\(\int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx\) [599]

Optimal result

Integrand size = 25, antiderivative size = 204 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\frac {6 a \left (a^2+2 b^2\right ) E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{5 d^2 f \sqrt {d \sec (e+f x)}}-\frac {6 a \left (a^2+2 b^2\right ) \tan (e+f x)}{5 d^2 f \sqrt {d \sec (e+f x)}}-\frac {2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{5 d^2 f \sqrt {d \sec (e+f x)}}-\frac {2 \left (2 b \left (a^2+2 b^2\right )-a \left (3 a^2+5 b^2\right ) \tan (e+f x)\right )}{5 d^2 f \sqrt {d \sec (e+f x)}} \]

[Out]

6/5*a*(a^2+2*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(t
an(f*x+e))),2^(1/2))*(sec(f*x+e)^2)^(1/4)/d^2/f/(d*sec(f*x+e))^(1/2)-6/5*a*(a^2+2*b^2)*tan(f*x+e)/d^2/f/(d*sec
(f*x+e))^(1/2)-2/5*cos(f*x+e)^2*(b-a*tan(f*x+e))*(a+b*tan(f*x+e))^2/d^2/f/(d*sec(f*x+e))^(1/2)-2/5*(2*b*(a^2+2
*b^2)-a*(3*a^2+5*b^2)*tan(f*x+e))/d^2/f/(d*sec(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 204, 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, 792, 233, 202} \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\frac {6 a \left (a^2+2 b^2\right ) \sqrt [4]{\sec ^2(e+f x)} E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )}{5 d^2 f \sqrt {d \sec (e+f x)}}-\frac {6 a \left (a^2+2 b^2\right ) \tan (e+f x)}{5 d^2 f \sqrt {d \sec (e+f x)}}-\frac {2 \left (2 b \left (a^2+2 b^2\right )-a \left (3 a^2+5 b^2\right ) \tan (e+f x)\right )}{5 d^2 f \sqrt {d \sec (e+f x)}}-\frac {2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{5 d^2 f \sqrt {d \sec (e+f x)}} \]

[In]

Int[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(5/2),x]

[Out]

(6*a*(a^2 + 2*b^2)*EllipticE[ArcTan[Tan[e + f*x]]/2, 2]*(Sec[e + f*x]^2)^(1/4))/(5*d^2*f*Sqrt[d*Sec[e + f*x]])
- (6*a*(a^2 + 2*b^2)*Tan[e + f*x])/(5*d^2*f*Sqrt[d*Sec[e + f*x]]) - (2*Cos[e + f*x]^2*(b - a*Tan[e + f*x])*(a
+ b*Tan[e + f*x])^2)/(5*d^2*f*Sqrt[d*Sec[e + f*x]]) - (2*(2*b*(a^2 + 2*b^2) - a*(3*a^2 + 5*b^2)*Tan[e + f*x])
)/(5*d^2*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 792

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a*(e*f + d*g) - (
c*d*f - a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[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 )^{9/4}} \, dx,x,b \tan (e+f x)\right )}{b d^2 f \sqrt {d \sec (e+f x)}} \\ & = -\frac {2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{5 d^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 {3 a^2}{b^2}\right )-\frac {a x}{2 b^2}\right )}{\left (1+\frac {x^2}{b^2}\right )^{5/4}} \, dx,x,b \tan (e+f x)\right )}{5 d^2 f \sqrt {d \sec (e+f x)}} \\ & = -\frac {2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{5 d^2 f \sqrt {d \sec (e+f x)}}-\frac {2 \left (2 b \left (a^2+2 b^2\right )-a \left (3 a^2+5 b^2\right ) \tan (e+f x)\right )}{5 d^2 f \sqrt {d \sec (e+f x)}}-\frac {\left (3 a \left (2+\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 )}{5 d^2 f \sqrt {d \sec (e+f x)}} \\ & = -\frac {6 a \left (a^2+2 b^2\right ) \tan (e+f x)}{5 d^2 f \sqrt {d \sec (e+f x)}}-\frac {2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{5 d^2 f \sqrt {d \sec (e+f x)}}-\frac {2 \left (2 b \left (a^2+2 b^2\right )-a \left (3 a^2+5 b^2\right ) \tan (e+f x)\right )}{5 d^2 f \sqrt {d \sec (e+f x)}}+\frac {\left (3 a \left (2+\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 )}{5 d^2 f \sqrt {d \sec (e+f x)}} \\ & = \frac {6 a \left (a^2+2 b^2\right ) E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{5 d^2 f \sqrt {d \sec (e+f x)}}-\frac {6 a \left (a^2+2 b^2\right ) \tan (e+f x)}{5 d^2 f \sqrt {d \sec (e+f x)}}-\frac {2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{5 d^2 f \sqrt {d \sec (e+f x)}}-\frac {2 \left (2 b \left (a^2+2 b^2\right )-a \left (3 a^2+5 b^2\right ) \tan (e+f x)\right )}{5 d^2 f \sqrt {d \sec (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.94 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.74 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\frac {\sqrt {d \sec (e+f x)} \left (-b \left (9 a^2+17 b^2\right ) \cos (e+f x)-3 a^2 b \cos (3 (e+f x))+b^3 \cos (3 (e+f x))+12 a \left (a^2+2 b^2\right ) \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+a^3 \sin (e+f x)-3 a b^2 \sin (e+f x)+a^3 \sin (3 (e+f x))-3 a b^2 \sin (3 (e+f x))\right )}{10 d^3 f} \]

[In]

Integrate[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(5/2),x]

[Out]

(Sqrt[d*Sec[e + f*x]]*(-(b*(9*a^2 + 17*b^2)*Cos[e + f*x]) - 3*a^2*b*Cos[3*(e + f*x)] + b^3*Cos[3*(e + f*x)] +
12*a*(a^2 + 2*b^2)*Sqrt[Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2] + a^3*Sin[e + f*x] - 3*a*b^2*Sin[e + f*x] + a^
3*Sin[3*(e + f*x)] - 3*a*b^2*Sin[3*(e + f*x)]))/(10*d^3*f)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 26.00 (sec) , antiderivative size = 1289, normalized size of antiderivative = 6.32

method	result	size
parts	\(\text {Expression too large to display}\)	\(1289\)
default	\(\text {Expression too large to display}\)	\(1602\)

[In]

int((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/5*a^3/f/(cos(f*x+e)+1)/(d*sec(f*x+e))^(1/2)/d^2*(3*I*cos(f*x+e)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(c
os(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-3*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)+6*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)-6*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)+3*I*sec(f*x+e)*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)-3*I*sec(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)+sin(f*x+e)*cos(f*x+e)^2+sin(f*x+e)*cos(f*x+e)+3*sin(f*x+e))-1/1
0*b^3/f/(d*sec(f*x+e))^(1/2)/d^2*(5*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*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))-5*(-cos(f*x+e)/(cos(f*x+
e)+1)^2)^(1/2)*ln(2*(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))-4*cos(f*x+e)^2+5*sec(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*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))-5*
sec(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln(2*(2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-co
s(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1))+20)-6/5*a*b^2/f/(cos(f*x+e)+1)/(d*sec(f*x+e))^(
1/2)/d^2*(-2*I*cos(f*x+e)*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)+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*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)+4*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)
+sin(f*x+e)*cos(f*x+e)^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)*cos(f*x+e)-2*sin(f*x+e))-6/5*a^2*b/f/(d*sec(f*x+e))^(5/2)

Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=-\frac {3 \, \sqrt {2} {\left (-i \, a^{3} - 2 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 \, \sqrt {2} {\left (i \, a^{3} + 2 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + 2 \, {\left (5 \, b^{3} \cos \left (f x + e\right ) + {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (f x + e\right )^{2} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{5 \, d^{3} f} \]

[In]

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/5*(3*sqrt(2)*(-I*a^3 - 2*I*a*b^2)*sqrt(d)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) +
I*sin(f*x + e))) + 3*sqrt(2)*(I*a^3 + 2*I*a*b^2)*sqrt(d)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos
(f*x + e) - I*sin(f*x + e))) + 2*(5*b^3*cos(f*x + e) + (3*a^2*b - b^3)*cos(f*x + e)^3 - (a^3 - 3*a*b^2)*cos(f*
x + e)^2*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(d^3*f)

Sympy [F]

\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{3}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((a+b*tan(f*x+e))**3/(d*sec(f*x+e))**(5/2),x)

[Out]

Integral((a + b*tan(e + f*x))**3/(d*sec(e + f*x))**(5/2), x)

Maxima [F]

\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)^3/(d*sec(f*x + e))^(5/2), x)

Giac [F]

\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)^3/(d*sec(f*x + e))^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int((a + b*tan(e + f*x))^3/(d/cos(e + f*x))^(5/2),x)

[Out]

int((a + b*tan(e + f*x))^3/(d/cos(e + f*x))^(5/2), x)