3.186 \(\int (d+e x) \log (c (a+b x^2)^p) \, dx\)

Optimal. Leaf size=99 \[ \frac {(d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 e}-\frac {p \left (b d^2-a e^2\right ) \log \left (a+b x^2\right )}{2 b e}+\frac {2 \sqrt {a} d p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b}}-2 d p x-\frac {1}{2} e p x^2 \]

[Out]

-2*d*p*x-1/2*e*p*x^2-1/2*(-a*e^2+b*d^2)*p*ln(b*x^2+a)/b/e+1/2*(e*x+d)^2*ln(c*(b*x^2+a)^p)/e+2*d*p*arctan(x*b^(
1/2)/a^(1/2))*a^(1/2)/b^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {2463, 801, 635, 205, 260} \[ \frac {(d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 e}-\frac {p \left (b d^2-a e^2\right ) \log \left (a+b x^2\right )}{2 b e}+\frac {2 \sqrt {a} d p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b}}-2 d p x-\frac {1}{2} e p x^2 \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*Log[c*(a + b*x^2)^p],x]

[Out]

-2*d*p*x - (e*p*x^2)/2 + (2*Sqrt[a]*d*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[b] - ((b*d^2 - a*e^2)*p*Log[a + b*x^
2])/(2*b*e) + ((d + e*x)^2*Log[c*(a + b*x^2)^p])/(2*e)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 2463

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.) + (g_.)*(x_))^(r_.), x_Symbol] :> Simp[((
f + g*x)^(r + 1)*(a + b*Log[c*(d + e*x^n)^p]))/(g*(r + 1)), x] - Dist[(b*e*n*p)/(g*(r + 1)), Int[(x^(n - 1)*(f
+ g*x)^(r + 1))/(d + e*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, r}, x] && (IGtQ[r, 0] || RationalQ[n
]) && NeQ[r, -1]

Rubi steps

\begin {align*} \int (d+e x) \log \left (c \left (a+b x^2\right )^p\right ) \, dx &=\frac {(d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 e}-\frac {(b p) \int \frac {x (d+e x)^2}{a+b x^2} \, dx}{e}\\ &=\frac {(d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 e}-\frac {(b p) \int \left (\frac {2 d e}{b}+\frac {e^2 x}{b}-\frac {2 a d e-\left (b d^2-a e^2\right ) x}{b \left (a+b x^2\right )}\right ) \, dx}{e}\\ &=-2 d p x-\frac {1}{2} e p x^2+\frac {(d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 e}+\frac {p \int \frac {2 a d e-\left (b d^2-a e^2\right ) x}{a+b x^2} \, dx}{e}\\ &=-2 d p x-\frac {1}{2} e p x^2+\frac {(d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 e}+(2 a d p) \int \frac {1}{a+b x^2} \, dx+\frac {\left (\left (-b d^2+a e^2\right ) p\right ) \int \frac {x}{a+b x^2} \, dx}{e}\\ &=-2 d p x-\frac {1}{2} e p x^2+\frac {2 \sqrt {a} d p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b}}-\frac {\left (b d^2-a e^2\right ) p \log \left (a+b x^2\right )}{2 b e}+\frac {(d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 e}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 83, normalized size = 0.84 \[ d x \log \left (c \left (a+b x^2\right )^p\right )+\frac {1}{2} e \left (\frac {\left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b}-p x^2\right )+\frac {2 \sqrt {a} d p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b}}-2 d p x \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2*d*p*x + (2*Sqrt[a]*d*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[b] + d*x*Log[c*(a + b*x^2)^p] + (e*(-(p*x^2) + ((a
+ b*x^2)*Log[c*(a + b*x^2)^p])/b))/2

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fricas [A]  time = 0.45, size = 198, normalized size = 2.00 \[ \left [-\frac {b e p x^{2} - 2 \, b d p \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 4 \, b d p x - {\left (b e p x^{2} + 2 \, b d p x + a e p\right )} \log \left (b x^{2} + a\right ) - {\left (b e x^{2} + 2 \, b d x\right )} \log \relax (c)}{2 \, b}, -\frac {b e p x^{2} - 4 \, b d p \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 4 \, b d p x - {\left (b e p x^{2} + 2 \, b d p x + a e p\right )} \log \left (b x^{2} + a\right ) - {\left (b e x^{2} + 2 \, b d x\right )} \log \relax (c)}{2 \, b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(b*e*p*x^2 - 2*b*d*p*sqrt(-a/b)*log((b*x^2 + 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)) + 4*b*d*p*x - (b*e*p*x^2
+ 2*b*d*p*x + a*e*p)*log(b*x^2 + a) - (b*e*x^2 + 2*b*d*x)*log(c))/b, -1/2*(b*e*p*x^2 - 4*b*d*p*sqrt(a/b)*arct
an(b*x*sqrt(a/b)/a) + 4*b*d*p*x - (b*e*p*x^2 + 2*b*d*p*x + a*e*p)*log(b*x^2 + a) - (b*e*x^2 + 2*b*d*x)*log(c))
/b]

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giac [A]  time = 0.18, size = 100, normalized size = 1.01 \[ \frac {2 \, a d p \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}} + \frac {b p x^{2} e \log \left (b x^{2} + a\right ) - b p x^{2} e + 2 \, b d p x \log \left (b x^{2} + a\right ) + b x^{2} e \log \relax (c) - 4 \, b d p x + a p e \log \left (b x^{2} + a\right ) + 2 \, b d x \log \relax (c)}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*d*p*arctan(b*x/sqrt(a*b))/sqrt(a*b) + 1/2*(b*p*x^2*e*log(b*x^2 + a) - b*p*x^2*e + 2*b*d*p*x*log(b*x^2 + a)
+ b*x^2*e*log(c) - 4*b*d*p*x + a*p*e*log(b*x^2 + a) + 2*b*d*x*log(c))/b

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maple [A]  time = 0.07, size = 93, normalized size = 0.94 \[ \frac {2 a d p \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}-\frac {e p \,x^{2}}{2}+\frac {e \,x^{2} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}{2}-2 d p x +d x \ln \left (c \left (b \,x^{2}+a \right )^{p}\right )-\frac {a e p}{2 b}+\frac {a e \ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*ln(c*(b*x^2+a)^p),x)

[Out]

d*x*ln(c*(b*x^2+a)^p)-2*d*p*x+2*d*p*a/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)+1/2*e*ln(c*(b*x^2+a)^p)*x^2-1/2*e*
p*x^2+1/2*e/b*ln(c*(b*x^2+a)^p)*a-1/2*a*p*e/b

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maxima [A]  time = 0.98, size = 80, normalized size = 0.81 \[ \frac {1}{2} \, {\left (\frac {4 \, a d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b} + \frac {a e \log \left (b x^{2} + a\right )}{b^{2}} - \frac {e x^{2} + 4 \, d x}{b}\right )} b p + \frac {1}{2} \, {\left (e x^{2} + 2 \, d x\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*a*d*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b) + a*e*log(b*x^2 + a)/b^2 - (e*x^2 + 4*d*x)/b)*b*p + 1/2*(e*x^2
+ 2*d*x)*log((b*x^2 + a)^p*c)

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mupad [B]  time = 1.13, size = 81, normalized size = 0.82 \[ d\,x\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )-\frac {e\,p\,x^2}{2}-2\,d\,p\,x+\frac {e\,x^2\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{2}+\frac {a\,e\,p\,\ln \left (b\,x^2+a\right )}{2\,b}+\frac {2\,\sqrt {a}\,d\,p\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b*x^2)^p)*(d + e*x),x)

[Out]

d*x*log(c*(a + b*x^2)^p) - (e*p*x^2)/2 - 2*d*p*x + (e*x^2*log(c*(a + b*x^2)^p))/2 + (a*e*p*log(a + b*x^2))/(2*
b) + (2*a^(1/2)*d*p*atan((b^(1/2)*x)/a^(1/2)))/b^(1/2)

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sympy [A]  time = 11.68, size = 160, normalized size = 1.62 \[ \begin {cases} \frac {i \sqrt {a} d p \log {\left (a + b x^{2} \right )}}{b \sqrt {\frac {1}{b}}} - \frac {2 i \sqrt {a} d p \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{b \sqrt {\frac {1}{b}}} + \frac {a e p \log {\left (a + b x^{2} \right )}}{2 b} + d p x \log {\left (a + b x^{2} \right )} - 2 d p x + d x \log {\relax (c )} + \frac {e p x^{2} \log {\left (a + b x^{2} \right )}}{2} - \frac {e p x^{2}}{2} + \frac {e x^{2} \log {\relax (c )}}{2} & \text {for}\: b \neq 0 \\\left (d x + \frac {e x^{2}}{2}\right ) \log {\left (a^{p} c \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*ln(c*(b*x**2+a)**p),x)

[Out]

Piecewise((I*sqrt(a)*d*p*log(a + b*x**2)/(b*sqrt(1/b)) - 2*I*sqrt(a)*d*p*log(-I*sqrt(a)*sqrt(1/b) + x)/(b*sqrt
(1/b)) + a*e*p*log(a + b*x**2)/(2*b) + d*p*x*log(a + b*x**2) - 2*d*p*x + d*x*log(c) + e*p*x**2*log(a + b*x**2)
/2 - e*p*x**2/2 + e*x**2*log(c)/2, Ne(b, 0)), ((d*x + e*x**2/2)*log(a**p*c), True))

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