\(\int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{3/2}} \, dx\) [598]

Optimal result

Integrand size = 25, antiderivative size = 146 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{3/2}} \, dx=\frac {2 a \left (a^2+6 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sec ^2(e+f x)^{3/4}}{3 f (d \sec (e+f x))^{3/2}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{3 f (d \sec (e+f x))^{3/2}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (a^2-2 b^2\right )+a b \tan (e+f x)\right )}{3 f (d \sec (e+f x))^{3/2}} \]

[Out]

2/3*a*(a^2+6*b^2)*(cos(1/2*arctan(tan(f*x+e)))^2)^(1/2)/cos(1/2*arctan(tan(f*x+e)))*EllipticF(sin(1/2*arctan(t
an(f*x+e))),2^(1/2))*(sec(f*x+e)^2)^(3/4)/f/(d*sec(f*x+e))^(3/2)-2/3*(b-a*tan(f*x+e))*(a+b*tan(f*x+e))^2/f/(d*
sec(f*x+e))^(3/2)-2/3*b*sec(f*x+e)^2*(2*a^2-4*b^2+a*b*tan(f*x+e))/f/(d*sec(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3593, 753, 794, 237} \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{3/2}} \, dx=\frac {2 a \left (a^2+6 b^2\right ) \sec ^2(e+f x)^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{3 f (d \sec (e+f x))^{3/2}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (a^2-2 b^2\right )+a b \tan (e+f x)\right )}{3 f (d \sec (e+f x))^{3/2}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{3 f (d \sec (e+f x))^{3/2}} \]

[In]

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

[Out]

(2*a*(a^2 + 6*b^2)*EllipticF[ArcTan[Tan[e + f*x]]/2, 2]*(Sec[e + f*x]^2)^(3/4))/(3*f*(d*Sec[e + f*x])^(3/2)) -
(2*(b - a*Tan[e + f*x])*(a + b*Tan[e + f*x])^2)/(3*f*(d*Sec[e + f*x])^(3/2)) - (2*b*Sec[e + f*x]^2*(2*(a^2 -
2*b^2) + a*b*Tan[e + f*x]))/(3*f*(d*Sec[e + f*x])^(3/2))

Rule 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 753

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(a*e - c*d*x)*((a
+ c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 3593

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

Rubi steps \begin{align*} \text {integral}& = \frac {\sec ^2(e+f x)^{3/4} \text {Subst}\left (\int \frac {(a+x)^3}{\left (1+\frac {x^2}{b^2}\right )^{7/4}} \, dx,x,b \tan (e+f x)\right )}{b f (d \sec (e+f x))^{3/2}} \\ & = -\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{3 f (d \sec (e+f x))^{3/2}}+\frac {\left (2 b \sec ^2(e+f x)^{3/4}\right ) \text {Subst}\left (\int \frac {(a+x) \left (\frac {1}{2} \left (4+\frac {a^2}{b^2}\right )-\frac {3 a x}{2 b^2}\right )}{\left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{3 f (d \sec (e+f x))^{3/2}} \\ & = -\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{3 f (d \sec (e+f x))^{3/2}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (a^2-2 b^2\right )+a b \tan (e+f x)\right )}{3 f (d \sec (e+f x))^{3/2}}+\frac {\left (a \left (6+\frac {a^2}{b^2}\right ) b \sec ^2(e+f x)^{3/4}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{3 f (d \sec (e+f x))^{3/2}} \\ & = \frac {2 a \left (a^2+6 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sec ^2(e+f x)^{3/4}}{3 f (d \sec (e+f x))^{3/2}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{3 f (d \sec (e+f x))^{3/2}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (a^2-2 b^2\right )+a b \tan (e+f x)\right )}{3 f (d \sec (e+f x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.80 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{3/2}} \, dx=\frac {\sec ^2(e+f x) \left (-3 a^2 b+7 b^3+\left (-3 a^2 b+b^3\right ) \cos (2 (e+f x))+2 a \left (a^2+6 b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )+a^3 \sin (2 (e+f x))-3 a b^2 \sin (2 (e+f x))\right )}{3 f (d \sec (e+f x))^{3/2}} \]

[In]

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

[Out]

(Sec[e + f*x]^2*(-3*a^2*b + 7*b^3 + (-3*a^2*b + b^3)*Cos[2*(e + f*x)] + 2*a*(a^2 + 6*b^2)*Sqrt[Cos[e + f*x]]*E
llipticF[(e + f*x)/2, 2] + a^3*Sin[2*(e + f*x)] - 3*a*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 = 23.96 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.21

method	result	size
default	\(-\frac {2 \left (i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a^{3}+6 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a \,b^{2}+i \sec \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) a^{3}+6 i \sec \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) a \,b^{2}+3 \cos \left (f x +e \right ) a^{2} b -\cos \left (f x +e \right ) b^{3}-a^{3} \sin \left (f x +e \right )+3 a \,b^{2} \sin \left (f x +e \right )-3 \sec \left (f x +e \right ) b^{3}\right )}{3 d f \sqrt {d \sec \left (f x +e \right )}}\)	\(323\)
parts	\(\text {Expression too large to display}\)	\(1146\)

[In]

int((a+b*tan(f*x+e))^3/(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*(csc(f
*x+e)-cot(f*x+e)),I)*a^3+6*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e
)-cot(f*x+e)),I)*a*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*(cs
c(f*x+e)-cot(f*x+e)),I)*a^3+6*I*sec(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*Elliptic
F(I*(csc(f*x+e)-cot(f*x+e)),I)*a*b^2+3*cos(f*x+e)*a^2*b-cos(f*x+e)*b^3-a^3*sin(f*x+e)+3*a*b^2*sin(f*x+e)-3*sec
(f*x+e)*b^3)

Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.01 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{3/2}} \, dx=\frac {\sqrt {2} {\left (-i \, a^{3} - 6 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + \sqrt {2} {\left (i \, a^{3} + 6 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (3 \, b^{3} - {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{3 \, d^{2} f} \]

[In]

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

[Out]

1/3*(sqrt(2)*(-I*a^3 - 6*I*a*b^2)*sqrt(d)*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + sqrt(2)*
(I*a^3 + 6*I*a*b^2)*sqrt(d)*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) + 2*(3*b^3 - (3*a^2*b -
b^3)*cos(f*x + e)^2 + (a^3 - 3*a*b^2)*cos(f*x + e)*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(d^2*f)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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