\(\int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx\) [586]

Optimal result

Integrand size = 25, antiderivative size = 143 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx=\frac {2 \left (7 a^2-2 b^2\right ) d^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{21 f}+\frac {18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac {2 \left (7 a^2-2 b^2\right ) d (d \sec (e+f x))^{3/2} \sin (e+f x)}{21 f}+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f} \]

[Out]

18/35*a*b*(d*sec(f*x+e))^(5/2)/f+2/21*(7*a^2-2*b^2)*d*(d*sec(f*x+e))^(3/2)*sin(f*x+e)/f+2/21*(7*a^2-2*b^2)*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+2/7*b*(d*sec(f*x+e))^(5/2)*(a+b*tan(f*x+e))/f

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 143, 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 = {3589, 3567, 3853, 3856, 2720} \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx=\frac {2 d^2 \left (7 a^2-2 b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{21 f}+\frac {2 d \left (7 a^2-2 b^2\right ) \sin (e+f x) (d \sec (e+f x))^{3/2}}{21 f}+\frac {18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f} \]

[In]

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

[Out]

(2*(7*a^2 - 2*b^2)*d^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]])/(21*f) + (18*a*b*(d*
Sec[e + f*x])^(5/2))/(35*f) + (2*(7*a^2 - 2*b^2)*d*(d*Sec[e + f*x])^(3/2)*Sin[e + f*x])/(21*f) + (2*b*(d*Sec[e
+ f*x])^(5/2)*(a + b*Tan[e + f*x]))/(7*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 3589

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

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} (a+b \tan (e+f x))}{7 f}+\frac {2}{7} \int (d \sec (e+f x))^{5/2} \left (\frac {7 a^2}{2}-b^2+\frac {9}{2} a b \tan (e+f x)\right ) \, dx \\ & = \frac {18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}+\frac {1}{7} \left (7 a^2-2 b^2\right ) \int (d \sec (e+f x))^{5/2} \, dx \\ & = \frac {18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac {2 \left (7 a^2-2 b^2\right ) d (d \sec (e+f x))^{3/2} \sin (e+f x)}{21 f}+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}+\frac {1}{21} \left (\left (7 a^2-2 b^2\right ) d^2\right ) \int \sqrt {d \sec (e+f x)} \, dx \\ & = \frac {18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac {2 \left (7 a^2-2 b^2\right ) d (d \sec (e+f x))^{3/2} \sin (e+f x)}{21 f}+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}+\frac {1}{21} \left (\left (7 a^2-2 b^2\right ) 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 \left (7 a^2-2 b^2\right ) d^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{21 f}+\frac {18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac {2 \left (7 a^2-2 b^2\right ) d (d \sec (e+f x))^{3/2} \sin (e+f x)}{21 f}+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.44 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.89 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx=\frac {2 d^2 \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2 \left (5 \left (7 a^2-2 b^2\right ) \cos ^{\frac {5}{2}}(e+f x) \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )+\frac {5}{2} \left (7 a^2-2 b^2\right ) \sin (2 (e+f x))+3 b (14 a+5 b \tan (e+f x))\right )}{105 f (a \cos (e+f x)+b \sin (e+f x))^2} \]

[In]

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

[Out]

(2*d^2*Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x])^2*(5*(7*a^2 - 2*b^2)*Cos[e + f*x]^(5/2)*EllipticF[(e + f*x)/2
, 2] + (5*(7*a^2 - 2*b^2)*Sin[2*(e + f*x)])/2 + 3*b*(14*a + 5*b*Tan[e + f*x])))/(105*f*(a*Cos[e + f*x] + b*Sin
[e + f*x])^2)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 146.29 (sec) , antiderivative size = 317, normalized size of antiderivative = 2.22

method	result	size
default	\(-\frac {2 d^{2} \sqrt {d \sec \left (f x +e \right )}\, \left (35 i \cos \left (f x +e \right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, a^{2}-10 i \cos \left (f x +e \right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, b^{2}+35 i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) a^{2}-10 i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) b^{2}-35 \tan \left (f x +e \right ) a^{2}+10 \tan \left (f x +e \right ) b^{2}-42 \left (\sec ^{2}\left (f x +e \right )\right ) a b -15 \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right ) b^{2}\right )}{105 f}\)	\(317\)
parts	\(-\frac {2 a^{2} \sqrt {d \sec \left (f x +e \right )}\, d^{2} \left (i \cos \left (f x +e \right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}-\tan \left (f x +e \right )\right )}{3 f}+\frac {2 i b^{2} \sqrt {d \sec \left (f x +e \right )}\, d^{2} \left (2 F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )+2 \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}+2 i \tan \left (f x +e \right )-3 i \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )\right )}{21 f}+\frac {4 a b \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}\)	\(328\)

[In]

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

[Out]

-2/105*d^2/f*(d*sec(f*x+e))^(1/2)*(35*I*cos(f*x+e)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(cos(f*x+e)/(cos(f*x
+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*a^2-10*I*cos(f*x+e)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(cos(f*x+e)/
(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*b^2+35*I*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(
1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*a^2-10*I*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2
)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*b^2-35*tan(f*x+e)*a^2+10*tan(f*x+e)*b^2-42*sec(f*x+e)^2*a*b-15*tan(f*
x+e)*sec(f*x+e)^2*b^2)

Fricas [C] (verification not implemented)

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

Time = 0.09 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.20 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx=\frac {-5 i \, \sqrt {2} {\left (7 \, a^{2} - 2 \, b^{2}\right )} d^{\frac {5}{2}} \cos \left (f x + e\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 5 i \, \sqrt {2} {\left (7 \, a^{2} - 2 \, b^{2}\right )} d^{\frac {5}{2}} \cos \left (f x + e\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (42 \, a b d^{2} \cos \left (f x + e\right ) + 5 \, {\left ({\left (7 \, a^{2} - 2 \, b^{2}\right )} d^{2} \cos \left (f x + e\right )^{2} + 3 \, b^{2} d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{105 \, f \cos \left (f x + e\right )^{3}} \]

[In]

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

[Out]

1/105*(-5*I*sqrt(2)*(7*a^2 - 2*b^2)*d^(5/2)*cos(f*x + e)^3*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x
+ e)) + 5*I*sqrt(2)*(7*a^2 - 2*b^2)*d^(5/2)*cos(f*x + e)^3*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*
x + e)) + 2*(42*a*b*d^2*cos(f*x + e) + 5*((7*a^2 - 2*b^2)*d^2*cos(f*x + e)^2 + 3*b^2*d^2)*sin(f*x + e))*sqrt(d
/cos(f*x + e)))/(f*cos(f*x + e)^3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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