3.70 \(\int x^m \log (d (a+b x+c x^2)^n) \, dx\)

Optimal. Leaf size=157 \[ -\frac {2 c n x^{m+2} \, _2F_1\left (1,m+2;m+3;-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{(m+1) (m+2) \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 c n x^{m+2} \, _2F_1\left (1,m+2;m+3;-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{(m+1) (m+2) \left (\sqrt {b^2-4 a c}+b\right )}+\frac {x^{m+1} \log \left (d \left (a+b x+c x^2\right )^n\right )}{m+1} \]

[Out]

x^(1+m)*ln(d*(c*x^2+b*x+a)^n)/(1+m)-2*c*n*x^(2+m)*hypergeom([1, 2+m],[3+m],-2*c*x/(b-(-4*a*c+b^2)^(1/2)))/(1+m
)/(2+m)/(b-(-4*a*c+b^2)^(1/2))-2*c*n*x^(2+m)*hypergeom([1, 2+m],[3+m],-2*c*x/(b+(-4*a*c+b^2)^(1/2)))/(1+m)/(2+
m)/(b+(-4*a*c+b^2)^(1/2))

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Rubi [A]  time = 0.22, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2525, 830, 64} \[ -\frac {2 c n x^{m+2} \, _2F_1\left (1,m+2;m+3;-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{(m+1) (m+2) \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 c n x^{m+2} \, _2F_1\left (1,m+2;m+3;-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{(m+1) (m+2) \left (\sqrt {b^2-4 a c}+b\right )}+\frac {x^{m+1} \log \left (d \left (a+b x+c x^2\right )^n\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[x^m*Log[d*(a + b*x + c*x^2)^n],x]

[Out]

(-2*c*n*x^(2 + m)*Hypergeometric2F1[1, 2 + m, 3 + m, (-2*c*x)/(b - Sqrt[b^2 - 4*a*c])])/((b - Sqrt[b^2 - 4*a*c
])*(1 + m)*(2 + m)) - (2*c*n*x^(2 + m)*Hypergeometric2F1[1, 2 + m, 3 + m, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(
(b + Sqrt[b^2 - 4*a*c])*(1 + m)*(2 + m)) + (x^(1 + m)*Log[d*(a + b*x + c*x^2)^n])/(1 + m)

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rule 830

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m, (f + g*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !RationalQ[m]

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rubi steps

\begin {align*} \int x^m \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx &=\frac {x^{1+m} \log \left (d \left (a+b x+c x^2\right )^n\right )}{1+m}-\frac {n \int \frac {x^{1+m} (b+2 c x)}{a+b x+c x^2} \, dx}{1+m}\\ &=\frac {x^{1+m} \log \left (d \left (a+b x+c x^2\right )^n\right )}{1+m}-\frac {n \int \left (\frac {2 c x^{1+m}}{b-\sqrt {b^2-4 a c}+2 c x}+\frac {2 c x^{1+m}}{b+\sqrt {b^2-4 a c}+2 c x}\right ) \, dx}{1+m}\\ &=\frac {x^{1+m} \log \left (d \left (a+b x+c x^2\right )^n\right )}{1+m}-\frac {(2 c n) \int \frac {x^{1+m}}{b-\sqrt {b^2-4 a c}+2 c x} \, dx}{1+m}-\frac {(2 c n) \int \frac {x^{1+m}}{b+\sqrt {b^2-4 a c}+2 c x} \, dx}{1+m}\\ &=-\frac {2 c n x^{2+m} \, _2F_1\left (1,2+m;3+m;-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) (1+m) (2+m)}-\frac {2 c n x^{2+m} \, _2F_1\left (1,2+m;3+m;-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) (1+m) (2+m)}+\frac {x^{1+m} \log \left (d \left (a+b x+c x^2\right )^n\right )}{1+m}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 137, normalized size = 0.87 \[ -\frac {x^{m+1} \left (n x \left (\sqrt {b^2-4 a c}+b\right ) \, _2F_1\left (1,m+2;m+3;\frac {2 c x}{\sqrt {b^2-4 a c}-b}\right )+n x \left (b-\sqrt {b^2-4 a c}\right ) \, _2F_1\left (1,m+2;m+3;-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )-2 a (m+2) \log \left (d (a+x (b+c x))^n\right )\right )}{2 a \left (m^2+3 m+2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*Log[d*(a + b*x + c*x^2)^n],x]

[Out]

-1/2*(x^(1 + m)*((b + Sqrt[b^2 - 4*a*c])*n*x*Hypergeometric2F1[1, 2 + m, 3 + m, (2*c*x)/(-b + Sqrt[b^2 - 4*a*c
])] + (b - Sqrt[b^2 - 4*a*c])*n*x*Hypergeometric2F1[1, 2 + m, 3 + m, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])] - 2*a*(
2 + m)*Log[d*(a + x*(b + c*x))^n]))/(a*(2 + 3*m + m^2))

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{m} \log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*log(d*(c*x^2+b*x+a)^n),x, algorithm="fricas")

[Out]

integral(x^m*log((c*x^2 + b*x + a)^n*d), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} \log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*log(d*(c*x^2+b*x+a)^n),x, algorithm="giac")

[Out]

integrate(x^m*log((c*x^2 + b*x + a)^n*d), x)

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maple [F]  time = 1.40, size = 0, normalized size = 0.00 \[ \int x^{m} \ln \left (d \left (c \,x^{2}+b x +a \right )^{n}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*ln(d*(c*x^2+b*x+a)^n),x)

[Out]

int(x^m*ln(d*(c*x^2+b*x+a)^n),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {x x^{m} \log \left ({\left (c x^{2} + b x + a\right )}^{n}\right )}{m + 1} + \int \frac {{\left ({\left ({\left (m + 1\right )} \log \relax (d) - 2 \, n\right )} c x^{2} + {\left ({\left (m + 1\right )} \log \relax (d) - n\right )} b x + a {\left (m + 1\right )} \log \relax (d)\right )} x^{m}}{c {\left (m + 1\right )} x^{2} + b {\left (m + 1\right )} x + a {\left (m + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*log(d*(c*x^2+b*x+a)^n),x, algorithm="maxima")

[Out]

x*x^m*log((c*x^2 + b*x + a)^n)/(m + 1) + integrate((((m + 1)*log(d) - 2*n)*c*x^2 + ((m + 1)*log(d) - n)*b*x +
a*(m + 1)*log(d))*x^m/(c*(m + 1)*x^2 + b*(m + 1)*x + a*(m + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^m\,\ln \left (d\,{\left (c\,x^2+b\,x+a\right )}^n\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*log(d*(a + b*x + c*x^2)^n),x)

[Out]

int(x^m*log(d*(a + b*x + c*x^2)^n), 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(x**m*ln(d*(c*x**2+b*x+a)**n),x)

[Out]

Timed out

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