\(\int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx\) [597]

Optimal result

Integrand size = 25, antiderivative size = 178 \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx=\frac {2 a \left (a^2-6 b^2\right ) E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{f \sqrt {d \sec (e+f x)}}-\frac {2 a \left (a^2-6 b^2\right ) \tan (e+f x)}{f \sqrt {d \sec (e+f x)}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{f \sqrt {d \sec (e+f x)}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (3 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{3 f \sqrt {d \sec (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 178, 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 = {3593, 753, 794, 233, 202} \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx=\frac {2 a \left (a^2-6 b^2\right ) \sqrt [4]{\sec ^2(e+f x)} E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )}{f \sqrt {d \sec (e+f x)}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (3 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{3 f \sqrt {d \sec (e+f x)}}-\frac {2 a \left (a^2-6 b^2\right ) \tan (e+f x)}{f \sqrt {d \sec (e+f x)}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{f \sqrt {d \sec (e+f x)}} \]

[In]

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

[Out]

(2*a*(a^2 - 6*b^2)*EllipticE[ArcTan[Tan[e + f*x]]/2, 2]*(Sec[e + f*x]^2)^(1/4))/(f*Sqrt[d*Sec[e + f*x]]) - (2*
a*(a^2 - 6*b^2)*Tan[e + f*x])/(f*Sqrt[d*Sec[e + f*x]]) - (2*(b - a*Tan[e + f*x])*(a + b*Tan[e + f*x])^2)/(f*Sq
rt[d*Sec[e + f*x]]) - (2*b*Sec[e + f*x]^2*(2*(3*a^2 - 2*b^2) + 3*a*b*Tan[e + f*x]))/(3*f*Sqrt[d*Sec[e + f*x]])

Rule 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 233

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], 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 {\sqrt [4]{\sec ^2(e+f x)} \text {Subst}\left (\int \frac {(a+x)^3}{\left (1+\frac {x^2}{b^2}\right )^{5/4}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt {d \sec (e+f x)}} \\ & = -\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{f \sqrt {d \sec (e+f x)}}+\frac {\left (2 b \sqrt [4]{\sec ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {(a+x) \left (\frac {1}{2} \left (4-\frac {a^2}{b^2}\right )-\frac {5 a x}{2 b^2}\right )}{\sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{f \sqrt {d \sec (e+f x)}} \\ & = -\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{f \sqrt {d \sec (e+f x)}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (3 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{3 f \sqrt {d \sec (e+f x)}}+\frac {\left (a \left (6-\frac {a^2}{b^2}\right ) b \sqrt [4]{\sec ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{f \sqrt {d \sec (e+f x)}} \\ & = -\frac {2 a \left (a^2-6 b^2\right ) \tan (e+f x)}{f \sqrt {d \sec (e+f x)}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{f \sqrt {d \sec (e+f x)}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (3 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{3 f \sqrt {d \sec (e+f x)}}-\frac {\left (a \left (6-\frac {a^2}{b^2}\right ) b \sqrt [4]{\sec ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {x^2}{b^2}\right )^{5/4}} \, dx,x,b \tan (e+f x)\right )}{f \sqrt {d \sec (e+f x)}} \\ & = \frac {2 a \left (a^2-6 b^2\right ) E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{f \sqrt {d \sec (e+f x)}}-\frac {2 a \left (a^2-6 b^2\right ) \tan (e+f x)}{f \sqrt {d \sec (e+f x)}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{f \sqrt {d \sec (e+f x)}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (3 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{3 f \sqrt {d \sec (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.90 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.73 \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx=\frac {d \left (6 a \left (a^2-6 b^2\right ) \cos ^{\frac {3}{2}}(e+f x) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+b \left (-9 a^2+5 b^2+\left (-9 a^2+3 b^2\right ) \cos (2 (e+f x))+9 a b \sin (2 (e+f x))\right )\right ) (a+b \tan (e+f x))^3}{3 f (d \sec (e+f x))^{3/2} (a \cos (e+f x)+b \sin (e+f x))^3} \]

[In]

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

[Out]

(d*(6*a*(a^2 - 6*b^2)*Cos[e + f*x]^(3/2)*EllipticE[(e + f*x)/2, 2] + b*(-9*a^2 + 5*b^2 + (-9*a^2 + 3*b^2)*Cos[
2*(e + f*x)] + 9*a*b*Sin[2*(e + f*x)]))*(a + b*Tan[e + f*x])^3)/(3*f*(d*Sec[e + f*x])^(3/2)*(a*Cos[e + f*x] +
b*Sin[e + f*x])^3)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 21.30 (sec) , antiderivative size = 1114, normalized size of antiderivative = 6.26

method	result	size
parts	\(\text {Expression too large to display}\)	\(1114\)
default	\(\text {Expression too large to display}\)	\(2626\)

[In]

int((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*a^3/f/(cos(f*x+e)+1)/(d*sec(f*x+e))^(1/2)*(I*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2)
*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)-I*cos(f*x+e)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(cos(f*x+
e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+2*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*
EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)-2*I*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e
)),I)*(1/(cos(f*x+e)+1))^(1/2)+I*sec(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*Ellipti
cE(I*(csc(f*x+e)-cot(f*x+e)),I)-I*sec(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x
+e)),I)*(1/(cos(f*x+e)+1))^(1/2)+sin(f*x+e))-1/6*b^3/f/(cos(f*x+e)+1)/(d*sec(f*x+e))^(1/2)/(-cos(f*x+e)/(cos(f
*x+e)+1)^2)^(1/2)*(-12*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+3*ln((2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*
x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1))-3*ln(2*(2*cos(f*x+e)*(-c
os(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1))-12*(-co
s(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-4*sec(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-4*sec(f*x+e)^2*(-cos(f*x+e)
/(cos(f*x+e)+1)^2)^(1/2))-6*a^2*b/(d*sec(f*x+e))^(1/2)/f-6*a*b^2/f/(cos(f*x+e)+1)/(d*sec(f*x+e))^(1/2)*(2*I*El
lipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)-2*I
*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)+
4*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)-4*I*(cos
(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2)+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)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)-2*I*sec(f*x+e
)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2)+sin(f*x+e)
-tan(f*x+e))

Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.95 \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx=-\frac {3 \, \sqrt {2} {\left (-i \, a^{3} + 6 i \, a b^{2}\right )} \sqrt {d} \cos \left (f x + e\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 \, \sqrt {2} {\left (i \, a^{3} - 6 i \, a b^{2}\right )} \sqrt {d} \cos \left (f x + e\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - 2 \, {\left (9 \, a b^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b^{3} - 3 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{3 \, d f \cos \left (f x + e\right )} \]

[In]

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/3*(3*sqrt(2)*(-I*a^3 + 6*I*a*b^2)*sqrt(d)*cos(f*x + e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co
s(f*x + e) + I*sin(f*x + e))) + 3*sqrt(2)*(I*a^3 - 6*I*a*b^2)*sqrt(d)*cos(f*x + e)*weierstrassZeta(-4, 0, weie
rstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) - 2*(9*a*b^2*cos(f*x + e)*sin(f*x + e) + b^3 - 3*(3*a^2
*b - b^3)*cos(f*x + e)^2)*sqrt(d/cos(f*x + e)))/(d*f*cos(f*x + e))

Sympy [F]

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

[In]

integrate((a+b*tan(f*x+e))**3/(d*sec(f*x+e))**(1/2),x)

[Out]

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

Maxima [F]

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

[In]

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int((a + b*tan(e + f*x))^3/(d/cos(e + f*x))^(1/2),x)

[Out]

int((a + b*tan(e + f*x))^3/(d/cos(e + f*x))^(1/2), x)