3.7.44 \(\int \frac {x^2 (a+b x^2)^2}{(c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=121 \[ -\frac {b (5 b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{7/2}}+\frac {2 b x (b c-a d)}{d^3 \sqrt {c+d x^2}}+\frac {x^3 (b c-a d)^2}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {b^2 x \sqrt {c+d x^2}}{2 d^3} \]

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Rubi [A]  time = 0.11, antiderivative size = 121, 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, 455, 388, 217, 206} \begin {gather*} \frac {x^3 (b c-a d)^2}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {2 b x (b c-a d)}{d^3 \sqrt {c+d x^2}}-\frac {b (5 b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{7/2}}+\frac {b^2 x \sqrt {c+d x^2}}{2 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*x^2)^2)/(c + d*x^2)^(5/2),x]

[Out]

((b*c - a*d)^2*x^3)/(3*c*d^2*(c + d*x^2)^(3/2)) + (2*b*(b*c - a*d)*x)/(d^3*Sqrt[c + d*x^2]) + (b^2*x*Sqrt[c +
d*x^2])/(2*d^3) - (b*(5*b*c - 4*a*d)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(2*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 388

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

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 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 )^{5/2}} \, dx &=\frac {(b c-a d)^2 x^3}{3 c d^2 \left (c+d x^2\right )^{3/2}}-\frac {\int \frac {x^2 \left (3 b c (b c-2 a d)-3 b^2 c d x^2\right )}{\left (c+d x^2\right )^{3/2}} \, dx}{3 c d^2}\\ &=\frac {(b c-a d)^2 x^3}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {2 b (b c-a d) x}{d^3 \sqrt {c+d x^2}}+\frac {\int \frac {-6 b c d (b c-a d)+3 b^2 c d^2 x^2}{\sqrt {c+d x^2}} \, dx}{3 c d^4}\\ &=\frac {(b c-a d)^2 x^3}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {2 b (b c-a d) x}{d^3 \sqrt {c+d x^2}}+\frac {b^2 x \sqrt {c+d x^2}}{2 d^3}-\frac {(b (5 b c-4 a d)) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{2 d^3}\\ &=\frac {(b c-a d)^2 x^3}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {2 b (b c-a d) x}{d^3 \sqrt {c+d x^2}}+\frac {b^2 x \sqrt {c+d x^2}}{2 d^3}-\frac {(b (5 b c-4 a d)) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{2 d^3}\\ &=\frac {(b c-a d)^2 x^3}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {2 b (b c-a d) x}{d^3 \sqrt {c+d x^2}}+\frac {b^2 x \sqrt {c+d x^2}}{2 d^3}-\frac {b (5 b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 118, normalized size = 0.98 \begin {gather*} \frac {x \left (2 a^2 d^3 x^2-4 a b c d \left (3 c+4 d x^2\right )+b^2 c \left (15 c^2+20 c d x^2+3 d^2 x^4\right )\right )}{6 c d^3 \left (c+d x^2\right )^{3/2}}+\frac {b (4 a d-5 b c) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{2 d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*x^2)^2)/(c + d*x^2)^(5/2),x]

[Out]

(x*(2*a^2*d^3*x^2 - 4*a*b*c*d*(3*c + 4*d*x^2) + b^2*c*(15*c^2 + 20*c*d*x^2 + 3*d^2*x^4)))/(6*c*d^3*(c + d*x^2)
^(3/2)) + (b*(-5*b*c + 4*a*d)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/(2*d^(7/2))

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IntegrateAlgebraic [A]  time = 0.25, size = 129, normalized size = 1.07 \begin {gather*} \frac {2 a^2 d^3 x^3-12 a b c^2 d x-16 a b c d^2 x^3+15 b^2 c^3 x+20 b^2 c^2 d x^3+3 b^2 c d^2 x^5}{6 c d^3 \left (c+d x^2\right )^{3/2}}+\frac {\left (5 b^2 c-4 a b d\right ) \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )}{2 d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^2*(a + b*x^2)^2)/(c + d*x^2)^(5/2),x]

[Out]

(15*b^2*c^3*x - 12*a*b*c^2*d*x + 20*b^2*c^2*d*x^3 - 16*a*b*c*d^2*x^3 + 2*a^2*d^3*x^3 + 3*b^2*c*d^2*x^5)/(6*c*d
^3*(c + d*x^2)^(3/2)) + ((5*b^2*c - 4*a*b*d)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/(2*d^(7/2))

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fricas [A]  time = 1.23, size = 409, normalized size = 3.38 \begin {gather*} \left [-\frac {3 \, {\left (5 \, b^{2} c^{4} - 4 \, a b c^{3} d + {\left (5 \, b^{2} c^{2} d^{2} - 4 \, a b c d^{3}\right )} x^{4} + 2 \, {\left (5 \, b^{2} c^{3} d - 4 \, a b c^{2} d^{2}\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 (3 \, b^{2} c d^{3} x^{5} + 2 \, {\left (10 \, b^{2} c^{2} d^{2} - 8 \, a b c d^{3} + a^{2} d^{4}\right )} x^{3} + 3 \, {\left (5 \, b^{2} c^{3} d - 4 \, a b c^{2} d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{12 \, {\left (c d^{6} x^{4} + 2 \, c^{2} d^{5} x^{2} + c^{3} d^{4}\right )}}, \frac {3 \, {\left (5 \, b^{2} c^{4} - 4 \, a b c^{3} d + {\left (5 \, b^{2} c^{2} d^{2} - 4 \, a b c d^{3}\right )} x^{4} + 2 \, {\left (5 \, b^{2} c^{3} d - 4 \, a b c^{2} d^{2}\right )} x^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (3 \, b^{2} c d^{3} x^{5} + 2 \, {\left (10 \, b^{2} c^{2} d^{2} - 8 \, a b c d^{3} + a^{2} d^{4}\right )} x^{3} + 3 \, {\left (5 \, b^{2} c^{3} d - 4 \, a b c^{2} d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{6 \, {\left (c d^{6} x^{4} + 2 \, c^{2} d^{5} x^{2} + c^{3} 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)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(5*b^2*c^4 - 4*a*b*c^3*d + (5*b^2*c^2*d^2 - 4*a*b*c*d^3)*x^4 + 2*(5*b^2*c^3*d - 4*a*b*c^2*d^2)*x^2)*
sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 2*(3*b^2*c*d^3*x^5 + 2*(10*b^2*c^2*d^2 - 8*a*b*c*d^3
+ a^2*d^4)*x^3 + 3*(5*b^2*c^3*d - 4*a*b*c^2*d^2)*x)*sqrt(d*x^2 + c))/(c*d^6*x^4 + 2*c^2*d^5*x^2 + c^3*d^4), 1
/6*(3*(5*b^2*c^4 - 4*a*b*c^3*d + (5*b^2*c^2*d^2 - 4*a*b*c*d^3)*x^4 + 2*(5*b^2*c^3*d - 4*a*b*c^2*d^2)*x^2)*sqrt
(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (3*b^2*c*d^3*x^5 + 2*(10*b^2*c^2*d^2 - 8*a*b*c*d^3 + a^2*d^4)*x^3 +
3*(5*b^2*c^3*d - 4*a*b*c^2*d^2)*x)*sqrt(d*x^2 + c))/(c*d^6*x^4 + 2*c^2*d^5*x^2 + c^3*d^4)]

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giac [A]  time = 0.47, size = 130, normalized size = 1.07 \begin {gather*} \frac {{\left ({\left (\frac {3 \, b^{2} x^{2}}{d} + \frac {2 \, {\left (10 \, b^{2} c^{2} d^{3} - 8 \, a b c d^{4} + a^{2} d^{5}\right )}}{c d^{5}}\right )} x^{2} + \frac {3 \, {\left (5 \, b^{2} c^{3} d^{2} - 4 \, a b c^{2} d^{3}\right )}}{c d^{5}}\right )} x}{6 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} + \frac {{\left (5 \, b^{2} c - 4 \, a b d\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{2 \, 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)^(5/2),x, algorithm="giac")

[Out]

1/6*((3*b^2*x^2/d + 2*(10*b^2*c^2*d^3 - 8*a*b*c*d^4 + a^2*d^5)/(c*d^5))*x^2 + 3*(5*b^2*c^3*d^2 - 4*a*b*c^2*d^3
)/(c*d^5))*x/(d*x^2 + c)^(3/2) + 1/2*(5*b^2*c - 4*a*b*d)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(7/2)

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maple [A]  time = 0.01, size = 185, normalized size = 1.53 \begin {gather*} \frac {b^{2} x^{5}}{2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} d}-\frac {2 a b \,x^{3}}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} d}+\frac {5 b^{2} c \,x^{3}}{6 \left (d \,x^{2}+c \right )^{\frac {3}{2}} d^{2}}-\frac {a^{2} x}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} d}+\frac {a^{2} x}{3 \sqrt {d \,x^{2}+c}\, c d}-\frac {2 a b x}{\sqrt {d \,x^{2}+c}\, d^{2}}+\frac {5 b^{2} c x}{2 \sqrt {d \,x^{2}+c}\, d^{3}}+\frac {2 a b \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{d^{\frac {5}{2}}}-\frac {5 b^{2} c \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{2 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)^(5/2),x)

[Out]

1/2*b^2*x^5/d/(d*x^2+c)^(3/2)+5/6*b^2*c/d^2*x^3/(d*x^2+c)^(3/2)+5/2*b^2*c/d^3*x/(d*x^2+c)^(1/2)-5/2*b^2*c/d^(7
/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))-2/3*a*b*x^3/d/(d*x^2+c)^(3/2)-2*a*b/d^2*x/(d*x^2+c)^(1/2)+2*a*b/d^(5/2)*ln(d
^(1/2)*x+(d*x^2+c)^(1/2))-1/3*a^2/d*x/(d*x^2+c)^(3/2)+1/3*a^2/c/d*x/(d*x^2+c)^(1/2)

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maxima [B]  time = 0.87, size = 211, normalized size = 1.74 \begin {gather*} \frac {b^{2} x^{5}}{2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d} - \frac {2}{3} \, a b x {\left (\frac {3 \, x^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d} + \frac {2 \, c}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{2}}\right )} + \frac {5 \, b^{2} c x {\left (\frac {3 \, x^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d} + \frac {2 \, c}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{2}}\right )}}{6 \, d} + \frac {5 \, b^{2} c x}{6 \, \sqrt {d x^{2} + c} d^{3}} - \frac {2 \, a b x}{3 \, \sqrt {d x^{2} + c} d^{2}} - \frac {a^{2} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d} + \frac {a^{2} x}{3 \, \sqrt {d x^{2} + c} c d} - \frac {5 \, b^{2} c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, d^{\frac {7}{2}}} + \frac {2 \, a b \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{d^{\frac {5}{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)^(5/2),x, algorithm="maxima")

[Out]

1/2*b^2*x^5/((d*x^2 + c)^(3/2)*d) - 2/3*a*b*x*(3*x^2/((d*x^2 + c)^(3/2)*d) + 2*c/((d*x^2 + c)^(3/2)*d^2)) + 5/
6*b^2*c*x*(3*x^2/((d*x^2 + c)^(3/2)*d) + 2*c/((d*x^2 + c)^(3/2)*d^2))/d + 5/6*b^2*c*x/(sqrt(d*x^2 + c)*d^3) -
2/3*a*b*x/(sqrt(d*x^2 + c)*d^2) - 1/3*a^2*x/((d*x^2 + c)^(3/2)*d) + 1/3*a^2*x/(sqrt(d*x^2 + c)*c*d) - 5/2*b^2*
c*arcsinh(d*x/sqrt(c*d))/d^(7/2) + 2*a*b*arcsinh(d*x/sqrt(c*d))/d^(5/2)

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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 )}^{5/2}} \,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)^(5/2),x)

[Out]

int((x^2*(a + b*x^2)^2)/(c + d*x^2)^(5/2), 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 {5}{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)**(5/2),x)

[Out]

Integral(x**2*(a + b*x**2)**2/(c + d*x**2)**(5/2), x)

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