\(\int \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx\) [31]

Optimal result

Integrand size = 26, antiderivative size = 660 \[ \int \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\frac {\sqrt {a-b+c} \arctan \left (\frac {\sqrt {a-b+c} \tan (d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e}+\frac {\sqrt {c} \tan (d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{e \left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right )}-\frac {\sqrt [4]{a} \sqrt [4]{c} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right ) \left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right ) \sqrt {\frac {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right )^2}}}{e \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}+\frac {\sqrt [4]{a} (b-2 c) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right ) \left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right ) \sqrt {\frac {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right )^2}}}{2 \left (\sqrt {a}-\sqrt {c}\right ) \sqrt [4]{c} e \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}+\frac {\left (\sqrt {a}+\sqrt {c}\right ) (a-b+c) \operatorname {EllipticPi}\left (-\frac {\left (\sqrt {a}-\sqrt {c}\right )^2}{4 \sqrt {a} \sqrt {c}},2 \arctan \left (\frac {\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right ) \left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right ) \sqrt {\frac {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right )^2}}}{4 \sqrt [4]{a} \left (\sqrt {a}-\sqrt {c}\right ) \sqrt [4]{c} e \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \] Output:

1/2*(a-b+c)^(1/2)*arctan((a-b+c)^(1/2)*tan(e*x+d)/(a+b*tan(e*x+d)^2+c*tan( 
e*x+d)^4)^(1/2))/e+c^(1/2)*tan(e*x+d)*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1 
/2)/e/(a^(1/2)+c^(1/2)*tan(e*x+d)^2)-a^(1/4)*c^(1/4)*EllipticE(sin(2*arcta 
n(c^(1/4)*tan(e*x+d)/a^(1/4))),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))*(a^(1/2)+c 
^(1/2)*tan(e*x+d)^2)*((a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)/(a^(1/2)+c^(1/2)*t 
an(e*x+d)^2)^2)^(1/2)/e/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)+1/2*a^(1/4 
)*(b-2*c)*InverseJacobiAM(2*arctan(c^(1/4)*tan(e*x+d)/a^(1/4)),1/2*(2-b/a^ 
(1/2)/c^(1/2))^(1/2))*(a^(1/2)+c^(1/2)*tan(e*x+d)^2)*((a+b*tan(e*x+d)^2+c* 
tan(e*x+d)^4)/(a^(1/2)+c^(1/2)*tan(e*x+d)^2)^2)^(1/2)/(a^(1/2)-c^(1/2))/c^ 
(1/4)/e/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)+1/4*(a^(1/2)+c^(1/2))*(a-b 
+c)*EllipticPi(sin(2*arctan(c^(1/4)*tan(e*x+d)/a^(1/4))),-1/4*(a^(1/2)-c^( 
1/2))^2/a^(1/2)/c^(1/2),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))*(a^(1/2)+c^(1/2)* 
tan(e*x+d)^2)*((a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)/(a^(1/2)+c^(1/2)*tan(e*x+ 
d)^2)^2)^(1/2)/a^(1/4)/(a^(1/2)-c^(1/2))/c^(1/4)/e/(a+b*tan(e*x+d)^2+c*tan 
(e*x+d)^4)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 11.37 (sec) , antiderivative size = 428, normalized size of antiderivative = 0.65 \[ \int \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\frac {i \left (\left (-b+\sqrt {b^2-4 a c}\right ) E\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \tan (d+e x)\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )-\left (b-2 c+\sqrt {b^2-4 a c}\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \tan (d+e x)\right ),\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )-2 (a-b+c) \operatorname {EllipticPi}\left (\frac {b+\sqrt {b^2-4 a c}}{2 c},i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \tan (d+e x)\right ),\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )\right ) \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c \tan ^2(d+e x)}{b+\sqrt {b^2-4 a c}}} \sqrt {1-\frac {2 c \tan ^2(d+e x)}{-b+\sqrt {b^2-4 a c}}}}{2 \sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} e \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \] Input:

Integrate[Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]
 

Output:

((I/2)*((-b + Sqrt[b^2 - 4*a*c])*EllipticE[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + S 
qrt[b^2 - 4*a*c])]*Tan[d + e*x]], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 
4*a*c])] - (b - 2*c + Sqrt[b^2 - 4*a*c])*EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[ 
c/(b + Sqrt[b^2 - 4*a*c])]*Tan[d + e*x]], (b + Sqrt[b^2 - 4*a*c])/(b - Sqr 
t[b^2 - 4*a*c])] - 2*(a - b + c)*EllipticPi[(b + Sqrt[b^2 - 4*a*c])/(2*c), 
 I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*Tan[d + e*x]], (b + Sqr 
t[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])])*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2* 
c*Tan[d + e*x]^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[1 - (2*c*Tan[d + e*x]^2)/( 
-b + Sqrt[b^2 - 4*a*c])])/(Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*e*Sqrt[ 
a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 694, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 4853, 1523, 25, 27, 1511, 27, 1416, 1509, 2220}

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.

\(\displaystyle \int \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {a+b \tan (d+e x)^2+c \tan (d+e x)^4}dx\)

\(\Big \downarrow \) 4853

\(\displaystyle \frac {\int \frac {\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}{\tan ^2(d+e x)+1}d\tan (d+e x)}{e}\)

\(\Big \downarrow \) 1523

\(\displaystyle \frac {\frac {(a-b+c) \int \frac {\sqrt {c} \tan ^2(d+e x)+\sqrt {a}}{\sqrt {a} \left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)}{1-\frac {\sqrt {c}}{\sqrt {a}}}-\frac {\int -\frac {\left (1-\frac {\sqrt {c}}{\sqrt {a}}\right ) c \tan ^2(d+e x)+b-c-\sqrt {a} \sqrt {c}}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)}{1-\frac {\sqrt {c}}{\sqrt {a}}}}{e}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {(a-b+c) \int \frac {\sqrt {c} \tan ^2(d+e x)+\sqrt {a}}{\sqrt {a} \left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)}{1-\frac {\sqrt {c}}{\sqrt {a}}}+\frac {\int \frac {\left (1-\frac {\sqrt {c}}{\sqrt {a}}\right ) c \tan ^2(d+e x)+b-c-\sqrt {a} \sqrt {c}}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)}{1-\frac {\sqrt {c}}{\sqrt {a}}}}{e}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {(a-b+c) \int \frac {\sqrt {c} \tan ^2(d+e x)+\sqrt {a}}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)}{\sqrt {a} \left (1-\frac {\sqrt {c}}{\sqrt {a}}\right )}+\frac {\int \frac {\left (1-\frac {\sqrt {c}}{\sqrt {a}}\right ) c \tan ^2(d+e x)+b-c-\sqrt {a} \sqrt {c}}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)}{1-\frac {\sqrt {c}}{\sqrt {a}}}}{e}\)

\(\Big \downarrow \) 1511

\(\displaystyle \frac {\frac {(b-2 c) \int \frac {1}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)-\sqrt {c} \left (\sqrt {a}-\sqrt {c}\right ) \int \frac {\sqrt {a}-\sqrt {c} \tan ^2(d+e x)}{\sqrt {a} \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)}{1-\frac {\sqrt {c}}{\sqrt {a}}}+\frac {(a-b+c) \int \frac {\sqrt {c} \tan ^2(d+e x)+\sqrt {a}}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)}{\sqrt {a} \left (1-\frac {\sqrt {c}}{\sqrt {a}}\right )}}{e}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {(b-2 c) \int \frac {1}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)-\frac {\sqrt {c} \left (\sqrt {a}-\sqrt {c}\right ) \int \frac {\sqrt {a}-\sqrt {c} \tan ^2(d+e x)}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)}{\sqrt {a}}}{1-\frac {\sqrt {c}}{\sqrt {a}}}+\frac {(a-b+c) \int \frac {\sqrt {c} \tan ^2(d+e x)+\sqrt {a}}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)}{\sqrt {a} \left (1-\frac {\sqrt {c}}{\sqrt {a}}\right )}}{e}\)

\(\Big \downarrow \) 1416

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

\(\Big \downarrow \) 1509

\(\displaystyle \frac {\frac {(a-b+c) \int \frac {\sqrt {c} \tan ^2(d+e x)+\sqrt {a}}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)}{\sqrt {a} \left (1-\frac {\sqrt {c}}{\sqrt {a}}\right )}+\frac {\frac {(b-2 c) \left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right ) \sqrt {\frac {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac {\sqrt {c} \left (\sqrt {a}-\sqrt {c}\right ) \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right ) \sqrt {\frac {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{\sqrt [4]{c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac {\tan (d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{\sqrt {a}+\sqrt {c} \tan ^2(d+e x)}\right )}{\sqrt {a}}}{1-\frac {\sqrt {c}}{\sqrt {a}}}}{e}\)

\(\Big \downarrow \) 2220

\(\displaystyle \frac {\frac {(a-b+c) \left (\frac {\left (\sqrt {a}-\sqrt {c}\right ) \arctan \left (\frac {\sqrt {a-b+c} \tan (d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt {a-b+c}}+\frac {\left (\sqrt {a}+\sqrt {c}\right ) \left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right ) \sqrt {\frac {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right )^2}} \operatorname {EllipticPi}\left (-\frac {\left (\sqrt {a}-\sqrt {c}\right )^2}{4 \sqrt {a} \sqrt {c}},2 \arctan \left (\frac {\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{4 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {a} \left (1-\frac {\sqrt {c}}{\sqrt {a}}\right )}+\frac {\frac {(b-2 c) \left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right ) \sqrt {\frac {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac {\sqrt {c} \left (\sqrt {a}-\sqrt {c}\right ) \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right ) \sqrt {\frac {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{\sqrt [4]{c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac {\tan (d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{\sqrt {a}+\sqrt {c} \tan ^2(d+e x)}\right )}{\sqrt {a}}}{1-\frac {\sqrt {c}}{\sqrt {a}}}}{e}\)

Input:

Int[Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]
 

Output:

(((a - b + c)*(((Sqrt[a] - Sqrt[c])*ArcTan[(Sqrt[a - b + c]*Tan[d + e*x])/ 
Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]])/(2*Sqrt[a - b + c]) + ((Sq 
rt[a] + Sqrt[c])*EllipticPi[-1/4*(Sqrt[a] - Sqrt[c])^2/(Sqrt[a]*Sqrt[c]), 
2*ArcTan[(c^(1/4)*Tan[d + e*x])/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4]*(Sq 
rt[a] + Sqrt[c]*Tan[d + e*x]^2)*Sqrt[(a + b*Tan[d + e*x]^2 + c*Tan[d + e*x 
]^4)/(Sqrt[a] + Sqrt[c]*Tan[d + e*x]^2)^2])/(4*a^(1/4)*c^(1/4)*Sqrt[a + b* 
Tan[d + e*x]^2 + c*Tan[d + e*x]^4])))/(Sqrt[a]*(1 - Sqrt[c]/Sqrt[a])) + (( 
(b - 2*c)*EllipticF[2*ArcTan[(c^(1/4)*Tan[d + e*x])/a^(1/4)], (2 - b/(Sqrt 
[a]*Sqrt[c]))/4]*(Sqrt[a] + Sqrt[c]*Tan[d + e*x]^2)*Sqrt[(a + b*Tan[d + e* 
x]^2 + c*Tan[d + e*x]^4)/(Sqrt[a] + Sqrt[c]*Tan[d + e*x]^2)^2])/(2*a^(1/4) 
*c^(1/4)*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]) - ((Sqrt[a] - Sqrt 
[c])*Sqrt[c]*(-((Tan[d + e*x]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4 
])/(Sqrt[a] + Sqrt[c]*Tan[d + e*x]^2)) + (a^(1/4)*EllipticE[2*ArcTan[(c^(1 
/4)*Tan[d + e*x])/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4]*(Sqrt[a] + Sqrt[c 
]*Tan[d + e*x]^2)*Sqrt[(a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4)/(Sqrt[a] 
+ Sqrt[c]*Tan[d + e*x]^2)^2])/(c^(1/4)*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d 
 + e*x]^4])))/Sqrt[a])/(1 - Sqrt[c]/Sqrt[a]))/e
 

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 1416

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c 
/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/ 
(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) 
], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
 

rule 1509

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

rule 1511

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q   Int[1/Sqrt[a + b*x^2 + c*x^ 
4], x], x] - Simp[e/q   Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x] /; 
NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && Pos 
Q[c/a]
 

rule 1523

Int[Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]/((d_) + (e_.)*(x_)^2), x_Symbo 
l] :> With[{q = Rt[c/a, 2]}, Simp[(c*d^2 - b*d*e + a*e^2)/(e*(e - d*q))   I 
nt[(1 + q*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x] - Simp[1/(e*(e 
 - d*q))   Int[(c*d - b*e + a*e*q - (c*e - a*d*q^3)*x^2)/Sqrt[a + b*x^2 + c 
*x^4], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c 
*d^2 - b*d*e + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a]
 

rule 2220

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + 
 (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]}, Simp[(-(B*d - A*e))*(A 
rcTan[Rt[-b + c*(d/e) + a*(e/d), 2]*(x/Sqrt[a + b*x^2 + c*x^4])]/(2*d*e*Rt[ 
-b + c*(d/e) + a*(e/d), 2])), x] + Simp[(B*d + A*e)*(1 + q^2*x^2)*(Sqrt[(a 
+ b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/(4*d*e*q*Sqrt[a + b*x^2 + c*x^4]))*El 
lipticPi[-(e - d*q^2)^2/(4*d*e*q^2), 2*ArcTan[q*x], 1/2 - b/(4*a*q^2)], x]] 
 /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] & 
& EqQ[c*A^2 - a*B^2, 0] && PosQ[B/A] && PosQ[-b + c*(d/e) + a*(e/d)]
 

rule 3042

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

rule 4853

Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, Simp[With[{d = FreeFa 
ctors[Tan[v], x]}, d/Coefficient[v, x, 1]   Subst[Int[SubstFor[1/(1 + d^2*x 
^2), Tan[v]/d, u, x], x], x, Tan[v]/d]], x] /;  !FalseQ[v] && FunctionOfQ[N 
onfreeFactors[Tan[v], x], u, x, True] && TryPureTanSubst[ActivateTrig[u], x 
]]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1496\) vs. \(2(548)=1096\).

Time = 0.55 (sec) , antiderivative size = 1497, normalized size of antiderivative = 2.27

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1497\)
default	\(\text {Expression too large to display}\)	\(1497\)

Input:

int((a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/e*(1/4*2^(1/2)/(-1/a*b+1/a*(-4*a*c+b^2)^(1/2))^(1/2)*(4+2/a*b*tan(e*x+d) 
^2-2/a*tan(e*x+d)^2*(-4*a*c+b^2)^(1/2))^(1/2)*(4+2/a*b*tan(e*x+d)^2+2/a*ta 
n(e*x+d)^2*(-4*a*c+b^2)^(1/2))^(1/2)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/ 
2)*EllipticF(1/2*tan(e*x+d)*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2* 
(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2))*b-1/4*2^(1/2)/(-1/a*b+1/a*(-4*a 
*c+b^2)^(1/2))^(1/2)*(4+2/a*b*tan(e*x+d)^2-2/a*tan(e*x+d)^2*(-4*a*c+b^2)^( 
1/2))^(1/2)*(4+2/a*b*tan(e*x+d)^2+2/a*tan(e*x+d)^2*(-4*a*c+b^2)^(1/2))^(1/ 
2)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)*EllipticF(1/2*tan(e*x+d)*2^(1/2 
)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c 
)^(1/2))*c-1/2*c*a*2^(1/2)/(-1/a*b+1/a*(-4*a*c+b^2)^(1/2))^(1/2)*(4+2/a*b* 
tan(e*x+d)^2-2/a*tan(e*x+d)^2*(-4*a*c+b^2)^(1/2))^(1/2)*(4+2/a*b*tan(e*x+d 
)^2+2/a*tan(e*x+d)^2*(-4*a*c+b^2)^(1/2))^(1/2)/(a+b*tan(e*x+d)^2+c*tan(e*x 
+d)^4)^(1/2)/(b+(-4*a*c+b^2)^(1/2))*EllipticF(1/2*tan(e*x+d)*2^(1/2)*((-b+ 
(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2) 
)+1/2*c*a*2^(1/2)/(-1/a*b+1/a*(-4*a*c+b^2)^(1/2))^(1/2)*(4+2/a*b*tan(e*x+d 
)^2-2/a*tan(e*x+d)^2*(-4*a*c+b^2)^(1/2))^(1/2)*(4+2/a*b*tan(e*x+d)^2+2/a*t 
an(e*x+d)^2*(-4*a*c+b^2)^(1/2))^(1/2)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1 
/2)/(b+(-4*a*c+b^2)^(1/2))*EllipticE(1/2*tan(e*x+d)*2^(1/2)*((-b+(-4*a*c+b 
^2)^(1/2))/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2))+a*2^(1/ 
2)/(-1/a*b+1/a*(-4*a*c+b^2)^(1/2))^(1/2)*(1+1/2/a*b*tan(e*x+d)^2-1/2/a*...
 

Fricas [F(-1)]

Timed out. \[ \int \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\text {Timed out} \] Input:

integrate((a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\int \sqrt {a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}}\, dx \] Input:

integrate((a+b*tan(e*x+d)**2+c*tan(e*x+d)**4)**(1/2),x)
 

Output:

Integral(sqrt(a + b*tan(d + e*x)**2 + c*tan(d + e*x)**4), x)
 

Maxima [F]

\[ \int \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\int { \sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a} \,d x } \] Input:

integrate((a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a), x)
 

Giac [F]

\[ \int \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\int { \sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a} \,d x } \] Input:

integrate((a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\int \sqrt {c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a} \,d x \] Input:

int((a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(1/2),x)
 

Output:

int((a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(1/2), x)
 

Reduce [F]

\[ \int \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\int \sqrt {\tan \left (e x +d \right )^{4} c +\tan \left (e x +d \right )^{2} b +a}d x \] Input:

int((a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x)
 

Output:

int(sqrt(tan(d + e*x)**4*c + tan(d + e*x)**2*b + a),x)