3.3.79 \(\int \frac {x^4 (d+e x)}{a+c x^2} \, dx\)

Optimal. Leaf size=87 \[ \frac {a^{3/2} d \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{c^{5/2}}+\frac {a^2 e \log \left (a+c x^2\right )}{2 c^3}-\frac {a d x}{c^2}-\frac {a e x^2}{2 c^2}+\frac {d x^3}{3 c}+\frac {e x^4}{4 c} \]

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Rubi [A]  time = 0.06, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {801, 635, 205, 260} \begin {gather*} \frac {a^{3/2} d \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{c^{5/2}}+\frac {a^2 e \log \left (a+c x^2\right )}{2 c^3}-\frac {a d x}{c^2}-\frac {a e x^2}{2 c^2}+\frac {d x^3}{3 c}+\frac {e x^4}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(d + e*x))/(a + c*x^2),x]

[Out]

-((a*d*x)/c^2) - (a*e*x^2)/(2*c^2) + (d*x^3)/(3*c) + (e*x^4)/(4*c) + (a^(3/2)*d*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/c
^(5/2) + (a^2*e*Log[a + c*x^2])/(2*c^3)

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]

Rubi steps

\begin {align*} \int \frac {x^4 (d+e x)}{a+c x^2} \, dx &=\int \left (-\frac {a d}{c^2}-\frac {a e x}{c^2}+\frac {d x^2}{c}+\frac {e x^3}{c}+\frac {a^2 d+a^2 e x}{c^2 \left (a+c x^2\right )}\right ) \, dx\\ &=-\frac {a d x}{c^2}-\frac {a e x^2}{2 c^2}+\frac {d x^3}{3 c}+\frac {e x^4}{4 c}+\frac {\int \frac {a^2 d+a^2 e x}{a+c x^2} \, dx}{c^2}\\ &=-\frac {a d x}{c^2}-\frac {a e x^2}{2 c^2}+\frac {d x^3}{3 c}+\frac {e x^4}{4 c}+\frac {\left (a^2 d\right ) \int \frac {1}{a+c x^2} \, dx}{c^2}+\frac {\left (a^2 e\right ) \int \frac {x}{a+c x^2} \, dx}{c^2}\\ &=-\frac {a d x}{c^2}-\frac {a e x^2}{2 c^2}+\frac {d x^3}{3 c}+\frac {e x^4}{4 c}+\frac {a^{3/2} d \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{c^{5/2}}+\frac {a^2 e \log \left (a+c x^2\right )}{2 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 75, normalized size = 0.86 \begin {gather*} \frac {12 a^{3/2} \sqrt {c} d \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )+6 a^2 e \log \left (a+c x^2\right )+c x \left (c x^2 (4 d+3 e x)-6 a (2 d+e x)\right )}{12 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(d + e*x))/(a + c*x^2),x]

[Out]

(c*x*(-6*a*(2*d + e*x) + c*x^2*(4*d + 3*e*x)) + 12*a^(3/2)*Sqrt[c]*d*ArcTan[(Sqrt[c]*x)/Sqrt[a]] + 6*a^2*e*Log
[a + c*x^2])/(12*c^3)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^4 (d+e x)}{a+c x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^4*(d + e*x))/(a + c*x^2),x]

[Out]

IntegrateAlgebraic[(x^4*(d + e*x))/(a + c*x^2), x]

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fricas [A]  time = 0.40, size = 176, normalized size = 2.02 \begin {gather*} \left [\frac {3 \, c^{2} e x^{4} + 4 \, c^{2} d x^{3} - 6 \, a c e x^{2} + 6 \, a c d \sqrt {-\frac {a}{c}} \log \left (\frac {c x^{2} + 2 \, c x \sqrt {-\frac {a}{c}} - a}{c x^{2} + a}\right ) - 12 \, a c d x + 6 \, a^{2} e \log \left (c x^{2} + a\right )}{12 \, c^{3}}, \frac {3 \, c^{2} e x^{4} + 4 \, c^{2} d x^{3} - 6 \, a c e x^{2} + 12 \, a c d \sqrt {\frac {a}{c}} \arctan \left (\frac {c x \sqrt {\frac {a}{c}}}{a}\right ) - 12 \, a c d x + 6 \, a^{2} e \log \left (c x^{2} + a\right )}{12 \, c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)/(c*x^2+a),x, algorithm="fricas")

[Out]

[1/12*(3*c^2*e*x^4 + 4*c^2*d*x^3 - 6*a*c*e*x^2 + 6*a*c*d*sqrt(-a/c)*log((c*x^2 + 2*c*x*sqrt(-a/c) - a)/(c*x^2
+ a)) - 12*a*c*d*x + 6*a^2*e*log(c*x^2 + a))/c^3, 1/12*(3*c^2*e*x^4 + 4*c^2*d*x^3 - 6*a*c*e*x^2 + 12*a*c*d*sqr
t(a/c)*arctan(c*x*sqrt(a/c)/a) - 12*a*c*d*x + 6*a^2*e*log(c*x^2 + a))/c^3]

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giac [A]  time = 0.15, size = 85, normalized size = 0.98 \begin {gather*} \frac {a^{2} d \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c^{2}} + \frac {a^{2} e \log \left (c x^{2} + a\right )}{2 \, c^{3}} + \frac {3 \, c^{3} x^{4} e + 4 \, c^{3} d x^{3} - 6 \, a c^{2} x^{2} e - 12 \, a c^{2} d x}{12 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)/(c*x^2+a),x, algorithm="giac")

[Out]

a^2*d*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*c^2) + 1/2*a^2*e*log(c*x^2 + a)/c^3 + 1/12*(3*c^3*x^4*e + 4*c^3*d*x^3 -
6*a*c^2*x^2*e - 12*a*c^2*d*x)/c^4

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maple [A]  time = 0.05, size = 77, normalized size = 0.89 \begin {gather*} \frac {e \,x^{4}}{4 c}+\frac {d \,x^{3}}{3 c}+\frac {a^{2} d \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c^{2}}-\frac {a e \,x^{2}}{2 c^{2}}+\frac {a^{2} e \ln \left (c \,x^{2}+a \right )}{2 c^{3}}-\frac {a d x}{c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(e*x+d)/(c*x^2+a),x)

[Out]

1/4/c*e*x^4+1/3/c*d*x^3-1/2*a*e*x^2/c^2-a*d*x/c^2+1/2*a^2*e*ln(c*x^2+a)/c^3+a^2/c^2*d/(a*c)^(1/2)*arctan(1/(a*
c)^(1/2)*c*x)

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maxima [A]  time = 1.19, size = 72, normalized size = 0.83 \begin {gather*} \frac {a^{2} d \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c^{2}} + \frac {a^{2} e \log \left (c x^{2} + a\right )}{2 \, c^{3}} + \frac {3 \, c e x^{4} + 4 \, c d x^{3} - 6 \, a e x^{2} - 12 \, a d x}{12 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)/(c*x^2+a),x, algorithm="maxima")

[Out]

a^2*d*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*c^2) + 1/2*a^2*e*log(c*x^2 + a)/c^3 + 1/12*(3*c*e*x^4 + 4*c*d*x^3 - 6*a
*e*x^2 - 12*a*d*x)/c^2

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mupad [B]  time = 0.06, size = 71, normalized size = 0.82 \begin {gather*} \frac {d\,x^3}{3\,c}+\frac {e\,x^4}{4\,c}+\frac {a^{3/2}\,d\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{c^{5/2}}+\frac {a^2\,e\,\ln \left (c\,x^2+a\right )}{2\,c^3}-\frac {a\,d\,x}{c^2}-\frac {a\,e\,x^2}{2\,c^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(d + e*x))/(a + c*x^2),x)

[Out]

(d*x^3)/(3*c) + (e*x^4)/(4*c) + (a^(3/2)*d*atan((c^(1/2)*x)/a^(1/2)))/c^(5/2) + (a^2*e*log(a + c*x^2))/(2*c^3)
- (a*d*x)/c^2 - (a*e*x^2)/(2*c^2)

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sympy [B]  time = 0.38, size = 189, normalized size = 2.17 \begin {gather*} - \frac {a d x}{c^{2}} - \frac {a e x^{2}}{2 c^{2}} + \left (\frac {a^{2} e}{2 c^{3}} - \frac {d \sqrt {- a^{3} c^{7}}}{2 c^{6}}\right ) \log {\left (x + \frac {- a^{2} e + 2 c^{3} \left (\frac {a^{2} e}{2 c^{3}} - \frac {d \sqrt {- a^{3} c^{7}}}{2 c^{6}}\right )}{a c d} \right )} + \left (\frac {a^{2} e}{2 c^{3}} + \frac {d \sqrt {- a^{3} c^{7}}}{2 c^{6}}\right ) \log {\left (x + \frac {- a^{2} e + 2 c^{3} \left (\frac {a^{2} e}{2 c^{3}} + \frac {d \sqrt {- a^{3} c^{7}}}{2 c^{6}}\right )}{a c d} \right )} + \frac {d x^{3}}{3 c} + \frac {e x^{4}}{4 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(e*x+d)/(c*x**2+a),x)

[Out]

-a*d*x/c**2 - a*e*x**2/(2*c**2) + (a**2*e/(2*c**3) - d*sqrt(-a**3*c**7)/(2*c**6))*log(x + (-a**2*e + 2*c**3*(a
**2*e/(2*c**3) - d*sqrt(-a**3*c**7)/(2*c**6)))/(a*c*d)) + (a**2*e/(2*c**3) + d*sqrt(-a**3*c**7)/(2*c**6))*log(
x + (-a**2*e + 2*c**3*(a**2*e/(2*c**3) + d*sqrt(-a**3*c**7)/(2*c**6)))/(a*c*d)) + d*x**3/(3*c) + e*x**4/(4*c)

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