Optimal. Leaf size=255 \[ \frac {\left (100 d^2+165 d e+81 e^2\right ) (d+e x)^{m+1}}{125 e^3 (m+1)}-\frac {(40 d+33 e) (d+e x)^{m+2}}{25 e^3 (m+2)}+\frac {4 (d+e x)^{m+3}}{5 e^3 (m+3)}-\frac {\left (-423 \sqrt {14}+6412 i\right ) (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {5 (d+e x)}{5 d+i \sqrt {14} e-e}\right )}{3500 (m+1) \left (5 i d-\left (\sqrt {14}+i\right ) e\right )}-\frac {\left (423 \sqrt {14}+6412 i\right ) (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {5 (d+e x)}{5 d-\left (1+i \sqrt {14}\right ) e}\right )}{3500 (m+1) \left (5 i d-\left (-\sqrt {14}+i\right ) e\right )} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.48, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {1628, 68} \[ \frac {\left (100 d^2+165 d e+81 e^2\right ) (d+e x)^{m+1}}{125 e^3 (m+1)}-\frac {(40 d+33 e) (d+e x)^{m+2}}{25 e^3 (m+2)}+\frac {4 (d+e x)^{m+3}}{5 e^3 (m+3)}-\frac {\left (-423 \sqrt {14}+6412 i\right ) (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {5 (d+e x)}{5 d+i \sqrt {14} e-e}\right )}{3500 (m+1) \left (5 i d-\left (\sqrt {14}+i\right ) e\right )}-\frac {\left (423 \sqrt {14}+6412 i\right ) (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {5 (d+e x)}{5 d-\left (1+i \sqrt {14}\right ) e}\right )}{3500 (m+1) \left (5 i d-\left (-\sqrt {14}+i\right ) e\right )} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 68
Rule 1628
Rubi steps
\begin {align*} \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx &=\int \left (\frac {\left (100 d^2+165 d e+81 e^2\right ) (d+e x)^m}{125 e^2}+\frac {\left (\frac {458}{125}+\frac {423 i}{125 \sqrt {14}}\right ) (d+e x)^m}{2-2 i \sqrt {14}+10 x}+\frac {\left (\frac {458}{125}-\frac {423 i}{125 \sqrt {14}}\right ) (d+e x)^m}{2+2 i \sqrt {14}+10 x}+\frac {(-40 d-33 e) (d+e x)^{1+m}}{25 e^2}+\frac {4 (d+e x)^{2+m}}{5 e^2}\right ) \, dx\\ &=\frac {\left (100 d^2+165 d e+81 e^2\right ) (d+e x)^{1+m}}{125 e^3 (1+m)}-\frac {(40 d+33 e) (d+e x)^{2+m}}{25 e^3 (2+m)}+\frac {4 (d+e x)^{3+m}}{5 e^3 (3+m)}+\frac {\left (6412-423 i \sqrt {14}\right ) \int \frac {(d+e x)^m}{2+2 i \sqrt {14}+10 x} \, dx}{1750}+\frac {\left (6412+423 i \sqrt {14}\right ) \int \frac {(d+e x)^m}{2-2 i \sqrt {14}+10 x} \, dx}{1750}\\ &=\frac {\left (100 d^2+165 d e+81 e^2\right ) (d+e x)^{1+m}}{125 e^3 (1+m)}-\frac {(40 d+33 e) (d+e x)^{2+m}}{25 e^3 (2+m)}+\frac {4 (d+e x)^{3+m}}{5 e^3 (3+m)}-\frac {\left (6412 i-423 \sqrt {14}\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {5 (d+e x)}{5 d-e+i \sqrt {14} e}\right )}{3500 \left (5 i d-\left (i+\sqrt {14}\right ) e\right ) (1+m)}-\frac {\left (6412 i+423 \sqrt {14}\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {5 (d+e x)}{5 d-\left (1+i \sqrt {14}\right ) e}\right )}{3500 \left (5 i d-\left (i-\sqrt {14}\right ) e\right ) (1+m)}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.73, size = 221, normalized size = 0.87 \[ \frac {(d+e x)^{m+1} \left (\frac {28 \left (100 d^2+165 d e+81 e^2\right )}{e^3 (m+1)}+\frac {2800 (d+e x)^2}{e^3 (m+3)}-\frac {140 (40 d+33 e) (d+e x)}{e^3 (m+2)}-\frac {\left (423 \sqrt {14}+6412 i\right ) \, _2F_1\left (1,m+1;m+2;\frac {5 (d+e x)}{5 d+\left (-1-i \sqrt {14}\right ) e}\right )}{(m+1) \left (5 i d+\left (\sqrt {14}-i\right ) e\right )}-\frac {\left (423 \sqrt {14}-6412 i\right ) \, _2F_1\left (1,m+1;m+2;\frac {5 (d+e x)}{5 d+i \left (i+\sqrt {14}\right ) e}\right )}{(m+1) \left (\left (\sqrt {14}+i\right ) e-5 i d\right )}\right )}{3500} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [F] time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (4 \, x^{4} - 5 \, x^{3} + 3 \, x^{2} + x + 2\right )} {\left (e x + d\right )}^{m}}{5 \, x^{2} + 2 \, x + 3}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (4 \, x^{4} - 5 \, x^{3} + 3 \, x^{2} + x + 2\right )} {\left (e x + d\right )}^{m}}{5 \, x^{2} + 2 \, x + 3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [F] time = 0.26, size = 0, normalized size = 0.00 \[ \int \frac {\left (4 x^{4}-5 x^{3}+3 x^{2}+x +2\right ) \left (e x +d \right )^{m}}{5 x^{2}+2 x +3}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (4 \, x^{4} - 5 \, x^{3} + 3 \, x^{2} + x + 2\right )} {\left (e x + d\right )}^{m}}{5 \, x^{2} + 2 \, x + 3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d+e\,x\right )}^m\,\left (4\,x^4-5\,x^3+3\,x^2+x+2\right )}{5\,x^2+2\,x+3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________