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

3.7.5.1 Optimal result

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

output

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)
 

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

input

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

output

(-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*Tan[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*AppellF 
1[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])))
 

3.7.5.3 Rubi [A] (warning: unable to verify)

Time = 0.61 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.74, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3994, 504, 310, 353, 73, 827, 218, 221, 993, 1537, 412}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3994

\(\displaystyle \frac {(d \sec (e+f x))^{3/2} \int \frac {1}{(a+b \tan (e+f x)) \sqrt [4]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 504

\(\displaystyle \frac {(d \sec (e+f x))^{3/2} \left (a \int \frac {1}{\sqrt [4]{\tan ^2(e+f x)+1} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))-\int \frac {b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 310

\(\displaystyle \frac {(d \sec (e+f x))^{3/2} \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {b^2 \tan ^2(e+f x)}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\tan ^2(e+f x)+1}}{b}-\int \frac {b \tan (e+f x)}{\sqrt [4]{\tan ^2(e+f x)+1} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 353

\(\displaystyle \frac {(d \sec (e+f x))^{3/2} \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {b^2 \tan ^2(e+f x)}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\tan ^2(e+f x)+1}}{b}-\frac {1}{2} \int \frac {1}{\sqrt [4]{\frac {\tan (e+f x)}{b}+1} \left (a^2-b^2 \tan ^2(e+f x)\right )}d\left (b^2 \tan ^2(e+f x)\right )\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(d \sec (e+f x))^{3/2} \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {b^2 \tan ^2(e+f x)}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\tan ^2(e+f x)+1}}{b}-2 b^2 \int \frac {\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{-\tan ^4(e+f x) b^6+b^2+a^2}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {(d \sec (e+f x))^{3/2} \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {b^2 \tan ^2(e+f x)}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\tan ^2(e+f x)+1}}{b}-2 b^2 \left (\frac {\int \frac {1}{\sqrt {a^2+b^2}-b^3 \tan ^2(e+f x)}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 b}-\frac {\int \frac {1}{\tan ^2(e+f x) b^3+\sqrt {a^2+b^2}}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 b}\right )\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {(d \sec (e+f x))^{3/2} \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {b^2 \tan ^2(e+f x)}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\tan ^2(e+f x)+1}}{b}-2 b^2 \left (\frac {\int \frac {1}{\sqrt {a^2+b^2}-b^3 \tan ^2(e+f x)}d\sqrt [4]{\frac {\tan (e+f x)}{b}+1}}{2 b}-\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}\right )\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(d \sec (e+f x))^{3/2} \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \int \frac {b^2 \tan ^2(e+f x)}{\sqrt {1-b^4 \tan ^4(e+f x)} \left (-b^4 \tan ^4(e+f x)+\frac {a^2}{b^2}+1\right )}d\sqrt [4]{\tan ^2(e+f x)+1}}{b}-2 b^2 \left (\frac {\text {arctanh}\left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}-\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}\right )\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 993

\(\displaystyle \frac {(d \sec (e+f x))^{3/2} \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \left (\frac {1}{2} b \int \frac {1}{\left (\sqrt {a^2+b^2}-b^3 \tan ^2(e+f x)\right ) \sqrt {1-b^4 \tan ^4(e+f x)}}d\sqrt [4]{\tan ^2(e+f x)+1}-\frac {1}{2} b \int \frac {1}{\left (\tan ^2(e+f x) b^3+\sqrt {a^2+b^2}\right ) \sqrt {1-b^4 \tan ^4(e+f x)}}d\sqrt [4]{\tan ^2(e+f x)+1}\right )}{b}-2 b^2 \left (\frac {\text {arctanh}\left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}-\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}\right )\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 1537

\(\displaystyle \frac {(d \sec (e+f x))^{3/2} \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \left (\frac {1}{2} b \int \frac {1}{\sqrt {1-b^2 \tan ^2(e+f x)} \sqrt {b^2 \tan ^2(e+f x)+1} \left (\sqrt {a^2+b^2}-b^3 \tan ^2(e+f x)\right )}d\sqrt [4]{\tan ^2(e+f x)+1}-\frac {1}{2} b \int \frac {1}{\sqrt {1-b^2 \tan ^2(e+f x)} \sqrt {b^2 \tan ^2(e+f x)+1} \left (\tan ^2(e+f x) b^3+\sqrt {a^2+b^2}\right )}d\sqrt [4]{\tan ^2(e+f x)+1}\right )}{b}-2 b^2 \left (\frac {\text {arctanh}\left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}-\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}\right )\right )}{b f \sec ^2(e+f x)^{3/4}}\)

\(\Big \downarrow \) 412

\(\displaystyle \frac {(d \sec (e+f x))^{3/2} \left (\frac {2 a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \left (\frac {b \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\tan ^2(e+f x)+1}\right ),-1\right )}{2 \sqrt {a^2+b^2}}-\frac {b \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\tan ^2(e+f x)+1}\right ),-1\right )}{2 \sqrt {a^2+b^2}}\right )}{b}-2 b^2 \left (\frac {\text {arctanh}\left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}-\frac {\arctan \left (\frac {b^{3/2} \tan (e+f x)}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} \sqrt [4]{a^2+b^2}}\right )\right )}{b f \sec ^2(e+f x)^{3/4}}\)

input

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

output

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

3.7.5.3.1 Defintions of rubi rules used

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 218

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

rule 221

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 310

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Sim 
p[2*(Sqrt[(-b)*(x^2/a)]/x)   Subst[Int[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 353

Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ 
{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
 

rule 412

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x 
_)^2]), x_Symbol] :> Simp[(1/(a*Sqrt[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] && S 
implerSqrtQ[-f/e, -d/c])
 

rule 504

Int[((a_) + (b_.)*(x_)^2)^(p_)/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[c   I 
nt[(a + b*x^2)^p/(c^2 - d^2*x^2), x], x] - Simp[d   Int[x*((a + b*x^2)^p/(c 
^2 - d^2*x^2)), x], x] /; FreeQ[{a, b, c, d, p}, x]
 

rule 827

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

rule 993

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]]}, Simp[s/(2* 
b)   Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Simp[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 1537

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3994

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_), x_Symbol] :> Simp[d^(2*IntPart[m/2])*((d*Sec[e + f*x])^(2*FracP 
art[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] && IntegerQ[n]
 

3.7.5.4 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	\(\text {Expression too large to display}\)	\(3634\)

input

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

output

-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)*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*(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)+...
 

3.7.5.5 Fricas [F(-2)]

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

input

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

output

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

3.7.5.6 Sympy [F]

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

input

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

output

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

3.7.5.7 Maxima [F]

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

input

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

output

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

3.7.5.8 Giac [F]

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

input

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

output

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

3.7.5.9 Mupad [F(-1)]

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

input

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

output

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