Optimal. Leaf size=340 \[ \frac{9 (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{3}{5} (4 x+1)\right )}{1445 (m+1)}-\frac{\left (\left (62+22 \sqrt{13}\right ) m+81\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{221 \sqrt{13} \left (13-2 \sqrt{13}\right ) (m+1)}+\frac{9 \left (13+9 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{7514 \left (13-2 \sqrt{13}\right ) (m+1)}+\frac{\left (\left (62-22 \sqrt{13}\right ) m+81\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{221 \sqrt{13} \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{9 \left (13-9 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{7514 \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{(43-33 x) (4 x+1)^{m+1}}{663 \left (3 x^2-5 x+1\right )} \]
[Out]
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Rubi [A] time = 1.00982, antiderivative size = 340, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148 \[ \frac{9 (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{3}{5} (4 x+1)\right )}{1445 (m+1)}-\frac{\left (\left (62+22 \sqrt{13}\right ) m+81\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{221 \sqrt{13} \left (13-2 \sqrt{13}\right ) (m+1)}+\frac{9 \left (13+9 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{7514 \left (13-2 \sqrt{13}\right ) (m+1)}+\frac{\left (\left (62-22 \sqrt{13}\right ) m+81\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{221 \sqrt{13} \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{9 \left (13-9 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{7514 \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{(43-33 x) (4 x+1)^{m+1}}{663 \left (3 x^2-5 x+1\right )} \]
Antiderivative was successfully verified.
[In] Int[(1 + 4*x)^m/((2 + 3*x)*(1 - 5*x + 3*x^2)^2),x]
[Out]
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Rubi in Sympy [A] time = 62.158, size = 274, normalized size = 0.81 \[ \frac{\left (- 429 x + 559\right ) \left (4 x + 1\right )^{m + 1}}{8619 \left (3 x^{2} - 5 x + 1\right )} - \frac{2 \left (286 m - \sqrt{13} \left (62 m + 81\right )\right ) \left (4 x + 1\right )^{m + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, m + 1 \\ m + 2 \end{matrix}\middle |{\frac{12 x + 3}{2 \sqrt{13} + 13}} \right )}}{2873 \left (4 \sqrt{13} + 26\right ) \left (m + 1\right )} - \frac{2 \left (286 m + \sqrt{13} \left (62 m + 81\right )\right ) \left (4 x + 1\right )^{m + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, m + 1 \\ m + 2 \end{matrix}\middle |{\frac{- 12 x - 3}{-13 + 2 \sqrt{13}}} \right )}}{2873 \left (- 4 \sqrt{13} + 26\right ) \left (m + 1\right )} + \frac{3 \left (3 + \frac{27 \sqrt{13}}{13}\right ) \left (4 x + 1\right )^{m + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, m + 1 \\ m + 2 \end{matrix}\middle |{\frac{- 12 x - 3}{-13 + 2 \sqrt{13}}} \right )}}{289 \left (- 4 \sqrt{13} + 26\right ) \left (m + 1\right )} + \frac{3 \left (- \frac{27 \sqrt{13}}{13} + 3\right ) \left (4 x + 1\right )^{m + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, m + 1 \\ m + 2 \end{matrix}\middle |{\frac{12 x + 3}{2 \sqrt{13} + 13}} \right )}}{289 \left (4 \sqrt{13} + 26\right ) \left (m + 1\right )} + \frac{9 \left (4 x + 1\right )^{m + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, m + 1 \\ m + 2 \end{matrix}\middle |{- \frac{12 x}{5} - \frac{3}{5}} \right )}}{1445 \left (m + 1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] rubi_integrate((1+4*x)**m/(2+3*x)/(3*x**2-5*x+1)**2,x)
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Mathematica [A] time = 0.139731, size = 0, normalized size = 0. \[ \int \frac{(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )^2} \, dx \]
Verification is Not applicable to the result.
[In] Integrate[(1 + 4*x)^m/((2 + 3*x)*(1 - 5*x + 3*x^2)^2),x]
[Out]
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Maple [F] time = 0.177, size = 0, normalized size = 0. \[ \int{\frac{ \left ( 1+4\,x \right ) ^{m}}{ \left ( 2+3\,x \right ) \left ( 3\,{x}^{2}-5\,x+1 \right ) ^{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In] int((1+4*x)^m/(2+3*x)/(3*x^2-5*x+1)^2,x)
[Out]
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Maxima [F] time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}{\left (3 \, x + 2\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate((4*x + 1)^m/((3*x^2 - 5*x + 1)^2*(3*x + 2)),x, algorithm="maxima")
[Out]
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Fricas [F] time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (4 \, x + 1\right )}^{m}}{27 \, x^{5} - 72 \, x^{4} + 33 \, x^{3} + 32 \, x^{2} - 17 \, x + 2}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate((4*x + 1)^m/((3*x^2 - 5*x + 1)^2*(3*x + 2)),x, algorithm="fricas")
[Out]
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Sympy [F] time = 0., size = 0, normalized size = 0. \[ \int \frac{\left (4 x + 1\right )^{m}}{\left (3 x + 2\right ) \left (3 x^{2} - 5 x + 1\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate((1+4*x)**m/(2+3*x)/(3*x**2-5*x+1)**2,x)
[Out]
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GIAC/XCAS [F] time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}{\left (3 \, x + 2\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate((4*x + 1)^m/((3*x^2 - 5*x + 1)^2*(3*x + 2)),x, algorithm="giac")
[Out]