Integrand size = 32, antiderivative size = 1 \[ \int \frac {A+B x+C x^2}{\sqrt {d+e x} \left (2+3 x+5 x^2\right )} \, dx=0 \] Output:
0
Result contains higher order function than in optimal. Order 3 vs. order 1 in optimal.
Time = 1.09 (sec) , antiderivative size = 259, normalized size of antiderivative = 259.00 \[ \int \frac {A+B x+C x^2}{\sqrt {d+e x} \left (2+3 x+5 x^2\right )} \, dx=\frac {1}{775} \left (\frac {310 C \sqrt {d+e x}}{e}+\frac {\left (50 i \sqrt {31} A+\left (-155-15 i \sqrt {31}\right ) B+\left (93-11 i \sqrt {31}\right ) C\right ) \arctan \left (\frac {\sqrt {10} \sqrt {-10 d+3 e-i \sqrt {31} e} \sqrt {d+e x}}{10 d-3 e+i \sqrt {31} e}\right )}{\sqrt {-d+\frac {1}{10} \left (3-i \sqrt {31}\right ) e}}+\frac {\left (-50 i \sqrt {31} A+5 i \left (31 i+3 \sqrt {31}\right ) B+\left (93+11 i \sqrt {31}\right ) C\right ) \arctan \left (\frac {\sqrt {10} \sqrt {-10 d+3 e+i \sqrt {31} e} \sqrt {d+e x}}{10 d-3 e-i \sqrt {31} e}\right )}{\sqrt {-d+\frac {1}{10} \left (3+i \sqrt {31}\right ) e}}\right ) \] Input:
Integrate[(A + B*x + C*x^2)/(Sqrt[d + e*x]*(2 + 3*x + 5*x^2)),x]
Output:
((310*C*Sqrt[d + e*x])/e + (((50*I)*Sqrt[31]*A + (-155 - (15*I)*Sqrt[31])* B + (93 - (11*I)*Sqrt[31])*C)*ArcTan[(Sqrt[10]*Sqrt[-10*d + 3*e - I*Sqrt[3 1]*e]*Sqrt[d + e*x])/(10*d - 3*e + I*Sqrt[31]*e)])/Sqrt[-d + ((3 - I*Sqrt[ 31])*e)/10] + (((-50*I)*Sqrt[31]*A + (5*I)*(31*I + 3*Sqrt[31])*B + (93 + ( 11*I)*Sqrt[31])*C)*ArcTan[(Sqrt[10]*Sqrt[-10*d + 3*e + I*Sqrt[31]*e]*Sqrt[ d + e*x])/(10*d - 3*e - I*Sqrt[31]*e)])/Sqrt[-d + ((3 + I*Sqrt[31])*e)/10] )/775
Result contains higher order function than in optimal. Order 3 vs. order 1 in optimal.
Time = 0.63 (sec) , antiderivative size = 228, normalized size of antiderivative = 228.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2159, 2009}
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.
\(\displaystyle \int \frac {A+B x+C x^2}{\left (5 x^2+3 x+2\right ) \sqrt {d+e x}} \, dx\) |
\(\Big \downarrow \) 2159 |
\(\displaystyle \int \left (\frac {-\frac {i (50 A-15 B-11 C)}{5 \sqrt {31}}+B-\frac {3 C}{5}}{\left (10 x-i \sqrt {31}+3\right ) \sqrt {d+e x}}+\frac {\frac {i (50 A-15 B-11 C)}{5 \sqrt {31}}+B-\frac {3 C}{5}}{\left (10 x+i \sqrt {31}+3\right ) \sqrt {d+e x}}+\frac {C}{5 \sqrt {d+e x}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\sqrt {\frac {2}{155}} \left (50 i A-5 \left (-\sqrt {31}+3 i\right ) B-\left (3 \sqrt {31}+11 i\right ) C\right ) \text {arctanh}\left (\frac {\sqrt {10} \sqrt {d+e x}}{\sqrt {10 d+i \left (-\sqrt {31}+3 i\right ) e}}\right )}{5 \sqrt {10 d-\left (3+i \sqrt {31}\right ) e}}+\frac {\sqrt {\frac {2}{155}} \left (50 i A-5 \left (\sqrt {31}+3 i\right ) B-\left (-3 \sqrt {31}+11 i\right ) C\right ) \text {arctanh}\left (\frac {\sqrt {10} \sqrt {d+e x}}{\sqrt {10 d+i \left (\sqrt {31}+3 i\right ) e}}\right )}{5 \sqrt {10 d+i \left (\sqrt {31}+3 i\right ) e}}+\frac {2 C \sqrt {d+e x}}{5 e}\) |
Input:
Int[(A + B*x + C*x^2)/(Sqrt[d + e*x]*(2 + 3*x + 5*x^2)),x]
Output:
(2*C*Sqrt[d + e*x])/(5*e) - (Sqrt[2/155]*((50*I)*A - 5*(3*I - Sqrt[31])*B - (11*I + 3*Sqrt[31])*C)*ArcTanh[(Sqrt[10]*Sqrt[d + e*x])/Sqrt[10*d + I*(3 *I - Sqrt[31])*e]])/(5*Sqrt[10*d - (3 + I*Sqrt[31])*e]) + (Sqrt[2/155]*((5 0*I)*A - 5*(3*I + Sqrt[31])*B - (11*I - 3*Sqrt[31])*C)*ArcTanh[(Sqrt[10]*S qrt[d + e*x])/Sqrt[10*d + I*(3*I + Sqrt[31])*e]])/(5*Sqrt[10*d + I*(3*I + Sqrt[31])*e])
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 2.02 (sec) , antiderivative size = 669, normalized size of antiderivative = 669.00
method | result | size |
pseudoelliptic | \(-\frac {-31 e \left (\frac {\left (B -\frac {3 C}{5}\right ) \sqrt {5}\, \sqrt {5 d^{2}-3 d e +2 e^{2}}}{5}+\left (A -\frac {2 C}{5}\right ) e -\left (B -\frac {3 C}{5}\right ) d \right ) \arctan \left (\frac {2 \sqrt {5}\, \sqrt {e x +d}-\sqrt {2 \sqrt {5}\, \sqrt {5 d^{2}-3 d e +2 e^{2}}+10 d -3 e}}{\sqrt {2 \sqrt {5}\, \sqrt {5 d^{2}-3 d e +2 e^{2}}-10 d +3 e}}\right )-31 e \left (\frac {\left (B -\frac {3 C}{5}\right ) \sqrt {5}\, \sqrt {5 d^{2}-3 d e +2 e^{2}}}{5}+\left (A -\frac {2 C}{5}\right ) e -\left (B -\frac {3 C}{5}\right ) d \right ) \arctan \left (\frac {2 \sqrt {5}\, \sqrt {e x +d}+\sqrt {2 \sqrt {5}\, \sqrt {5 d^{2}-3 d e +2 e^{2}}+10 d -3 e}}{\sqrt {2 \sqrt {5}\, \sqrt {5 d^{2}-3 d e +2 e^{2}}-10 d +3 e}}\right )+\sqrt {2 \sqrt {5}\, \sqrt {5 d^{2}-3 d e +2 e^{2}}-10 d +3 e}\, \left (\left (\sqrt {5}\, \left (A -\frac {3 B}{10}-\frac {11 C}{50}\right ) \sqrt {5 d^{2}-3 d e +2 e^{2}}+\left (-2 B +\frac {3 C}{5}+\frac {3 A}{2}\right ) e -5 \left (A -\frac {3 B}{10}-\frac {11 C}{50}\right ) d \right ) \sqrt {2 \sqrt {5}\, \sqrt {5 d^{2}-3 d e +2 e^{2}}+10 d -3 e}\, \ln \left (\sqrt {5}\, \left (e x +d \right )-\sqrt {e x +d}\, \sqrt {2 \sqrt {5}\, \sqrt {5 d^{2}-3 d e +2 e^{2}}+10 d -3 e}+\sqrt {5 d^{2}-3 d e +2 e^{2}}\right )-\left (\sqrt {5}\, \left (A -\frac {3 B}{10}-\frac {11 C}{50}\right ) \sqrt {5 d^{2}-3 d e +2 e^{2}}+\left (-2 B +\frac {3 C}{5}+\frac {3 A}{2}\right ) e -5 \left (A -\frac {3 B}{10}-\frac {11 C}{50}\right ) d \right ) \sqrt {2 \sqrt {5}\, \sqrt {5 d^{2}-3 d e +2 e^{2}}+10 d -3 e}\, \ln \left (\sqrt {5}\, \left (e x +d \right )+\sqrt {e x +d}\, \sqrt {2 \sqrt {5}\, \sqrt {5 d^{2}-3 d e +2 e^{2}}+10 d -3 e}+\sqrt {5 d^{2}-3 d e +2 e^{2}}\right )-\frac {62 C \sqrt {e x +d}\, \sqrt {5 d^{2}-3 d e +2 e^{2}}}{5}\right )}{31 \sqrt {2 \sqrt {5}\, \sqrt {5 d^{2}-3 d e +2 e^{2}}-10 d +3 e}\, \sqrt {5 d^{2}-3 d e +2 e^{2}}\, e}\) | \(669\) |
derivativedivides | \(\text {Expression too large to display}\) | \(5722\) |
default | \(\text {Expression too large to display}\) | \(5722\) |
risch | \(\text {Expression too large to display}\) | \(21229\) |
Input:
int((C*x^2+B*x+A)/(e*x+d)^(1/2)/(5*x^2+3*x+2),x,method=_RETURNVERBOSE)
Output:
-1/31*(-31*e*(1/5*(B-3/5*C)*5^(1/2)*(5*d^2-3*d*e+2*e^2)^(1/2)+(A-2/5*C)*e- (B-3/5*C)*d)*arctan((2*5^(1/2)*(e*x+d)^(1/2)-(2*5^(1/2)*(5*d^2-3*d*e+2*e^2 )^(1/2)+10*d-3*e)^(1/2))/(2*5^(1/2)*(5*d^2-3*d*e+2*e^2)^(1/2)-10*d+3*e)^(1 /2))-31*e*(1/5*(B-3/5*C)*5^(1/2)*(5*d^2-3*d*e+2*e^2)^(1/2)+(A-2/5*C)*e-(B- 3/5*C)*d)*arctan((2*5^(1/2)*(e*x+d)^(1/2)+(2*5^(1/2)*(5*d^2-3*d*e+2*e^2)^( 1/2)+10*d-3*e)^(1/2))/(2*5^(1/2)*(5*d^2-3*d*e+2*e^2)^(1/2)-10*d+3*e)^(1/2) )+(2*5^(1/2)*(5*d^2-3*d*e+2*e^2)^(1/2)-10*d+3*e)^(1/2)*((5^(1/2)*(A-3/10*B -11/50*C)*(5*d^2-3*d*e+2*e^2)^(1/2)+(-2*B+3/5*C+3/2*A)*e-5*(A-3/10*B-11/50 *C)*d)*(2*5^(1/2)*(5*d^2-3*d*e+2*e^2)^(1/2)+10*d-3*e)^(1/2)*ln(5^(1/2)*(e* x+d)-(e*x+d)^(1/2)*(2*5^(1/2)*(5*d^2-3*d*e+2*e^2)^(1/2)+10*d-3*e)^(1/2)+(5 *d^2-3*d*e+2*e^2)^(1/2))-(5^(1/2)*(A-3/10*B-11/50*C)*(5*d^2-3*d*e+2*e^2)^( 1/2)+(-2*B+3/5*C+3/2*A)*e-5*(A-3/10*B-11/50*C)*d)*(2*5^(1/2)*(5*d^2-3*d*e+ 2*e^2)^(1/2)+10*d-3*e)^(1/2)*ln(5^(1/2)*(e*x+d)+(e*x+d)^(1/2)*(2*5^(1/2)*( 5*d^2-3*d*e+2*e^2)^(1/2)+10*d-3*e)^(1/2)+(5*d^2-3*d*e+2*e^2)^(1/2))-62/5*C *(e*x+d)^(1/2)*(5*d^2-3*d*e+2*e^2)^(1/2)))/(2*5^(1/2)*(5*d^2-3*d*e+2*e^2)^ (1/2)-10*d+3*e)^(1/2)/(5*d^2-3*d*e+2*e^2)^(1/2)/e
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.36 (sec) , antiderivative size = 6132, normalized size of antiderivative = 6132.00 \[ \int \frac {A+B x+C x^2}{\sqrt {d+e x} \left (2+3 x+5 x^2\right )} \, dx=\text {Too large to display} \] Input:
integrate((C*x^2+B*x+A)/(e*x+d)^(1/2)/(5*x^2+3*x+2),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {A+B x+C x^2}{\sqrt {d+e x} \left (2+3 x+5 x^2\right )} \, dx=\int \frac {A + B x + C x^{2}}{\sqrt {d + e x} \left (5 x^{2} + 3 x + 2\right )}\, dx \] Input:
integrate((C*x**2+B*x+A)/(e*x+d)**(1/2)/(5*x**2+3*x+2),x)
Output:
Integral((A + B*x + C*x**2)/(sqrt(d + e*x)*(5*x**2 + 3*x + 2)), x)
\[ \int \frac {A+B x+C x^2}{\sqrt {d+e x} \left (2+3 x+5 x^2\right )} \, dx=\int { \frac {C x^{2} + B x + A}{\sqrt {e x + d} {\left (5 \, x^{2} + 3 \, x + 2\right )}} \,d x } \] Input:
integrate((C*x^2+B*x+A)/(e*x+d)^(1/2)/(5*x^2+3*x+2),x, algorithm="maxima")
Output:
integrate((C*x^2 + B*x + A)/(sqrt(e*x + d)*(5*x^2 + 3*x + 2)), x)
\[ \int \frac {A+B x+C x^2}{\sqrt {d+e x} \left (2+3 x+5 x^2\right )} \, dx=\int { \frac {C x^{2} + B x + A}{\sqrt {e x + d} {\left (5 \, x^{2} + 3 \, x + 2\right )}} \,d x } \] Input:
integrate((C*x^2+B*x+A)/(e*x+d)^(1/2)/(5*x^2+3*x+2),x, algorithm="giac")
Output:
sage0*x
Time = 17.59 (sec) , antiderivative size = 3714, normalized size of antiderivative = 3714.00 \[ \int \frac {A+B x+C x^2}{\sqrt {d+e x} \left (2+3 x+5 x^2\right )} \, dx=\text {Too large to display} \] Input:
int((A + B*x + C*x^2)/((d + e*x)^(1/2)*(3*x + 5*x^2 + 2)),x)
Output:
log(((d + e*x)^(1/2)*(440*B^2*e^2 - 2000*A^2*e^2 + (632*C^2*e^2)/5 + 1200* A*B*e^2 + 880*A*C*e^2 - 1008*B*C*e^2) - ((A^2*e*375i - A^2*d*1250i + B^2*d *275i + B^2*e*150i + C^2*d*79i - C^2*e*126i + A*B*d*750i - A*B*e*1000i + A *C*d*550i + A*C*e*300i - B*C*d*630i + B*C*e*220i - 125*31^(1/2)*A^2*e - 75 *31^(1/2)*B^2*d + 50*31^(1/2)*B^2*e + 33*31^(1/2)*C^2*d - 2*31^(1/2)*C^2*e + 250*31^(1/2)*A*B*d - 150*31^(1/2)*A*C*d + 100*31^(1/2)*A*C*e - 10*31^(1 /2)*B*C*d - 60*31^(1/2)*B*C*e)/(7750*(d^2*5i - d*e*3i + e^2*2i)))^(1/2)*(6 200*A*e^3 - 2480*C*e^3 + (62000*d*e^2 - 18600*e^3)*(d + e*x)^(1/2)*((A^2*e *375i - A^2*d*1250i + B^2*d*275i + B^2*e*150i + C^2*d*79i - C^2*e*126i + A *B*d*750i - A*B*e*1000i + A*C*d*550i + A*C*e*300i - B*C*d*630i + B*C*e*220 i - 125*31^(1/2)*A^2*e - 75*31^(1/2)*B^2*d + 50*31^(1/2)*B^2*e + 33*31^(1/ 2)*C^2*d - 2*31^(1/2)*C^2*e + 250*31^(1/2)*A*B*d - 150*31^(1/2)*A*C*d + 10 0*31^(1/2)*A*C*e - 10*31^(1/2)*B*C*d - 60*31^(1/2)*B*C*e)/(7750*(d^2*5i - d*e*3i + e^2*2i)))^(1/2) - 6200*B*d*e^2 + 3720*C*d*e^2))*((A^2*e*375i - A^ 2*d*1250i + B^2*d*275i + B^2*e*150i + C^2*d*79i - C^2*e*126i + A*B*d*750i - A*B*e*1000i + A*C*d*550i + A*C*e*300i - B*C*d*630i + B*C*e*220i - 125*31 ^(1/2)*A^2*e - 75*31^(1/2)*B^2*d + 50*31^(1/2)*B^2*e + 33*31^(1/2)*C^2*d - 2*31^(1/2)*C^2*e + 250*31^(1/2)*A*B*d - 150*31^(1/2)*A*C*d + 100*31^(1/2) *A*C*e - 10*31^(1/2)*B*C*d - 60*31^(1/2)*B*C*e)/(7750*(d^2*5i - d*e*3i + e ^2*2i)))^(1/2) - 80*B^3*e^2 + (96*C^3*e^2)/5 + 120*A*B^2*e^2 - 200*A^2*...
Time = 0.19 (sec) , antiderivative size = 6499, normalized size of antiderivative = 6499.00 \[ \int \frac {A+B x+C x^2}{\sqrt {d+e x} \left (2+3 x+5 x^2\right )} \, dx =\text {Too large to display} \] Input:
int((C*x^2+B*x+A)/(e*x+d)^(1/2)/(5*x^2+3*x+2),x)
Output:
( - 500*sqrt(2*sqrt(5*d**2 - 3*d*e + 2*e**2)*sqrt(5) - 10*d + 3*e)*sqrt(5* d**2 - 3*d*e + 2*e**2)*atan((sqrt(2*sqrt(5*d**2 - 3*d*e + 2*e**2)*sqrt(5) + 10*d - 3*e) - 2*sqrt(d + e*x)*sqrt(5))/sqrt(2*sqrt(5*d**2 - 3*d*e + 2*e* *2)*sqrt(5) - 10*d + 3*e))*a*d + 150*sqrt(2*sqrt(5*d**2 - 3*d*e + 2*e**2)* sqrt(5) - 10*d + 3*e)*sqrt(5*d**2 - 3*d*e + 2*e**2)*atan((sqrt(2*sqrt(5*d* *2 - 3*d*e + 2*e**2)*sqrt(5) + 10*d - 3*e) - 2*sqrt(d + e*x)*sqrt(5))/sqrt (2*sqrt(5*d**2 - 3*d*e + 2*e**2)*sqrt(5) - 10*d + 3*e))*a*e + 150*sqrt(2*s qrt(5*d**2 - 3*d*e + 2*e**2)*sqrt(5) - 10*d + 3*e)*sqrt(5*d**2 - 3*d*e + 2 *e**2)*atan((sqrt(2*sqrt(5*d**2 - 3*d*e + 2*e**2)*sqrt(5) + 10*d - 3*e) - 2*sqrt(d + e*x)*sqrt(5))/sqrt(2*sqrt(5*d**2 - 3*d*e + 2*e**2)*sqrt(5) - 10 *d + 3*e))*b*d - 200*sqrt(2*sqrt(5*d**2 - 3*d*e + 2*e**2)*sqrt(5) - 10*d + 3*e)*sqrt(5*d**2 - 3*d*e + 2*e**2)*atan((sqrt(2*sqrt(5*d**2 - 3*d*e + 2*e **2)*sqrt(5) + 10*d - 3*e) - 2*sqrt(d + e*x)*sqrt(5))/sqrt(2*sqrt(5*d**2 - 3*d*e + 2*e**2)*sqrt(5) - 10*d + 3*e))*b*e + 110*sqrt(2*sqrt(5*d**2 - 3*d *e + 2*e**2)*sqrt(5) - 10*d + 3*e)*sqrt(5*d**2 - 3*d*e + 2*e**2)*atan((sqr t(2*sqrt(5*d**2 - 3*d*e + 2*e**2)*sqrt(5) + 10*d - 3*e) - 2*sqrt(d + e*x)* sqrt(5))/sqrt(2*sqrt(5*d**2 - 3*d*e + 2*e**2)*sqrt(5) - 10*d + 3*e))*c*d + 60*sqrt(2*sqrt(5*d**2 - 3*d*e + 2*e**2)*sqrt(5) - 10*d + 3*e)*sqrt(5*d**2 - 3*d*e + 2*e**2)*atan((sqrt(2*sqrt(5*d**2 - 3*d*e + 2*e**2)*sqrt(5) + 10 *d - 3*e) - 2*sqrt(d + e*x)*sqrt(5))/sqrt(2*sqrt(5*d**2 - 3*d*e + 2*e**...