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

Optimal. Leaf size=77 \[ \frac {a^{3/2} (b c-a d) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2}}-\frac {a x (b c-a d)}{b^3}+\frac {x^3 (b c-a d)}{3 b^2}+\frac {d x^5}{5 b} \]

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Rubi [A]  time = 0.05, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {459, 302, 205} \begin {gather*} \frac {a^{3/2} (b c-a d) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2}}+\frac {x^3 (b c-a d)}{3 b^2}-\frac {a x (b c-a d)}{b^3}+\frac {d x^5}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(c + d*x^2))/(a + b*x^2),x]

[Out]

-((a*(b*c - a*d)*x)/b^3) + ((b*c - a*d)*x^3)/(3*b^2) + (d*x^5)/(5*b) + (a^(3/2)*(b*c - a*d)*ArcTan[(Sqrt[b]*x)
/Sqrt[a]])/b^(7/2)

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rubi steps

\begin {align*} \int \frac {x^4 \left (c+d x^2\right )}{a+b x^2} \, dx &=\frac {d x^5}{5 b}-\frac {(-5 b c+5 a d) \int \frac {x^4}{a+b x^2} \, dx}{5 b}\\ &=\frac {d x^5}{5 b}-\frac {(-5 b c+5 a d) \int \left (-\frac {a}{b^2}+\frac {x^2}{b}+\frac {a^2}{b^2 \left (a+b x^2\right )}\right ) \, dx}{5 b}\\ &=-\frac {a (b c-a d) x}{b^3}+\frac {(b c-a d) x^3}{3 b^2}+\frac {d x^5}{5 b}+\frac {\left (a^2 (b c-a d)\right ) \int \frac {1}{a+b x^2} \, dx}{b^3}\\ &=-\frac {a (b c-a d) x}{b^3}+\frac {(b c-a d) x^3}{3 b^2}+\frac {d x^5}{5 b}+\frac {a^{3/2} (b c-a d) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 77, normalized size = 1.00 \begin {gather*} -\frac {a^{3/2} (a d-b c) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2}}+\frac {a x (a d-b c)}{b^3}+\frac {x^3 (b c-a d)}{3 b^2}+\frac {d x^5}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-(b*c) + a*d)*x)/b^3 + ((b*c - a*d)*x^3)/(3*b^2) + (d*x^5)/(5*b) - (a^(3/2)*(-(b*c) + a*d)*ArcTan[(Sqrt[b]
*x)/Sqrt[a]])/b^(7/2)

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.94, size = 178, normalized size = 2.31 \begin {gather*} \left [\frac {6 \, b^{2} d x^{5} + 10 \, {\left (b^{2} c - a b d\right )} x^{3} - 15 \, {\left (a b c - a^{2} d\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) - 30 \, {\left (a b c - a^{2} d\right )} x}{30 \, b^{3}}, \frac {3 \, b^{2} d x^{5} + 5 \, {\left (b^{2} c - a b d\right )} x^{3} + 15 \, {\left (a b c - a^{2} d\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) - 15 \, {\left (a b c - a^{2} d\right )} x}{15 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(6*b^2*d*x^5 + 10*(b^2*c - a*b*d)*x^3 - 15*(a*b*c - a^2*d)*sqrt(-a/b)*log((b*x^2 - 2*b*x*sqrt(-a/b) - a)
/(b*x^2 + a)) - 30*(a*b*c - a^2*d)*x)/b^3, 1/15*(3*b^2*d*x^5 + 5*(b^2*c - a*b*d)*x^3 + 15*(a*b*c - a^2*d)*sqrt
(a/b)*arctan(b*x*sqrt(a/b)/a) - 15*(a*b*c - a^2*d)*x)/b^3]

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giac [A]  time = 0.39, size = 84, normalized size = 1.09 \begin {gather*} \frac {{\left (a^{2} b c - a^{3} d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {3 \, b^{4} d x^{5} + 5 \, b^{4} c x^{3} - 5 \, a b^{3} d x^{3} - 15 \, a b^{3} c x + 15 \, a^{2} b^{2} d x}{15 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*b*c - a^3*d)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/15*(3*b^4*d*x^5 + 5*b^4*c*x^3 - 5*a*b^3*d*x^3 - 15
*a*b^3*c*x + 15*a^2*b^2*d*x)/b^5

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maple [A]  time = 0.00, size = 92, normalized size = 1.19 \begin {gather*} \frac {d \,x^{5}}{5 b}-\frac {a d \,x^{3}}{3 b^{2}}+\frac {c \,x^{3}}{3 b}-\frac {a^{3} d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{3}}+\frac {a^{2} c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{2}}+\frac {a^{2} d x}{b^{3}}-\frac {a c x}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*d*x^5/b-1/3/b^2*x^3*a*d+1/3/b*x^3*c+1/b^3*a^2*d*x-1/b^2*a*c*x-a^3/b^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x
)*d+a^2/b^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)*c

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maxima [A]  time = 2.53, size = 77, normalized size = 1.00 \begin {gather*} \frac {{\left (a^{2} b c - a^{3} d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {3 \, b^{2} d x^{5} + 5 \, {\left (b^{2} c - a b d\right )} x^{3} - 15 \, {\left (a b c - a^{2} d\right )} x}{15 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*b*c - a^3*d)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/15*(3*b^2*d*x^5 + 5*(b^2*c - a*b*d)*x^3 - 15*(a*b*
c - a^2*d)*x)/b^3

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mupad [B]  time = 0.12, size = 96, normalized size = 1.25 \begin {gather*} x^3\,\left (\frac {c}{3\,b}-\frac {a\,d}{3\,b^2}\right )+\frac {d\,x^5}{5\,b}-\frac {a^{3/2}\,\mathrm {atan}\left (\frac {a^{3/2}\,\sqrt {b}\,x\,\left (a\,d-b\,c\right )}{a^3\,d-a^2\,b\,c}\right )\,\left (a\,d-b\,c\right )}{b^{7/2}}-\frac {a\,x\,\left (\frac {c}{b}-\frac {a\,d}{b^2}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(c + d*x^2))/(a + b*x^2),x)

[Out]

x^3*(c/(3*b) - (a*d)/(3*b^2)) + (d*x^5)/(5*b) - (a^(3/2)*atan((a^(3/2)*b^(1/2)*x*(a*d - b*c))/(a^3*d - a^2*b*c
))*(a*d - b*c))/b^(7/2) - (a*x*(c/b - (a*d)/b^2))/b

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sympy [B]  time = 0.36, size = 153, normalized size = 1.99 \begin {gather*} x^{3} \left (- \frac {a d}{3 b^{2}} + \frac {c}{3 b}\right ) + x \left (\frac {a^{2} d}{b^{3}} - \frac {a c}{b^{2}}\right ) + \frac {\sqrt {- \frac {a^{3}}{b^{7}}} \left (a d - b c\right ) \log {\left (- \frac {b^{3} \sqrt {- \frac {a^{3}}{b^{7}}} \left (a d - b c\right )}{a^{2} d - a b c} + x \right )}}{2} - \frac {\sqrt {- \frac {a^{3}}{b^{7}}} \left (a d - b c\right ) \log {\left (\frac {b^{3} \sqrt {- \frac {a^{3}}{b^{7}}} \left (a d - b c\right )}{a^{2} d - a b c} + x \right )}}{2} + \frac {d x^{5}}{5 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3*(-a*d/(3*b**2) + c/(3*b)) + x*(a**2*d/b**3 - a*c/b**2) + sqrt(-a**3/b**7)*(a*d - b*c)*log(-b**3*sqrt(-a**
3/b**7)*(a*d - b*c)/(a**2*d - a*b*c) + x)/2 - sqrt(-a**3/b**7)*(a*d - b*c)*log(b**3*sqrt(-a**3/b**7)*(a*d - b*
c)/(a**2*d - a*b*c) + x)/2 + d*x**5/(5*b)

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