\(\int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx\) [594]

Optimal result

Integrand size = 25, antiderivative size = 198 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=\frac {2 a \left (7 a^2-6 b^2\right ) d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{21 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 a \left (7 a^2-6 b^2\right ) d^2 \sqrt {d \sec (e+f x)} \tan (e+f x)}{21 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f} \]

[Out]

2/21*a*(7*a^2-6*b^2)*d^2*(cos(1/2*arctan(tan(f*x+e)))^2)^(1/2)/cos(1/2*arctan(tan(f*x+e)))*EllipticF(sin(1/2*a
rctan(tan(f*x+e))),2^(1/2))*(d*sec(f*x+e))^(1/2)/f/(sec(f*x+e)^2)^(1/4)+2/21*a*(7*a^2-6*b^2)*d^2*(d*sec(f*x+e)
)^(1/2)*tan(f*x+e)/f+2/9*b*d^2*sec(f*x+e)^2*(d*sec(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2/f+2/315*b*d^2*sec(f*x+e)^2
*(d*sec(f*x+e))^(1/2)*(154*a^2-28*b^2+65*a*b*tan(f*x+e))/f

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 198, 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, 757, 794, 201, 237} \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=\frac {2 a d^2 \left (7 a^2-6 b^2\right ) \sqrt {d \sec (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{21 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f}+\frac {2 a d^2 \left (7 a^2-6 b^2\right ) \tan (e+f x) \sqrt {d \sec (e+f x)}}{21 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f} \]

[In]

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

[Out]

(2*a*(7*a^2 - 6*b^2)*d^2*EllipticF[ArcTan[Tan[e + f*x]]/2, 2]*Sqrt[d*Sec[e + f*x]])/(21*f*(Sec[e + f*x]^2)^(1/
4)) + (2*a*(7*a^2 - 6*b^2)*d^2*Sqrt[d*Sec[e + f*x]]*Tan[e + f*x])/(21*f) + (2*b*d^2*Sec[e + f*x]^2*Sqrt[d*Sec[
e + f*x]]*(a + b*Tan[e + f*x])^2)/(9*f) + (2*b*d^2*Sec[e + f*x]^2*Sqrt[d*Sec[e + f*x]]*(14*(11*a^2 - 2*b^2) +
65*a*b*Tan[e + f*x]))/(315*f)

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 757

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*((a + c*x^2)^(p
+ 1)/(c*(m + 2*p + 1))), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && 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 {\left (d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int (a+x)^3 \sqrt [4]{1+\frac {x^2}{b^2}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac {\left (2 b d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int (a+x) \left (\frac {1}{2} \left (-4+\frac {9 a^2}{b^2}\right )+\frac {13 a x}{2 b^2}\right ) \sqrt [4]{1+\frac {x^2}{b^2}} \, dx,x,b \tan (e+f x)\right )}{9 f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f}-\frac {\left (a \left (6-\frac {7 a^2}{b^2}\right ) b d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \sqrt [4]{1+\frac {x^2}{b^2}} \, dx,x,b \tan (e+f x)\right )}{7 f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 a \left (7 a^2-6 b^2\right ) d^2 \sqrt {d \sec (e+f x)} \tan (e+f x)}{21 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f}-\frac {\left (a \left (6-\frac {7 a^2}{b^2}\right ) b d^2 \sqrt {d \sec (e+f x)}\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 )}{21 f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 a \left (7 a^2-6 b^2\right ) d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{21 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 a \left (7 a^2-6 b^2\right ) d^2 \sqrt {d \sec (e+f x)} \tan (e+f x)}{21 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.21 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.79 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=-\frac {2 d (d \sec (e+f x))^{3/2} \left (63 b \left (-3 a^2+b^2\right ) \cos ^2(e+f x)-15 a \left (7 a^2-6 b^2\right ) \cos ^{\frac {9}{2}}(e+f x) \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )-15 a \left (7 a^2-6 b^2\right ) \cos ^3(e+f x) \sin (e+f x)-\frac {5}{2} b^2 (14 b+27 a \sin (2 (e+f x)))\right ) (a+b \tan (e+f x))^3}{315 f (a \cos (e+f x)+b \sin (e+f x))^3} \]

[In]

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

[Out]

(-2*d*(d*Sec[e + f*x])^(3/2)*(63*b*(-3*a^2 + b^2)*Cos[e + f*x]^2 - 15*a*(7*a^2 - 6*b^2)*Cos[e + f*x]^(9/2)*Ell
ipticF[(e + f*x)/2, 2] - 15*a*(7*a^2 - 6*b^2)*Cos[e + f*x]^3*Sin[e + f*x] - (5*b^2*(14*b + 27*a*Sin[2*(e + f*x
)]))/2)*(a + b*Tan[e + f*x])^3)/(315*f*(a*Cos[e + f*x] + b*Sin[e + f*x])^3)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 619.10 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.76

method	result	size
default	\(-\frac {2 d^{2} \sqrt {d \sec \left (f x +e \right )}\, \left (105 i \cos \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) a^{3}-90 i \cos \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) a \,b^{2}+105 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a^{3}-90 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a \,b^{2}-105 \tan \left (f x +e \right ) a^{3}+90 \tan \left (f x +e \right ) a \,b^{2}-189 \left (\sec ^{2}\left (f x +e \right )\right ) a^{2} b -135 \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right ) a \,b^{2}+63 b^{3} \left (\sec ^{2}\left (f x +e \right )\right )-35 \left (\sec ^{4}\left (f x +e \right )\right ) b^{3}\right )}{315 f}\)	\(349\)
parts	\(-\frac {2 a^{3} \sqrt {d \sec \left (f x +e \right )}\, d^{2} \left (i \cos \left (f x +e \right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}-\tan \left (f x +e \right )\right )}{3 f}+\frac {2 b^{3} \left (\frac {\left (d \sec \left (f x +e \right )\right )^{\frac {9}{2}}}{9}-\frac {d^{2} \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}{5}\right )}{f \,d^{2}}+\frac {6 a^{2} b \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}+\frac {2 i a \,b^{2} \sqrt {d \sec \left (f x +e \right )}\, d^{2} \left (2 F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )+2 \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}+2 i \tan \left (f x +e \right )-3 i \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )\right )}{7 f}\)	\(370\)

[In]

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

[Out]

-2/315*d^2/f*(d*sec(f*x+e))^(1/2)*(105*I*cos(f*x+e)*(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^3-90*I*cos(f*x+e)*(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+105*I*(1/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-c
ot(f*x+e)),I)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a^3-90*I*(1/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(
f*x+e)),I)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a*b^2-105*tan(f*x+e)*a^3+90*tan(f*x+e)*a*b^2-189*sec(f*x+e)^2*a^2
*b-135*tan(f*x+e)*sec(f*x+e)^2*a*b^2+63*b^3*sec(f*x+e)^2-35*sec(f*x+e)^4*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 = 202, normalized size of antiderivative = 1.02 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx=\frac {-15 i \, \sqrt {2} {\left (7 \, a^{3} - 6 \, a b^{2}\right )} d^{\frac {5}{2}} \cos \left (f x + e\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 15 i \, \sqrt {2} {\left (7 \, a^{3} - 6 \, a b^{2}\right )} d^{\frac {5}{2}} \cos \left (f x + e\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (35 \, b^{3} d^{2} + 63 \, {\left (3 \, a^{2} b - b^{3}\right )} d^{2} \cos \left (f x + e\right )^{2} + 15 \, {\left (9 \, a b^{2} d^{2} \cos \left (f x + e\right ) + {\left (7 \, a^{3} - 6 \, a b^{2}\right )} d^{2} \cos \left (f x + e\right )^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{315 \, f \cos \left (f x + e\right )^{4}} \]

[In]

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

[Out]

1/315*(-15*I*sqrt(2)*(7*a^3 - 6*a*b^2)*d^(5/2)*cos(f*x + e)^4*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(
f*x + e)) + 15*I*sqrt(2)*(7*a^3 - 6*a*b^2)*d^(5/2)*cos(f*x + e)^4*weierstrassPInverse(-4, 0, cos(f*x + e) - I*
sin(f*x + e)) + 2*(35*b^3*d^2 + 63*(3*a^2*b - b^3)*d^2*cos(f*x + e)^2 + 15*(9*a*b^2*d^2*cos(f*x + e) + (7*a^3
- 6*a*b^2)*d^2*cos(f*x + e)^3)*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(f*cos(f*x + e)^4)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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