\(\int \frac {(d \sec (e+f x))^{3/2}}{a+b \tan (e+f x)} \, dx\) [605]

Optimal result

Integrand size = 25, antiderivative size = 334 \[ \int \frac {(d \sec (e+f x))^{3/2}}{a+b \tan (e+f x)} \, dx=\frac {\arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) (d \sec (e+f x))^{3/2}}{\sqrt {b} \sqrt [4]{a^2+b^2} f \sec ^2(e+f x)^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) (d \sec (e+f x))^{3/2}}{\sqrt {b} \sqrt [4]{a^2+b^2} f \sec ^2(e+f x)^{3/4}}-\frac {a \cot (e+f x) \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}}{b \sqrt {a^2+b^2} f \sec ^2(e+f x)^{3/4}}+\frac {a \cot (e+f x) \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}}{b \sqrt {a^2+b^2} f \sec ^2(e+f x)^{3/4}} \]

[Out]

arctan((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1/4))*(d*sec(f*x+e))^(3/2)/(a^2+b^2)^(1/4)/f/(sec(f*x+e)^2)^(3/
4)/b^(1/2)-arctanh((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1/4))*(d*sec(f*x+e))^(3/2)/(a^2+b^2)^(1/4)/f/(sec(f
*x+e)^2)^(3/4)/b^(1/2)-a*cot(f*x+e)*EllipticPi((sec(f*x+e)^2)^(1/4),-b/(a^2+b^2)^(1/2),I)*(d*sec(f*x+e))^(3/2)
*(-tan(f*x+e)^2)^(1/2)/b/f/(sec(f*x+e)^2)^(3/4)/(a^2+b^2)^(1/2)+a*cot(f*x+e)*EllipticPi((sec(f*x+e)^2)^(1/4),b
/(a^2+b^2)^(1/2),I)*(d*sec(f*x+e))^(3/2)*(-tan(f*x+e)^2)^(1/2)/b/f/(sec(f*x+e)^2)^(3/4)/(a^2+b^2)^(1/2)

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3593, 760, 408, 504, 1227, 551, 455, 65, 304, 211, 214} \[ \int \frac {(d \sec (e+f x))^{3/2}}{a+b \tan (e+f x)} \, dx=-\frac {a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) (d \sec (e+f x))^{3/2} \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right )}{b f \sqrt {a^2+b^2} \sec ^2(e+f x)^{3/4}}+\frac {a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) (d \sec (e+f x))^{3/2} \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right )}{b f \sqrt {a^2+b^2} \sec ^2(e+f x)^{3/4}}+\frac {(d \sec (e+f x))^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{\sqrt {b} f \sqrt [4]{a^2+b^2} \sec ^2(e+f x)^{3/4}}-\frac {(d \sec (e+f x))^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{\sqrt {b} f \sqrt [4]{a^2+b^2} \sec ^2(e+f x)^{3/4}} \]

[In]

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

[Out]

(ArcTan[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*(d*Sec[e + f*x])^(3/2))/(Sqrt[b]*(a^2 + b^2)^(1/4)
*f*(Sec[e + f*x]^2)^(3/4)) - (ArcTanh[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*(d*Sec[e + f*x])^(3/
2))/(Sqrt[b]*(a^2 + b^2)^(1/4)*f*(Sec[e + f*x]^2)^(3/4)) - (a*Cot[e + f*x]*EllipticPi[-(b/Sqrt[a^2 + b^2]), Ar
cSin[(Sec[e + f*x]^2)^(1/4)], -1]*(d*Sec[e + f*x])^(3/2)*Sqrt[-Tan[e + f*x]^2])/(b*Sqrt[a^2 + b^2]*f*(Sec[e +
f*x]^2)^(3/4)) + (a*Cot[e + f*x]*EllipticPi[b/Sqrt[a^2 + b^2], ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*(d*Sec[e +
f*x])^(3/2)*Sqrt[-Tan[e + f*x]^2])/(b*Sqrt[a^2 + b^2]*f*(Sec[e + f*x]^2)^(3/4))

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 408

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[2*(Sqrt[(-b)*(x^2/a)]/x), Subst[I
nt[x^2/(Sqrt[1 - x^4/a]*(b*c - a*d + d*x^4)), x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0]

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 504

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s
= Denominator[Rt[-a/b, 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), Int[1/
((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 551

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1/(a*Sqr
t[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b*(c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], x] /; FreeQ[{a, b, c,
d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-f/e, -d/c])

Rule 760

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(1/4)), x_Symbol] :> Dist[d, Int[1/((d^2 - e^2*x^2)*(a + c*x^
2)^(1/4)), x], x] - Dist[e, Int[x/((d^2 - e^2*x^2)*(a + c*x^2)^(1/4)), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ
[c*d^2 + a*e^2, 0]

Rule 1227

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[Sqrt[-c],
Int[1/((d + e*x^2)*Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c,
0]

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 {(d \sec (e+f x))^{3/2} \text {Subst}\left (\int \frac {1}{(a+x) \sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{b f \sec ^2(e+f x)^{3/4}} \\ & = -\frac {(d \sec (e+f x))^{3/2} \text {Subst}\left (\int \frac {x}{\left (a^2-x^2\right ) \sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{b f \sec ^2(e+f x)^{3/4}}+\frac {\left (a (d \sec (e+f x))^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (a^2-x^2\right ) \sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{b f \sec ^2(e+f x)^{3/4}} \\ & = -\frac {(d \sec (e+f x))^{3/2} \text {Subst}\left (\int \frac {1}{\left (a^2-x\right ) \sqrt [4]{1+\frac {x}{b^2}}} \, dx,x,b^2 \tan ^2(e+f x)\right )}{2 b f \sec ^2(e+f x)^{3/4}}+\frac {\left (2 a \cot (e+f x) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1-x^4} \left (1+\frac {a^2}{b^2}-x^4\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{b^2 f \sec ^2(e+f x)^{3/4}} \\ & = -\frac {\left (2 b (d \sec (e+f x))^{3/2}\right ) \text {Subst}\left (\int \frac {x^2}{a^2+b^2-b^2 x^4} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{f \sec ^2(e+f x)^{3/4}}+\frac {\left (a \cot (e+f x) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+b^2}-b x^2\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}-\frac {\left (a \cot (e+f x) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+b^2}+b x^2\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}} \\ & = -\frac {(d \sec (e+f x))^{3/2} \text {Subst}\left (\int \frac {1}{\sqrt {a^2+b^2}-b x^2} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{f \sec ^2(e+f x)^{3/4}}+\frac {(d \sec (e+f x))^{3/2} \text {Subst}\left (\int \frac {1}{\sqrt {a^2+b^2}+b x^2} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{f \sec ^2(e+f x)^{3/4}}+\frac {\left (a \cot (e+f x) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (\sqrt {a^2+b^2}-b x^2\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}-\frac {\left (a \cot (e+f x) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (\sqrt {a^2+b^2}+b x^2\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}} \\ & = \frac {\arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) (d \sec (e+f x))^{3/2}}{\sqrt {b} \sqrt [4]{a^2+b^2} f \sec ^2(e+f x)^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) (d \sec (e+f x))^{3/2}}{\sqrt {b} \sqrt [4]{a^2+b^2} f \sec ^2(e+f x)^{3/4}}-\frac {a \cot (e+f x) \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}}{b \sqrt {a^2+b^2} f \sec ^2(e+f x)^{3/4}}+\frac {a \cot (e+f x) \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}}{b \sqrt {a^2+b^2} f \sec ^2(e+f x)^{3/4}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 3.43 (sec) , antiderivative size = 276, normalized size of antiderivative = 0.83 \[ \int \frac {(d \sec (e+f x))^{3/2}}{a+b \tan (e+f x)} \, dx=-\frac {12 d^2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},\frac {1}{4},\frac {3}{2},\frac {a-i b}{a+b \tan (e+f x)},\frac {a+i b}{a+b \tan (e+f x)}\right ) (a+b \tan (e+f x))}{b f \sqrt {d \sec (e+f x)} \left ((a+i b) \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {5}{4},\frac {5}{2},\frac {a-i b}{a+b \tan (e+f x)},\frac {a+i b}{a+b \tan (e+f x)}\right )+(a-i b) \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},\frac {1}{4},\frac {5}{2},\frac {a-i b}{a+b \tan (e+f x)},\frac {a+i b}{a+b \tan (e+f x)}\right )+6 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},\frac {1}{4},\frac {3}{2},\frac {a-i b}{a+b \tan (e+f x)},\frac {a+i b}{a+b \tan (e+f x)}\right ) (a+b \tan (e+f x))\right )} \]

[In]

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

[Out]

(-12*d^2*AppellF1[1/2, 1/4, 1/4, 3/2, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b*Tan[e + f*x])]*(a + b*T
an[e + f*x]))/(b*f*Sqrt[d*Sec[e + f*x]]*((a + I*b)*AppellF1[3/2, 1/4, 5/4, 5/2, (a - I*b)/(a + b*Tan[e + f*x])
, (a + I*b)/(a + b*Tan[e + f*x])] + (a - I*b)*AppellF1[3/2, 5/4, 1/4, 5/2, (a - I*b)/(a + b*Tan[e + f*x]), (a
+ I*b)/(a + b*Tan[e + f*x])] + 6*AppellF1[1/2, 1/4, 1/4, 3/2, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b
*Tan[e + f*x])]*(a + b*Tan[e + f*x])))

Maple [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3633 vs. \(2 (282 ) = 564\).

Time = 9.43 (sec) , antiderivative size = 3634, normalized size of antiderivative = 10.88

method	result	size
default	\(\text {Expression too large to display}\)	\(3634\)

[In]

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

[Out]

-1/2*I*d/f*(cos(f*x+e)+1)*(-I*(a^2+b^2)^(3/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^
4)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*ln(2)*(-cos(f*x+e)/(cos(f*x+e
)+1)^2)^(1/2)*a^2-I*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*arctanh(1/2*(cos(
f*x+e)*(a^2+b^2)^(1/2)*b+a^2*cos(f*x+e)+b^2*cos(f*x+e)-b*(a^2+b^2)^(1/2)-b^2)/(cos(f*x+e)+1)/(-cos(f*x+e)/(cos
(f*x+e)+1)^2)^(1/2)/(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)/a^2)*(-cos(f*x+e)/
(cos(f*x+e)+1)^2)^(1/2)*a^2*b^3-I*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*arct
anh(1/2*(cos(f*x+e)*(a^2+b^2)^(1/2)*b-a^2*cos(f*x+e)-b^2*cos(f*x+e)-b*(a^2+b^2)^(1/2)+b^2)/(cos(f*x+e)+1)/(-co
s(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)/a^2)
*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*a^4*b-I*(a^2+b^2)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*
a^2*b+2*b^3)/a^4)^(1/2)*arctanh(1/2*(cos(f*x+e)*(a^2+b^2)^(1/2)*b-a^2*cos(f*x+e)-b^2*cos(f*x+e)-b*(a^2+b^2)^(1
/2)+b^2)/(cos(f*x+e)+1)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*
a^2*b-2*b^3)/a^4)^(1/2)/a^2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*b^4+I*(a^2+b^2)^(3/2)*(b*((a^2+b^2)^(1/2)*a^
2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*arctanh(1/2*(cos(f*x+e)*(a^2+b^2)^(1/2)*b-a^2*cos(f*x+e)-b^2
*cos(f*x+e)-b*(a^2+b^2)^(1/2)+b^2)/(cos(f*x+e)+1)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(-b*((a^2+b^2)^(1/2)*a^
2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)/a^2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*b^2+I*(a^2+b^2)^(1
/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^
2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(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)/(c
os(f*x+e)+1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1))*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*a^2*b^2-4*(a^2+b^2)^(
3/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b
^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*(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+4*(a^2+b^2)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3
)/a^4)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*(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^3+4*b*a^3*(1/(cos(f*x+e)+1))^(1/2
)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticPi(I*(csc(f*x+e)-cot(f*x+e)),-1/(-b+(a^2+b^2)^(1/2))^2*a^2,I)*(b*(
(a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)
*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*(a^2+b^2)^(1/2)+4*b*a^3*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1
/2)*EllipticPi(I*(csc(f*x+e)-cot(f*x+e)),-1/(b+(a^2+b^2)^(1/2))^2*a^2,I)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(
1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*(a
^2+b^2)^(1/2)-I*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*arctanh(1/2*(cos(f*x+
e)*(a^2+b^2)^(1/2)*b+a^2*cos(f*x+e)+b^2*cos(f*x+e)-b*(a^2+b^2)^(1/2)-b^2)/(cos(f*x+e)+1)/(-cos(f*x+e)/(cos(f*x
+e)+1)^2)^(1/2)/(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)/a^2)*(-cos(f*x+e)/(cos
(f*x+e)+1)^2)^(1/2)*a^4*b-I*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*arctanh(1/
2*(cos(f*x+e)*(a^2+b^2)^(1/2)*b-a^2*cos(f*x+e)-b^2*cos(f*x+e)-b*(a^2+b^2)^(1/2)+b^2)/(cos(f*x+e)+1)/(-cos(f*x+
e)/(cos(f*x+e)+1)^2)^(1/2)/(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)/a^2)*(-cos
(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*a^2*b^3+I*(a^2+b^2)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2
*b-2*b^3)/a^4)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*ln(2)*(-cos(f*x+e
)/(cos(f*x+e)+1)^2)^(1/2)*a^2*b^2-I*(a^2+b^2)^(3/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b
^3)/a^4)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(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))*(-cos(f*x+
e)/(cos(f*x+e)+1)^2)^(1/2)*a^2-I*(a^2+b^2)^(3/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)
/a^4)^(1/2)*arctanh(1/2*(cos(f*x+e)*(a^2+b^2)^(1/2)*b+a^2*cos(f*x+e)+b^2*cos(f*x+e)-b*(a^2+b^2)^(1/2)-b^2)/(co
s(f*x+e)+1)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/
a^4)^(1/2)/a^2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*b^2-I*(a^2+b^2)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2
)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(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))*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*a^2*b^2+I*(a^2+b^2)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+
b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*arctanh(1/2*(cos(f*x+e)*(a^2+b^2)^(1/2)*b+a^2*cos(f*x+e)+b^2*cos(f*x+
e)-b*(a^2+b^2)^(1/2)-b^2)/(cos(f*x+e)+1)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b
^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)/a^2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*b^4+I*(a^2+b^2)^(3/2)*(-b*((
a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b
^2+2*a^2*b+2*b^3)/a^4)^(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))*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*a^2)*(d*sec(f*x+e))^(1/2)/b/(
a^2+b^2)^(1/2)/(b+(a^2+b^2)^(1/2))/a^2/(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)
/(-b+(a^2+b^2)^(1/2))/(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)

Fricas [F(-2)]

Exception generated. \[ \int \frac {(d \sec (e+f x))^{3/2}}{a+b \tan (e+f x)} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   catdef: division by zero

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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