3.7.32 \(\int \frac {x^2 (a+b x^2)^2}{(c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=152 \[ \frac {\left (8 a^2 d^2-24 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{7/2}}-\frac {x \sqrt {c+d x^2} \left (8 a^2 d^2-24 a b c d+15 b^2 c^2\right )}{8 c d^3}+\frac {x^3 (b c-a d)^2}{c d^2 \sqrt {c+d x^2}}+\frac {b^2 x^3 \sqrt {c+d x^2}}{4 d^2} \]

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Rubi [A]  time = 0.12, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {463, 459, 321, 217, 206} \begin {gather*} -\frac {x \sqrt {c+d x^2} \left (8 a^2 d^2-24 a b c d+15 b^2 c^2\right )}{8 c d^3}+\frac {\left (8 a^2 d^2-24 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{7/2}}+\frac {x^3 (b c-a d)^2}{c d^2 \sqrt {c+d x^2}}+\frac {b^2 x^3 \sqrt {c+d x^2}}{4 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*x^2)^2)/(c + d*x^2)^(3/2),x]

[Out]

((b*c - a*d)^2*x^3)/(c*d^2*Sqrt[c + d*x^2]) - ((15*b^2*c^2 - 24*a*b*c*d + 8*a^2*d^2)*x*Sqrt[c + d*x^2])/(8*c*d
^3) + (b^2*x^3*Sqrt[c + d*x^2])/(4*d^2) + ((15*b^2*c^2 - 24*a*b*c*d + 8*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c +
d*x^2]])/(8*d^(7/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 321

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

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]

Rule 463

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> -Simp[((b*c - a*
d)^2*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b^2*e*n*(p + 1)), x] + Dist[1/(a*b^2*n*(p + 1)), Int[(e*x)^m*(a + b
*x^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a,
b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx &=\frac {(b c-a d)^2 x^3}{c d^2 \sqrt {c+d x^2}}-\frac {\int \frac {x^2 \left (-a^2 d^2+3 (b c-a d)^2-b^2 c d x^2\right )}{\sqrt {c+d x^2}} \, dx}{c d^2}\\ &=\frac {(b c-a d)^2 x^3}{c d^2 \sqrt {c+d x^2}}+\frac {b^2 x^3 \sqrt {c+d x^2}}{4 d^2}-\frac {\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) \int \frac {x^2}{\sqrt {c+d x^2}} \, dx}{4 c d^2}\\ &=\frac {(b c-a d)^2 x^3}{c d^2 \sqrt {c+d x^2}}-\frac {\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{8 c d^3}+\frac {b^2 x^3 \sqrt {c+d x^2}}{4 d^2}+\frac {\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{8 d^3}\\ &=\frac {(b c-a d)^2 x^3}{c d^2 \sqrt {c+d x^2}}-\frac {\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{8 c d^3}+\frac {b^2 x^3 \sqrt {c+d x^2}}{4 d^2}+\frac {\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{8 d^3}\\ &=\frac {(b c-a d)^2 x^3}{c d^2 \sqrt {c+d x^2}}-\frac {\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{8 c d^3}+\frac {b^2 x^3 \sqrt {c+d x^2}}{4 d^2}+\frac {\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 124, normalized size = 0.82 \begin {gather*} \frac {\left (8 a^2 d^2-24 a b c d+15 b^2 c^2\right ) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{8 d^{7/2}}+\sqrt {c+d x^2} \left (-\frac {x (a d-b c)^2}{d^3 \left (c+d x^2\right )}-\frac {b x (7 b c-8 a d)}{8 d^3}+\frac {b^2 x^3}{4 d^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*x^2)^2)/(c + d*x^2)^(3/2),x]

[Out]

Sqrt[c + d*x^2]*(-1/8*(b*(7*b*c - 8*a*d)*x)/d^3 + (b^2*x^3)/(4*d^2) - ((-(b*c) + a*d)^2*x)/(d^3*(c + d*x^2)))
+ ((15*b^2*c^2 - 24*a*b*c*d + 8*a^2*d^2)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/(8*d^(7/2))

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IntegrateAlgebraic [A]  time = 0.22, size = 129, normalized size = 0.85 \begin {gather*} \frac {\left (-8 a^2 d^2+24 a b c d-15 b^2 c^2\right ) \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )}{8 d^{7/2}}+\frac {-8 a^2 d^2 x+24 a b c d x+8 a b d^2 x^3-15 b^2 c^2 x-5 b^2 c d x^3+2 b^2 d^2 x^5}{8 d^3 \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^2*(a + b*x^2)^2)/(c + d*x^2)^(3/2),x]

[Out]

(-15*b^2*c^2*x + 24*a*b*c*d*x - 8*a^2*d^2*x - 5*b^2*c*d*x^3 + 8*a*b*d^2*x^3 + 2*b^2*d^2*x^5)/(8*d^3*Sqrt[c + d
*x^2]) + ((-15*b^2*c^2 + 24*a*b*c*d - 8*a^2*d^2)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/(8*d^(7/2))

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fricas [A]  time = 1.72, size = 350, normalized size = 2.30 \begin {gather*} \left [\frac {{\left (15 \, b^{2} c^{3} - 24 \, a b c^{2} d + 8 \, a^{2} c d^{2} + {\left (15 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (2 \, b^{2} d^{3} x^{5} - {\left (5 \, b^{2} c d^{2} - 8 \, a b d^{3}\right )} x^{3} - {\left (15 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{16 \, {\left (d^{5} x^{2} + c d^{4}\right )}}, -\frac {{\left (15 \, b^{2} c^{3} - 24 \, a b c^{2} d + 8 \, a^{2} c d^{2} + {\left (15 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (2 \, b^{2} d^{3} x^{5} - {\left (5 \, b^{2} c d^{2} - 8 \, a b d^{3}\right )} x^{3} - {\left (15 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{8 \, {\left (d^{5} x^{2} + c d^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

[1/16*((15*b^2*c^3 - 24*a*b*c^2*d + 8*a^2*c*d^2 + (15*b^2*c^2*d - 24*a*b*c*d^2 + 8*a^2*d^3)*x^2)*sqrt(d)*log(-
2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(2*b^2*d^3*x^5 - (5*b^2*c*d^2 - 8*a*b*d^3)*x^3 - (15*b^2*c^2*d
- 24*a*b*c*d^2 + 8*a^2*d^3)*x)*sqrt(d*x^2 + c))/(d^5*x^2 + c*d^4), -1/8*((15*b^2*c^3 - 24*a*b*c^2*d + 8*a^2*c*
d^2 + (15*b^2*c^2*d - 24*a*b*c*d^2 + 8*a^2*d^3)*x^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (2*b^2*d^3*
x^5 - (5*b^2*c*d^2 - 8*a*b*d^3)*x^3 - (15*b^2*c^2*d - 24*a*b*c*d^2 + 8*a^2*d^3)*x)*sqrt(d*x^2 + c))/(d^5*x^2 +
c*d^4)]

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giac [A]  time = 0.43, size = 131, normalized size = 0.86 \begin {gather*} \frac {{\left ({\left (\frac {2 \, b^{2} x^{2}}{d} - \frac {5 \, b^{2} c d^{3} - 8 \, a b d^{4}}{d^{5}}\right )} x^{2} - \frac {15 \, b^{2} c^{2} d^{2} - 24 \, a b c d^{3} + 8 \, a^{2} d^{4}}{d^{5}}\right )} x}{8 \, \sqrt {d x^{2} + c}} - \frac {{\left (15 \, b^{2} c^{2} - 24 \, a b c d + 8 \, a^{2} d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{8 \, d^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

1/8*((2*b^2*x^2/d - (5*b^2*c*d^3 - 8*a*b*d^4)/d^5)*x^2 - (15*b^2*c^2*d^2 - 24*a*b*c*d^3 + 8*a^2*d^4)/d^5)*x/sq
rt(d*x^2 + c) - 1/8*(15*b^2*c^2 - 24*a*b*c*d + 8*a^2*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(7/2)

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maple [A]  time = 0.01, size = 192, normalized size = 1.26 \begin {gather*} \frac {b^{2} x^{5}}{4 \sqrt {d \,x^{2}+c}\, d}+\frac {a b \,x^{3}}{\sqrt {d \,x^{2}+c}\, d}-\frac {5 b^{2} c \,x^{3}}{8 \sqrt {d \,x^{2}+c}\, d^{2}}-\frac {a^{2} x}{\sqrt {d \,x^{2}+c}\, d}+\frac {3 a b c x}{\sqrt {d \,x^{2}+c}\, d^{2}}-\frac {15 b^{2} c^{2} x}{8 \sqrt {d \,x^{2}+c}\, d^{3}}+\frac {a^{2} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}-\frac {3 a b c \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{d^{\frac {5}{2}}}+\frac {15 b^{2} c^{2} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{8 d^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x^2+a)^2/(d*x^2+c)^(3/2),x)

[Out]

1/4*b^2*x^5/d/(d*x^2+c)^(1/2)-5/8*b^2*c/d^2*x^3/(d*x^2+c)^(1/2)-15/8*b^2*c^2/d^3*x/(d*x^2+c)^(1/2)+15/8*b^2*c^
2/d^(7/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))+a*b*x^3/d/(d*x^2+c)^(1/2)+3*a*b*c/d^2*x/(d*x^2+c)^(1/2)-3*a*b*c/d^(5/2
)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))-a^2*x/d/(d*x^2+c)^(1/2)+a^2/d^(3/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))

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maxima [A]  time = 0.87, size = 170, normalized size = 1.12 \begin {gather*} \frac {b^{2} x^{5}}{4 \, \sqrt {d x^{2} + c} d} - \frac {5 \, b^{2} c x^{3}}{8 \, \sqrt {d x^{2} + c} d^{2}} + \frac {a b x^{3}}{\sqrt {d x^{2} + c} d} - \frac {15 \, b^{2} c^{2} x}{8 \, \sqrt {d x^{2} + c} d^{3}} + \frac {3 \, a b c x}{\sqrt {d x^{2} + c} d^{2}} - \frac {a^{2} x}{\sqrt {d x^{2} + c} d} + \frac {15 \, b^{2} c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {7}{2}}} - \frac {3 \, a b c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{d^{\frac {5}{2}}} + \frac {a^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{d^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

1/4*b^2*x^5/(sqrt(d*x^2 + c)*d) - 5/8*b^2*c*x^3/(sqrt(d*x^2 + c)*d^2) + a*b*x^3/(sqrt(d*x^2 + c)*d) - 15/8*b^2
*c^2*x/(sqrt(d*x^2 + c)*d^3) + 3*a*b*c*x/(sqrt(d*x^2 + c)*d^2) - a^2*x/(sqrt(d*x^2 + c)*d) + 15/8*b^2*c^2*arcs
inh(d*x/sqrt(c*d))/d^(7/2) - 3*a*b*c*arcsinh(d*x/sqrt(c*d))/d^(5/2) + a^2*arcsinh(d*x/sqrt(c*d))/d^(3/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,{\left (b\,x^2+a\right )}^2}{{\left (d\,x^2+c\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(a + b*x^2)^2)/(c + d*x^2)^(3/2),x)

[Out]

int((x^2*(a + b*x^2)^2)/(c + d*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \left (a + b x^{2}\right )^{2}}{\left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x**2+a)**2/(d*x**2+c)**(3/2),x)

[Out]

Integral(x**2*(a + b*x**2)**2/(c + d*x**2)**(3/2), x)

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