\(\int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2 \, dx\) [587]

Optimal result

Integrand size = 25, antiderivative size = 143 \[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2 \, dx=-\frac {2 \left (5 a^2-2 b^2\right ) d^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac {2 \left (5 a^2-2 b^2\right ) d \sqrt {d \sec (e+f x)} \sin (e+f x)}{5 f}+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f} \]

[Out]

14/15*a*b*(d*sec(f*x+e))^(3/2)/f-2/5*(5*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)*Ellipti
cE(sin(1/2*f*x+1/2*e),2^(1/2))/f/cos(f*x+e)^(1/2)/(d*sec(f*x+e))^(1/2)+2/5*(5*a^2-2*b^2)*d*sin(f*x+e)*(d*sec(f
*x+e))^(1/2)/f+2/5*b*(d*sec(f*x+e))^(3/2)*(a+b*tan(f*x+e))/f

Rubi [A] (verified)

Time = 0.19 (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, 2719} \[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2 \, dx=-\frac {2 d^2 \left (5 a^2-2 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 d \left (5 a^2-2 b^2\right ) \sin (e+f x) \sqrt {d \sec (e+f x)}}{5 f}+\frac {14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f} \]

[In]

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

[Out]

(-2*(5*a^2 - 2*b^2)*d^2*EllipticE[(e + f*x)/2, 2])/(5*f*Sqrt[Cos[e + f*x]]*Sqrt[d*Sec[e + f*x]]) + (14*a*b*(d*
Sec[e + f*x])^(3/2))/(15*f) + (2*(5*a^2 - 2*b^2)*d*Sqrt[d*Sec[e + f*x]]*Sin[e + f*x])/(5*f) + (2*b*(d*Sec[e +
f*x])^(3/2)*(a + b*Tan[e + f*x]))/(5*f)

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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))^{3/2} (a+b \tan (e+f x))}{5 f}+\frac {2}{5} \int (d \sec (e+f x))^{3/2} \left (\frac {5 a^2}{2}-b^2+\frac {7}{2} a b \tan (e+f x)\right ) \, dx \\ & = \frac {14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}+\frac {1}{5} \left (5 a^2-2 b^2\right ) \int (d \sec (e+f x))^{3/2} \, dx \\ & = \frac {14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac {2 \left (5 a^2-2 b^2\right ) d \sqrt {d \sec (e+f x)} \sin (e+f x)}{5 f}+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}-\frac {1}{5} \left (\left (5 a^2-2 b^2\right ) d^2\right ) \int \frac {1}{\sqrt {d \sec (e+f x)}} \, dx \\ & = \frac {14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac {2 \left (5 a^2-2 b^2\right ) d \sqrt {d \sec (e+f x)} \sin (e+f x)}{5 f}+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}-\frac {\left (\left (5 a^2-2 b^2\right ) d^2\right ) \int \sqrt {\cos (e+f x)} \, dx}{5 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}} \\ & = -\frac {2 \left (5 a^2-2 b^2\right ) d^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac {2 \left (5 a^2-2 b^2\right ) d \sqrt {d \sec (e+f x)} \sin (e+f x)}{5 f}+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.85 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.88 \[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2 \, dx=-\frac {2 d^2 (a+b \tan (e+f x))^2 \left (3 \left (5 a^2-2 b^2\right ) \cos ^{\frac {3}{2}}(e+f x) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+\left (-\frac {15 a^2}{2}+3 b^2\right ) \sin (2 (e+f x))-b (10 a+3 b \tan (e+f x))\right )}{15 f \sqrt {d \sec (e+f x)} (a \cos (e+f x)+b \sin (e+f x))^2} \]

[In]

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

[Out]

(-2*d^2*(a + b*Tan[e + f*x])^2*(3*(5*a^2 - 2*b^2)*Cos[e + f*x]^(3/2)*EllipticE[(e + f*x)/2, 2] + ((-15*a^2)/2
+ 3*b^2)*Sin[2*(e + f*x)] - b*(10*a + 3*b*Tan[e + f*x])))/(15*f*Sqrt[d*Sec[e + f*x]]*(a*Cos[e + f*x] + b*Sin[e
+ f*x])^2)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 21.00 (sec) , antiderivative size = 833, normalized size of antiderivative = 5.83

method	result	size
parts	\(\text {Expression too large to display}\)	\(833\)
default	\(\text {Expression too large to display}\)	\(844\)

[In]

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

[Out]

-2*a^2/f*(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)^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)^2+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)-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*(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)-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
)-sin(f*x+e))*(d*sec(f*x+e))^(1/2)*d/(cos(f*x+e)+1)-2/5*b^2/f*(d*sec(f*x+e))^(1/2)*d/(cos(f*x+e)+1)*(2*I*Ellip
ticF(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)^2-2*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)^2
+4*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)-4*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*(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*(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*sin(f*x
+e)-tan(f*x+e)-sec(f*x+e)*tan(f*x+e))+4/3*a*b*(d*sec(f*x+e))^(3/2)/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 = 171, normalized size of antiderivative = 1.20 \[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2 \, dx=\frac {-3 i \, \sqrt {2} {\left (5 \, a^{2} - 2 \, b^{2}\right )} d^{\frac {3}{2}} \cos \left (f x + e\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, a^{2} - 2 \, b^{2}\right )} d^{\frac {3}{2}} \cos \left (f x + e\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + 2 \, {\left (10 \, a b d \cos \left (f x + e\right ) + 3 \, {\left ({\left (5 \, a^{2} - 2 \, b^{2}\right )} d \cos \left (f x + e\right )^{2} + b^{2} d\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{15 \, f \cos \left (f x + e\right )^{2}} \]

[In]

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

[Out]

1/15*(-3*I*sqrt(2)*(5*a^2 - 2*b^2)*d^(3/2)*cos(f*x + e)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co
s(f*x + e) + I*sin(f*x + e))) + 3*I*sqrt(2)*(5*a^2 - 2*b^2)*d^(3/2)*cos(f*x + e)^2*weierstrassZeta(-4, 0, weie
rstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) + 2*(10*a*b*d*cos(f*x + e) + 3*((5*a^2 - 2*b^2)*d*cos(f
*x + e)^2 + b^2*d)*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(f*cos(f*x + e)^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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