*(d*Sec[e + f*x])^(3/2))

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 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3589

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[b*(d*Sec
[e + f*x])^m*((a + b*Tan[e + f*x])/(f*(m + 1))), x] + Dist[1/(m + 1), Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^
2 + a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 + b^2, 0] && NeQ[m, -1]

Rule 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

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

Mathematica [A] (verified)

Time = 2.43 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.73 \[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{3/2}} \, dx=\frac {\sec ^2(e+f x) \left (-2 a b-2 a b \cos (2 (e+f x))+2 \left (a^2+2 b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )+a^2 \sin (2 (e+f x))-b^2 \sin (2 (e+f x))\right )}{3 f (d \sec (e+f x))^{3/2}} \]

[In]

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

[Out]

(Sec[e + f*x]^2*(-2*a*b - 2*a*b*Cos[2*(e + f*x)] + 2*(a^2 + 2*b^2)*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2
] + a^2*Sin[2*(e + f*x)] - b^2*Sin[2*(e + f*x)]))/(3*f*(d*Sec[e + f*x])^(3/2))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 15.54 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.12

method	result	size
default	\(\frac {\frac {2 i F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, a^{2}}{3}+\frac {4 i F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, b^{2}}{3}+\frac {2 i \sec \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) a^{2}}{3}+\frac {4 i \sec \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) b^{2}}{3}-\frac {4 \cos \left (f x +e \right ) a b}{3}+\frac {2 a^{2} \sin \left (f x +e \right )}{3}-\frac {2 \sin \left (f x +e \right ) b^{2}}{3}}{d f \sqrt {d \sec \left (f x +e \right )}}\)	\(295\)
parts	\(-\frac {2 a^{2} \left (i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}+i \sec \left (f x +e \right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}-\sin \left (f x +e \right )\right )}{3 f \sqrt {d \sec \left (f x +e \right )}\, d}-\frac {2 b^{2} \left (2 i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}+2 i \sec \left (f x +e \right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}+\sin \left (f x +e \right )\right )}{3 f \sqrt {d \sec \left (f x +e \right )}\, d}-\frac {4 a b}{3 f \left (d \sec \left (f x +e \right )\right )^{\frac {3}{2}}}\)	\(309\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.94 \[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{3/2}} \, dx=\frac {\sqrt {2} {\left (-i \, a^{2} - 2 i \, b^{2}\right )} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + \sqrt {2} {\left (i \, a^{2} + 2 i \, b^{2}\right )} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \, {\left (2 \, a b \cos \left (f x + e\right )^{2} - {\left (a^{2} - b^{2}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{3 \, d^{2} f} \]

[In]

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

[Out]

1/3*(sqrt(2)*(-I*a^2 - 2*I*b^2)*sqrt(d)*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + sqrt(2)*(I
*a^2 + 2*I*b^2)*sqrt(d)*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) - 2*(2*a*b*cos(f*x + e)^2 -
(a^2 - b^2)*cos(f*x + e)*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(d^2*f)

Sympy [F]

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

[In]

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

[Out]

Integral((a + b*tan(e + f*x))**2/(d*sec(e + f*x))**(3/2), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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