\(\int \frac {a+a \sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx\) [107]

Optimal result

Integrand size = 23, antiderivative size = 187 \[ \int \frac {a+a \sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx=-\frac {a \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d \sqrt {e}}+\frac {a \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d \sqrt {e}}+\frac {a \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}+\sqrt {e} \tan (c+d x)}\right )}{\sqrt {2} d \sqrt {e}}+\frac {a \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{d \sqrt {e \tan (c+d x)}} \] Output:

-1/2*a*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/d/e^(1/2)+1/ 
2*a*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/d/e^(1/2)+1/2*a 
*arctanh(2^(1/2)*(e*tan(d*x+c))^(1/2)/(e^(1/2)+e^(1/2)*tan(d*x+c)))*2^(1/2 
)/d/e^(1/2)+a*InverseJacobiAM(c-1/4*Pi+d*x,2^(1/2))*sec(d*x+c)*sin(2*d*x+2 
*c)^(1/2)/d/(e*tan(d*x+c))^(1/2)
 

Mathematica [C] (warning: unable to verify)

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

Time = 0.85 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.18 \[ \int \frac {a+a \sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx=\frac {20 a \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x)) \sin (c+d x)}{d \left (2 \left (2 \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},2,\frac {9}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-\operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},1,\frac {9}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+5 \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right ) \sqrt {e \tan (c+d x)}} \] Input:

Integrate[(a + a*Sec[c + d*x])/Sqrt[e*Tan[c + d*x]],x]
 

Output:

(20*a*AppellF1[1/4, 1/2, 1, 5/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]* 
Cos[(c + d*x)/2]^2*(1 + Sec[c + d*x])*Sin[c + d*x])/(d*(2*(2*AppellF1[5/4, 
 1/2, 2, 9/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[5/4, 3/2 
, 1, 9/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(-1 + Cos[c + d*x]) + 
5*AppellF1[1/4, 1/2, 1, 5/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + 
 Cos[c + d*x]))*Sqrt[e*Tan[c + d*x]])
 

Rubi [A] (warning: unable to verify)

Time = 0.73 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.23, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.783, Rules used = {3042, 4372, 3042, 3094, 3042, 3053, 3042, 3120, 3957, 266, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

\(\displaystyle \int \frac {a \sec (c+d x)+a}{\sqrt {e \tan (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 4372

\(\displaystyle a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx+a \int \frac {\sec (c+d x)}{\sqrt {e \tan (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx+a \int \frac {\sec (c+d x)}{\sqrt {e \tan (c+d x)}}dx\)

\(\Big \downarrow \) 3094

\(\displaystyle a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx+\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}}dx}{\sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx+\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}}dx}{\sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}}\)

\(\Big \downarrow \) 3053

\(\displaystyle a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx+\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \int \frac {1}{\sqrt {\sin (2 c+2 d x)}}dx}{\sqrt {e \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx+\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \int \frac {1}{\sqrt {\sin (2 c+2 d x)}}dx}{\sqrt {e \tan (c+d x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx+\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {a e \int \frac {1}{\sqrt {e \tan (c+d x)} \left (\tan ^2(c+d x) e^2+e^2\right )}d(e \tan (c+d x))}{d}+\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 a e \int \frac {1}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{d}+\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {2 a e \left (\frac {\int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}+\frac {\int \frac {e^2 \tan ^2(c+d x)+e}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}\right )}{d}+\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {2 a e \left (\frac {\frac {1}{2} \int \frac {1}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}+\frac {1}{2} \int \frac {1}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 e}+\frac {\int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}\right )}{d}+\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {2 a e \left (\frac {\frac {\int \frac {1}{-e^2 \tan ^2(c+d x)-1}d\left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}-\frac {\int \frac {1}{-e^2 \tan ^2(c+d x)-1}d\left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}\right )}{d}+\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {2 a e \left (\frac {\int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}\right )}{d}+\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {2 a e \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}\right )}{d}+\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 a e \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}\right )}{d}+\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 a e \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}+\frac {\int \frac {\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {e}}}{2 e}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}\right )}{d}+\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {2 a e \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\log \left (\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\log \left (-\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}}{2 e}\right )}{d}+\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\)

Input:

Int[(a + a*Sec[c + d*x])/Sqrt[e*Tan[c + d*x]],x]
 

Output:

(2*a*e*((-(ArcTan[1 - Sqrt[2]*Sqrt[e]*Tan[c + d*x]]/(Sqrt[2]*Sqrt[e])) + A 
rcTan[1 + Sqrt[2]*Sqrt[e]*Tan[c + d*x]]/(Sqrt[2]*Sqrt[e]))/(2*e) + (-1/2*L 
og[e - Sqrt[2]*e^(3/2)*Tan[c + d*x] + e^2*Tan[c + d*x]^2]/(Sqrt[2]*Sqrt[e] 
) + Log[e + Sqrt[2]*e^(3/2)*Tan[c + d*x] + e^2*Tan[c + d*x]^2]/(2*Sqrt[2]* 
Sqrt[e]))/(2*e)))/d + (a*EllipticF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*Sqrt[Si 
n[2*c + 2*d*x]])/(d*Sqrt[e*Tan[c + d*x]])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 755

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

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

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

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 3042

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

rule 3053

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ 
)]]), x_Symbol] :> Simp[Sqrt[Sin[2*e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b 
*Cos[e + f*x]])   Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f 
}, x]
 

rule 3094

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] 
:> Simp[Sqrt[Sin[e + f*x]]/(Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]])   Int[ 
1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x]
 

rule 3120

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 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

rule 4372

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(e*Cot[c + d*x])^m, x], x] + Simp[b   Int[ 
(e*Cot[c + d*x])^m*Csc[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, m}, x]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.67 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.75

method	result	size
default	\(\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) a \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \left (\operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+i \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )\right ) \left (1+\sec \left (d x +c \right )\right )}{d \sqrt {e \tan \left (d x +c \right )}}\)	\(140\)
parts	\(\frac {a \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \tan \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \tan \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d e}+\frac {a \operatorname {EllipticF}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \left (1+\sec \left (d x +c \right )\right )}{d \sqrt {e \tan \left (d x +c \right )}}\)	\(242\)

Input:

int((a+a*sec(d*x+c))/(e*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

(1/2-1/2*I)*a/d*(-csc(d*x+c)+cot(d*x+c))^(1/2)*(2*cot(d*x+c)-2*csc(d*x+c)+ 
2)^(1/2)*(-cot(d*x+c)+csc(d*x+c)+1)^(1/2)*(EllipticPi((-cot(d*x+c)+csc(d*x 
+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))+I*EllipticPi((-cot(d*x+c)+csc(d*x+c)+1 
)^(1/2),1/2-1/2*I,1/2*2^(1/2)))/(e*tan(d*x+c))^(1/2)*(1+sec(d*x+c))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {a+a \sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sec(d*x+c))/(e*tan(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {a+a \sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx=a \left (\int \frac {1}{\sqrt {e \tan {\left (c + d x \right )}}}\, dx + \int \frac {\sec {\left (c + d x \right )}}{\sqrt {e \tan {\left (c + d x \right )}}}\, dx\right ) \] Input:

integrate((a+a*sec(d*x+c))/(e*tan(d*x+c))**(1/2),x)
 

Output:

a*(Integral(1/sqrt(e*tan(c + d*x)), x) + Integral(sec(c + d*x)/sqrt(e*tan( 
c + d*x)), x))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+a \sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a+a*sec(d*x+c))/(e*tan(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [F]

\[ \int \frac {a+a \sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx=\int { \frac {a \sec \left (d x + c\right ) + a}{\sqrt {e \tan \left (d x + c\right )}} \,d x } \] Input:

integrate((a+a*sec(d*x+c))/(e*tan(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

integrate((a*sec(d*x + c) + a)/sqrt(e*tan(d*x + c)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+a \sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx=\int \frac {a+\frac {a}{\cos \left (c+d\,x\right )}}{\sqrt {e\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \] Input:

int((a + a/cos(c + d*x))/(e*tan(c + d*x))^(1/2),x)
 

Output:

int((a + a/cos(c + d*x))/(e*tan(c + d*x))^(1/2), x)
 

Reduce [F]

\[ \int \frac {a+a \sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx=\frac {\sqrt {e}\, a \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )}d x +\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sec \left (d x +c \right )}{\tan \left (d x +c \right )}d x \right )}{e} \] Input:

int((a+a*sec(d*x+c))/(e*tan(d*x+c))^(1/2),x)
 

Output:

(sqrt(e)*a*(int(sqrt(tan(c + d*x))/tan(c + d*x),x) + int((sqrt(tan(c + d*x 
))*sec(c + d*x))/tan(c + d*x),x)))/e