3.370 \(\int \frac {(d+e x)^m (2+x+3 x^2-5 x^3+4 x^4)}{3+2 x+5 x^2} \, dx\)

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]

1/125*(100*d^2+165*d*e+81*e^2)*(e*x+d)^(1+m)/e^3/(1+m)-1/25*(40*d+33*e)*(e*x+d)^(2+m)/e^3/(2+m)+4/5*(e*x+d)^(3
+m)/e^3/(3+m)-1/3500*(e*x+d)^(1+m)*hypergeom([1, 1+m],[2+m],5*(e*x+d)/(5*d-e*(1+I*14^(1/2))))*(6412*I+423*14^(
1/2))/(1+m)/(5*I*d-e*(I-14^(1/2)))-1/3500*(e*x+d)^(1+m)*hypergeom([1, 1+m],[2+m],5*(e*x+d)/(5*d-e+I*14^(1/2)*e
))*(6412*I-423*14^(1/2))/(1+m)/(5*I*d-e*(I+14^(1/2)))

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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]

Int[((d + e*x)^m*(2 + x + 3*x^2 - 5*x^3 + 4*x^4))/(3 + 2*x + 5*x^2),x]

[Out]

((100*d^2 + 165*d*e + 81*e^2)*(d + e*x)^(1 + m))/(125*e^3*(1 + m)) - ((40*d + 33*e)*(d + e*x)^(2 + m))/(25*e^3
*(2 + m)) + (4*(d + e*x)^(3 + m))/(5*e^3*(3 + m)) - ((6412*I - 423*Sqrt[14])*(d + e*x)^(1 + m)*Hypergeometric2
F1[1, 1 + m, 2 + m, (5*(d + e*x))/(5*d - e + I*Sqrt[14]*e)])/(3500*((5*I)*d - (I + Sqrt[14])*e)*(1 + m)) - ((6
412*I + 423*Sqrt[14])*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (5*(d + e*x))/(5*d - (1 + I*Sqrt[14
])*e)])/(3500*((5*I)*d - (I - Sqrt[14])*e)*(1 + m))

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(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]

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*}

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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]

Integrate[((d + e*x)^m*(2 + x + 3*x^2 - 5*x^3 + 4*x^4))/(3 + 2*x + 5*x^2),x]

[Out]

((d + e*x)^(1 + m)*((28*(100*d^2 + 165*d*e + 81*e^2))/(e^3*(1 + m)) - (140*(40*d + 33*e)*(d + e*x))/(e^3*(2 +
m)) + (2800*(d + e*x)^2)/(e^3*(3 + m)) - ((6412*I + 423*Sqrt[14])*Hypergeometric2F1[1, 1 + m, 2 + m, (5*(d + e
*x))/(5*d + (-1 - I*Sqrt[14])*e)])/(((5*I)*d + (-I + Sqrt[14])*e)*(1 + m)) - ((-6412*I + 423*Sqrt[14])*Hyperge
ometric2F1[1, 1 + m, 2 + m, (5*(d + e*x))/(5*d + I*(I + Sqrt[14])*e)])/(((-5*I)*d + (I + Sqrt[14])*e)*(1 + m))
))/3500

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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]

integrate((e*x+d)^m*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3),x, algorithm="fricas")

[Out]

integral((4*x^4 - 5*x^3 + 3*x^2 + x + 2)*(e*x + d)^m/(5*x^2 + 2*x + 3), x)

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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]

integrate((e*x+d)^m*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3),x, algorithm="giac")

[Out]

integrate((4*x^4 - 5*x^3 + 3*x^2 + x + 2)*(e*x + d)^m/(5*x^2 + 2*x + 3), x)

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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]

int((e*x+d)^m*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3),x)

[Out]

int((e*x+d)^m*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3),x)

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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]

integrate((e*x+d)^m*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3),x, algorithm="maxima")

[Out]

integrate((4*x^4 - 5*x^3 + 3*x^2 + x + 2)*(e*x + d)^m/(5*x^2 + 2*x + 3), x)

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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]

int(((d + e*x)^m*(x + 3*x^2 - 5*x^3 + 4*x^4 + 2))/(2*x + 5*x^2 + 3),x)

[Out]

int(((d + e*x)^m*(x + 3*x^2 - 5*x^3 + 4*x^4 + 2))/(2*x + 5*x^2 + 3), x)

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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]

integrate((e*x+d)**m*(4*x**4-5*x**3+3*x**2+x+2)/(5*x**2+2*x+3),x)

[Out]

Timed out

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