\(\int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx\) [579]

Optimal result

Integrand size = 23, antiderivative size = 92 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx=\frac {2 a d^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{3 f}+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \sec (e+f x))^{3/2} \sin (e+f x)}{3 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3567, 3853, 3856, 2720} \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx=\frac {2 a d^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{3 f}+\frac {2 a d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f} \]

[In]

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

[Out]

(2*a*d^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]])/(3*f) + (2*b*(d*Sec[e + f*x])^(5/2
))/(5*f) + (2*a*d*(d*Sec[e + f*x])^(3/2)*Sin[e + f*x])/(3*f)

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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[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 (d \sec (e+f x))^{5/2}}{5 f}+a \int (d \sec (e+f x))^{5/2} \, dx \\ & = \frac {2 b (d \sec (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \sec (e+f x))^{3/2} \sin (e+f x)}{3 f}+\frac {1}{3} \left (a d^2\right ) \int \sqrt {d \sec (e+f x)} \, dx \\ & = \frac {2 b (d \sec (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \sec (e+f x))^{3/2} \sin (e+f x)}{3 f}+\frac {1}{3} \left (a d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx \\ & = \frac {2 a d^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{3 f}+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \sec (e+f x))^{3/2} \sin (e+f x)}{3 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.90 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.63 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx=\frac {(d \sec (e+f x))^{5/2} \left (6 b+10 a \cos ^{\frac {5}{2}}(e+f x) \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )+5 a \sin (2 (e+f x))\right )}{15 f} \]

[In]

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

[Out]

((d*Sec[e + f*x])^(5/2)*(6*b + 10*a*Cos[e + f*x]^(5/2)*EllipticF[(e + f*x)/2, 2] + 5*a*Sin[2*(e + f*x)]))/(15*
f)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 24.30 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.76

method	result	size
default	\(-\frac {2 a \sqrt {d \sec \left (f x +e \right )}\, d^{2} \left (i \cos \left (f x +e \right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}-\tan \left (f x +e \right )\right )}{3 f}+\frac {2 b \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}\)	\(162\)
parts	\(-\frac {2 a \sqrt {d \sec \left (f x +e \right )}\, d^{2} \left (i \cos \left (f x +e \right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}-\tan \left (f x +e \right )\right )}{3 f}+\frac {2 b \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}\)	\(162\)

[In]

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

[Out]

-2/3*a/f*(d*sec(f*x+e))^(1/2)*d^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)+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)-tan(f*x+e))+2/5*b*(d*sec(f*x+e))^(5/2)/f

Fricas [C] (verification not implemented)

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

Time = 0.09 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.34 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx=\frac {-5 i \, \sqrt {2} a d^{\frac {5}{2}} \cos \left (f x + e\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 5 i \, \sqrt {2} a d^{\frac {5}{2}} \cos \left (f x + e\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (5 \, a d^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 3 \, b d^{2}\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{15 \, f \cos \left (f x + e\right )^{2}} \]

[In]

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

[Out]

1/15*(-5*I*sqrt(2)*a*d^(5/2)*cos(f*x + e)^2*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + 5*I*sq
rt(2)*a*d^(5/2)*cos(f*x + e)^2*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) + 2*(5*a*d^2*cos(f*x
+ e)*sin(f*x + e) + 3*b*d^2)*sqrt(d/cos(f*x + e)))/(f*cos(f*x + e)^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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