\(\int \frac {1}{\sqrt {e x} (c+d x) \sqrt {a+b x^2}} \, dx\) [1389]

Optimal result

Integrand size = 26, antiderivative size = 371 \[ \int \frac {1}{\sqrt {e x} (c+d x) \sqrt {a+b x^2}} \, dx=\frac {\sqrt {d} \arctan \left (\frac {\sqrt {b c^2+a d^2} \sqrt {e x}}{\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a+b x^2}}\right )}{\sqrt {c} \sqrt {b c^2+a d^2} \sqrt {e}}+\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{\sqrt [4]{a} \left (\sqrt {b} c-\sqrt {a} d\right ) \sqrt {e} \sqrt {a+b x^2}}-\frac {\left (\sqrt {b} c+\sqrt {a} d\right ) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticPi}\left (-\frac {\left (\sqrt {b} c-\sqrt {a} d\right )^2}{4 \sqrt {a} \sqrt {b} c d},2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} c \left (\sqrt {b} c-\sqrt {a} d\right ) \sqrt {e} \sqrt {a+b x^2}} \] Output:

d^(1/2)*arctan((a*d^2+b*c^2)^(1/2)*(e*x)^(1/2)/c^(1/2)/d^(1/2)/e^(1/2)/(b* 
x^2+a)^(1/2))/c^(1/2)/(a*d^2+b*c^2)^(1/2)/e^(1/2)+b^(1/4)*(a^(1/2)+b^(1/2) 
*x)*((b*x^2+a)/(a^(1/2)+b^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/ 
4)*(e*x)^(1/2)/a^(1/4)/e^(1/2)),1/2*2^(1/2))/a^(1/4)/(b^(1/2)*c-a^(1/2)*d) 
/e^(1/2)/(b*x^2+a)^(1/2)-1/2*(b^(1/2)*c+a^(1/2)*d)*(a^(1/2)+b^(1/2)*x)*((b 
*x^2+a)/(a^(1/2)+b^(1/2)*x)^2)^(1/2)*EllipticPi(sin(2*arctan(b^(1/4)*(e*x) 
^(1/2)/a^(1/4)/e^(1/2))),-1/4*(b^(1/2)*c-a^(1/2)*d)^2/a^(1/2)/b^(1/2)/c/d, 
1/2*2^(1/2))/a^(1/4)/b^(1/4)/c/(b^(1/2)*c-a^(1/2)*d)/e^(1/2)/(b*x^2+a)^(1/ 
2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 22.55 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.39 \[ \int \frac {1}{\sqrt {e x} (c+d x) \sqrt {a+b x^2}} \, dx=\frac {2 i \sqrt {1+\frac {a}{b x^2}} x^{3/2} \left (\operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}{\sqrt {x}}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {i \sqrt {b} c}{\sqrt {a} d},i \text {arcsinh}\left (\frac {\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}{\sqrt {x}}\right ),-1\right )\right )}{\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}} c \sqrt {e x} \sqrt {a+b x^2}} \] Input:

Integrate[1/(Sqrt[e*x]*(c + d*x)*Sqrt[a + b*x^2]),x]
 

Output:

((2*I)*Sqrt[1 + a/(b*x^2)]*x^(3/2)*(EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[a])/S 
qrt[b]]/Sqrt[x]], -1] - EllipticPi[((-I)*Sqrt[b]*c)/(Sqrt[a]*d), I*ArcSinh 
[Sqrt[(I*Sqrt[a])/Sqrt[b]]/Sqrt[x]], -1]))/(Sqrt[(I*Sqrt[a])/Sqrt[b]]*c*Sq 
rt[e*x]*Sqrt[a + b*x^2])
 

Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {616, 27, 1541, 27, 761, 2221}

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.

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

\(\Big \downarrow \) 616

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1541

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 2221

\(\displaystyle 2 \left (\frac {\sqrt [4]{b} \left (\sqrt {a} e+\sqrt {b} e x\right ) \sqrt {\frac {a e^2+b e^2 x^2}{\left (\sqrt {a} e+\sqrt {b} e x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} e^{3/2} \sqrt {a+b x^2} \left (\sqrt {b} c-\sqrt {a} d\right )}-\frac {d \left (\frac {\left (\sqrt {a} d+\sqrt {b} c\right ) \left (\sqrt {a} e+\sqrt {b} e x\right ) \sqrt {\frac {a e^2+b e^2 x^2}{\left (\sqrt {a} e+\sqrt {b} e x\right )^2}} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \left (\frac {\sqrt {b} c}{\sqrt {a}}-d\right )^2}{4 \sqrt {b} c d},2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{b} c d \sqrt {e} \sqrt {a+b x^2}}-\frac {\sqrt {e} \left (\sqrt {b} c-\sqrt {a} d\right ) \arctan \left (\frac {\sqrt {e x} \sqrt {a d^2+b c^2}}{\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a+b x^2}}\right )}{2 \sqrt {c} \sqrt {d} \sqrt {a d^2+b c^2}}\right )}{e \left (\sqrt {b} c-\sqrt {a} d\right )}\right )\)

Input:

Int[1/(Sqrt[e*x]*(c + d*x)*Sqrt[a + b*x^2]),x]
 

Output:

2*((b^(1/4)*(Sqrt[a]*e + Sqrt[b]*e*x)*Sqrt[(a*e^2 + b*e^2*x^2)/(Sqrt[a]*e 
+ Sqrt[b]*e*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e]) 
], 1/2])/(2*a^(1/4)*(Sqrt[b]*c - Sqrt[a]*d)*e^(3/2)*Sqrt[a + b*x^2]) - (d* 
(-1/2*((Sqrt[b]*c - Sqrt[a]*d)*Sqrt[e]*ArcTan[(Sqrt[b*c^2 + a*d^2]*Sqrt[e* 
x])/(Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a + b*x^2])])/(Sqrt[c]*Sqrt[d]*Sqrt[b*c^ 
2 + a*d^2]) + ((Sqrt[b]*c + Sqrt[a]*d)*(Sqrt[a]*e + Sqrt[b]*e*x)*Sqrt[(a*e 
^2 + b*e^2*x^2)/(Sqrt[a]*e + Sqrt[b]*e*x)^2]*EllipticPi[-1/4*(Sqrt[a]*((Sq 
rt[b]*c)/Sqrt[a] - d)^2)/(Sqrt[b]*c*d), 2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1 
/4)*Sqrt[e])], 1/2])/(4*a^(1/4)*b^(1/4)*c*d*Sqrt[e]*Sqrt[a + b*x^2])))/((S 
qrt[b]*c - Sqrt[a]*d)*e))
 

Defintions of rubi rules used

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 616

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e   Subst[Int[x^(k*(m + 1) - 
 1)*(c + d*(x^k/e))^n*(a + b*(x^(2*k)/e^2))^p, x], x, (e*x)^(1/k)], x]] /; 
FreeQ[{a, b, c, d, e, p}, x] && ILtQ[n, 0] && FractionQ[m]
 

rule 761

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

rule 1541

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

rule 2221

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

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.37

method	result	size
default	\(\frac {\sqrt {-a b}\, \operatorname {EllipticPi}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {-a b}\, d}{\sqrt {-a b}\, d -b c}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}}{\sqrt {b \,x^{2}+a}\, \left (b c -\sqrt {-a b}\, d \right ) \sqrt {e x}}\)	\(139\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) e x}\, \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticPi}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, -\frac {\sqrt {-a b}}{b \left (-\frac {\sqrt {-a b}}{b}+\frac {c}{d}\right )}, \frac {\sqrt {2}}{2}\right )}{\sqrt {e x}\, \sqrt {b \,x^{2}+a}\, d b \sqrt {b e \,x^{3}+a e x}\, \left (-\frac {\sqrt {-a b}}{b}+\frac {c}{d}\right )}\)	\(188\)

Input:

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

Output:

(-a*b)^(1/2)*EllipticPi(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),(-a*b)^(1/ 
2)*d/((-a*b)^(1/2)*d-b*c),1/2*2^(1/2))*(-1/(-a*b)^(1/2)*b*x)^(1/2)*((-b*x+ 
(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)* 
2^(1/2)/(b*x^2+a)^(1/2)/(b*c-(-a*b)^(1/2)*d)/(e*x)^(1/2)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {e x} (c+d x) \sqrt {a+b x^2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{\sqrt {e x} (c+d x) \sqrt {a+b x^2}} \, dx=\int \frac {1}{\sqrt {e x} \sqrt {a + b x^{2}} \left (c + d x\right )}\, dx \] Input:

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

Output:

Integral(1/(sqrt(e*x)*sqrt(a + b*x**2)*(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt {e x} (c+d x) \sqrt {a+b x^2}} \, dx=\int { \frac {1}{\sqrt {b x^{2} + a} {\left (d x + c\right )} \sqrt {e x}} \,d x } \] Input:

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

Output:

integrate(1/(sqrt(b*x^2 + a)*(d*x + c)*sqrt(e*x)), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt {e x} (c+d x) \sqrt {a+b x^2}} \, dx=\int { \frac {1}{\sqrt {b x^{2} + a} {\left (d x + c\right )} \sqrt {e x}} \,d x } \] Input:

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

Output:

integrate(1/(sqrt(b*x^2 + a)*(d*x + c)*sqrt(e*x)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {e x} (c+d x) \sqrt {a+b x^2}} \, dx=\int \frac {1}{\sqrt {e\,x}\,\sqrt {b\,x^2+a}\,\left (c+d\,x\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt {e x} (c+d x) \sqrt {a+b x^2}} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b d \,x^{4}+b c \,x^{3}+a d \,x^{2}+a c x}d x \right )}{e} \] Input:

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

Output:

(sqrt(e)*int((sqrt(x)*sqrt(a + b*x**2))/(a*c*x + a*d*x**2 + b*c*x**3 + b*d 
*x**4),x))/e