Integrand size = 28, antiderivative size = 428 \[ \int \frac {1}{\sqrt {2+3 x} \sqrt {4+5 x} \left (1+x^2\right )^{3/2}} \, dx=\frac {(22-7 x) \sqrt {2+3 x} \sqrt {4+5 x}}{533 \sqrt {1+x^2}}+\frac {7 \left (123+5 \sqrt {533}\right ) \sqrt {2+3 x} \sqrt {4+5 x} \sqrt {1+x^2}}{533 \left (82+4 \sqrt {533}+\left (123+5 \sqrt {533}\right ) x\right )}+\frac {7 (2+3 x) \sqrt {\frac {1+x^2}{(2+3 x)^2}} \left (41+\frac {\sqrt {533} (4+5 x)}{2+3 x}\right ) \sqrt {\frac {41-\frac {46 (4+5 x)}{2+3 x}+\frac {13 (4+5 x)^2}{(2+3 x)^2}}{\left (41+\frac {\sqrt {533} (4+5 x)}{2+3 x}\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{\frac {13}{41}} \sqrt {4+5 x}}{\sqrt {2+3 x}}\right )|\frac {533+23 \sqrt {533}}{1066}\right )}{533^{3/4} \sqrt {1+x^2} \sqrt {41-\frac {46 (4+5 x)}{2+3 x}+\frac {13 (4+5 x)^2}{(2+3 x)^2}}}-\frac {\left (1763+79 \sqrt {533}\right ) \left (82+4 \sqrt {533}+\left (123+5 \sqrt {533}\right ) x\right ) \sqrt {\frac {1+x^2}{\left (82+4 \sqrt {533}+\left (123+5 \sqrt {533}\right ) x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{\frac {13}{41}} \sqrt {4+5 x}}{\sqrt {2+3 x}}\right ),\frac {1}{2}+\frac {23}{2 \sqrt {533}}\right )}{41 \sqrt [4]{533} \left (65+3 \sqrt {533}\right ) \sqrt {1+x^2}} \] Output:
1/533*(22-7*x)*(2+3*x)^(1/2)*(4+5*x)^(1/2)/(x^2+1)^(1/2)+7*(5*533^(1/2)+12 3)*(2+3*x)^(1/2)*(4+5*x)^(1/2)*(x^2+1)^(1/2)/(43706+2132*533^(1/2)+533*(5* 533^(1/2)+123)*x)+7/533*(2+3*x)*((x^2+1)/(2+3*x)^2)^(1/2)*(41+533^(1/2)*(4 +5*x)/(2+3*x))*((41-46*(4+5*x)/(2+3*x)+13*(4+5*x)^2/(2+3*x)^2)/(41+533^(1/ 2)*(4+5*x)/(2+3*x))^2)^(1/2)*EllipticE(sin(2*arctan(1/41*13^(1/4)*41^(3/4) *(4+5*x)^(1/2)/(2+3*x)^(1/2))),1/1066*(568178+24518*533^(1/2))^(1/2))*533^ (1/4)/(x^2+1)^(1/2)/(41-46*(4+5*x)/(2+3*x)+13*(4+5*x)^2/(2+3*x)^2)^(1/2)-1 /21853*(1763+79*533^(1/2))*((5*533^(1/2)+123)*x+4*533^(1/2)+82)*((x^2+1)/( (5*533^(1/2)+123)*x+4*533^(1/2)+82)^2)^(1/2)*InverseJacobiAM(2*arctan(1/41 *13^(1/4)*41^(3/4)*(4+5*x)^(1/2)/(2+3*x)^(1/2)),1/1066*(568178+24518*533^( 1/2))^(1/2))*533^(3/4)/(65+3*533^(1/2))/(x^2+1)^(1/2)
Result contains complex when optimal does not.
Time = 18.58 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.61 \[ \int \frac {1}{\sqrt {2+3 x} \sqrt {4+5 x} \left (1+x^2\right )^{3/2}} \, dx=\frac {\sqrt {2+3 x} \sqrt {\frac {(23+2 i) (4+5 x)}{2+3 x}} \left ((23-2 i) \sqrt {\frac {(4-5 i) (-i+x)}{2+3 x}} (5+4 x) \sqrt {\frac {(23+2 i) (4+5 x)}{2+3 x}}+(28-35 i) \sqrt {41} (-i+x) \sqrt {\frac {(4+5 i) (i+x)}{2+3 x}} E\left (\arcsin \left (\frac {\sqrt {\frac {(23+2 i) (4+5 x)}{2+3 x}}}{\sqrt {41}}\right )|\frac {525}{533}-\frac {92 i}{533}\right )+(158+110 i) \sqrt {41} (1+i x) \sqrt {\frac {(4+5 i) (i+x)}{2+3 x}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(23+2 i) (4+5 x)}{2+3 x}}}{\sqrt {41}}\right ),\frac {525}{533}-\frac {92 i}{533}\right )\right )}{21853 \sqrt {\frac {(4-5 i) (-i+x)}{2+3 x}} \sqrt {4+5 x} \sqrt {1+x^2}} \] Input:
Integrate[1/(Sqrt[2 + 3*x]*Sqrt[4 + 5*x]*(1 + x^2)^(3/2)),x]
Output:
(Sqrt[2 + 3*x]*Sqrt[((23 + 2*I)*(4 + 5*x))/(2 + 3*x)]*((23 - 2*I)*Sqrt[((4 - 5*I)*(-I + x))/(2 + 3*x)]*(5 + 4*x)*Sqrt[((23 + 2*I)*(4 + 5*x))/(2 + 3* x)] + (28 - 35*I)*Sqrt[41]*(-I + x)*Sqrt[((4 + 5*I)*(I + x))/(2 + 3*x)]*El lipticE[ArcSin[Sqrt[((23 + 2*I)*(4 + 5*x))/(2 + 3*x)]/Sqrt[41]], 525/533 - (92*I)/533] + (158 + 110*I)*Sqrt[41]*(1 + I*x)*Sqrt[((4 + 5*I)*(I + x))/( 2 + 3*x)]*EllipticF[ArcSin[Sqrt[((23 + 2*I)*(4 + 5*x))/(2 + 3*x)]/Sqrt[41] ], 525/533 - (92*I)/533]))/(21853*Sqrt[((4 - 5*I)*(-I + x))/(2 + 3*x)]*Sqr t[4 + 5*x]*Sqrt[1 + x^2])
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.
| \(\displaystyle \int \frac {1}{\sqrt {3 x+2} \sqrt {5 x+4} \left (x^2+1\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 744 |
| \(\displaystyle \int \frac {1}{\sqrt {3 x+2} \sqrt {5 x+4} \left (x^2+1\right )^{3/2}}dx\) |
Input:
Int[1/(Sqrt[2 + 3*x]*Sqrt[4 + 5*x]*(1 + x^2)^(3/2)),x]
Output:
$Aborted
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_ )^2)^(p_), x_Symbol] :> Unintegrable[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n, p}, x]
Result contains complex when optimal does not.
Time = 1.20 (sec) , antiderivative size = 497, normalized size of antiderivative = 1.16
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(497\) |
| default | \(\text {Expression too large to display}\) | \(1281\) |
Input:
int(1/(3*x+2)^(1/2)/(5*x+4)^(1/2)/(x^2+1)^(3/2),x,method=_RETURNVERBOSE)
Output:
((3*x+2)*(5*x+4)*(x^2+1))^(1/2)/(3*x+2)^(1/2)/(5*x+4)^(1/2)/(x^2+1)^(1/2)* (-2*(15*x^2+22*x+8)*(-11/533+7/1066*x)/((x^2+1)*(15*x^2+22*x+8))^(1/2)+(39 905/65559-3470/65559*I)*((69/65+6/65*I)*(x+2/3)/(x+4/5))^(1/2)*(x+4/5)^2*( (-4/65+6/65*I)*(x-I)/(x+4/5))^(1/2)*((-4/65-6/65*I)*(x+I)/(x+4/5))^(1/2)*1 5^(1/2)/((x+2/3)*(x+4/5)*(x-I)*(x+I))^(1/2)*EllipticF(((69/65+6/65*I)*(x+2 /3)/(x+4/5))^(1/2),23/533*533^(1/2)-2/533*I*533^(1/2))+(8855/65559-770/655 59*I)*((69/65+6/65*I)*(x+2/3)/(x+4/5))^(1/2)*(x+4/5)^2*((-4/65+6/65*I)*(x- I)/(x+4/5))^(1/2)*((-4/65-6/65*I)*(x+I)/(x+4/5))^(1/2)*15^(1/2)/((x+2/3)*( x+4/5)*(x-I)*(x+I))^(1/2)*(-4/5*EllipticF(((69/65+6/65*I)*(x+2/3)/(x+4/5)) ^(1/2),23/533*533^(1/2)-2/533*I*533^(1/2))+2/15*EllipticPi(((69/65+6/65*I) *(x+2/3)/(x+4/5))^(1/2),115/123-10/123*I,23/533*533^(1/2)-2/533*I*533^(1/2 )))+7/533*((x+2/3)*(x-I)*(x+I)+(-2/3+I)*((69/65+6/65*I)*(x+2/3)/(x+4/5))^( 1/2)*(x+4/5)^2*((-4/65+6/65*I)*(x-I)/(x+4/5))^(1/2)*((-4/65-6/65*I)*(x+I)/ (x+4/5))^(1/2)*((-151/41-240/41*I)*EllipticF(((69/65+6/65*I)*(x+2/3)/(x+4/ 5))^(1/2),23/533*533^(1/2)-2/533*I*533^(1/2))+(5+15/2*I)*EllipticE(((69/65 +6/65*I)*(x+2/3)/(x+4/5))^(1/2),23/533*533^(1/2)-2/533*I*533^(1/2))+(88/12 3+110/123*I)*EllipticPi(((69/65+6/65*I)*(x+2/3)/(x+4/5))^(1/2),115/123-10/ 123*I,23/533*533^(1/2)-2/533*I*533^(1/2))))*15^(1/2)/((x+2/3)*(x+4/5)*(x-I )*(x+I))^(1/2))
\[ \int \frac {1}{\sqrt {2+3 x} \sqrt {4+5 x} \left (1+x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{2} + 1\right )}^{\frac {3}{2}} \sqrt {5 \, x + 4} \sqrt {3 \, x + 2}} \,d x } \] Input:
integrate(1/(2+3*x)^(1/2)/(4+5*x)^(1/2)/(x^2+1)^(3/2),x, algorithm="fricas ")
Output:
integral(sqrt(x^2 + 1)*sqrt(5*x + 4)*sqrt(3*x + 2)/(15*x^6 + 22*x^5 + 38*x ^4 + 44*x^3 + 31*x^2 + 22*x + 8), x)
\[ \int \frac {1}{\sqrt {2+3 x} \sqrt {4+5 x} \left (1+x^2\right )^{3/2}} \, dx=\int \frac {1}{\sqrt {3 x + 2} \sqrt {5 x + 4} \left (x^{2} + 1\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(2+3*x)**(1/2)/(4+5*x)**(1/2)/(x**2+1)**(3/2),x)
Output:
Integral(1/(sqrt(3*x + 2)*sqrt(5*x + 4)*(x**2 + 1)**(3/2)), x)
\[ \int \frac {1}{\sqrt {2+3 x} \sqrt {4+5 x} \left (1+x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{2} + 1\right )}^{\frac {3}{2}} \sqrt {5 \, x + 4} \sqrt {3 \, x + 2}} \,d x } \] Input:
integrate(1/(2+3*x)^(1/2)/(4+5*x)^(1/2)/(x^2+1)^(3/2),x, algorithm="maxima ")
Output:
integrate(1/((x^2 + 1)^(3/2)*sqrt(5*x + 4)*sqrt(3*x + 2)), x)
\[ \int \frac {1}{\sqrt {2+3 x} \sqrt {4+5 x} \left (1+x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{2} + 1\right )}^{\frac {3}{2}} \sqrt {5 \, x + 4} \sqrt {3 \, x + 2}} \,d x } \] Input:
integrate(1/(2+3*x)^(1/2)/(4+5*x)^(1/2)/(x^2+1)^(3/2),x, algorithm="giac")
Output:
integrate(1/((x^2 + 1)^(3/2)*sqrt(5*x + 4)*sqrt(3*x + 2)), x)
Timed out. \[ \int \frac {1}{\sqrt {2+3 x} \sqrt {4+5 x} \left (1+x^2\right )^{3/2}} \, dx=\int \frac {1}{\sqrt {3\,x+2}\,\sqrt {5\,x+4}\,{\left (x^2+1\right )}^{3/2}} \,d x \] Input:
int(1/((3*x + 2)^(1/2)*(5*x + 4)^(1/2)*(x^2 + 1)^(3/2)),x)
Output:
int(1/((3*x + 2)^(1/2)*(5*x + 4)^(1/2)*(x^2 + 1)^(3/2)), x)
\[ \int \frac {1}{\sqrt {2+3 x} \sqrt {4+5 x} \left (1+x^2\right )^{3/2}} \, dx=\int \frac {\sqrt {3 x +2}\, \sqrt {5 x +4}\, \sqrt {x^{2}+1}}{15 x^{6}+22 x^{5}+38 x^{4}+44 x^{3}+31 x^{2}+22 x +8}d x \] Input:
int(1/(2+3*x)^(1/2)/(4+5*x)^(1/2)/(x^2+1)^(3/2),x)
Output:
int((sqrt(3*x + 2)*sqrt(5*x + 4)*sqrt(x**2 + 1))/(15*x**6 + 22*x**5 + 38*x **4 + 44*x**3 + 31*x**2 + 22*x + 8),x)