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

Optimal. Leaf size=146 \[ -\frac {c \left (8 a^2 d^2+b c (5 b c-12 a d)\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{7/2}}+\frac {x \sqrt {c+d x^2} \left (8 a^2 d^2+b c (5 b c-12 a d)\right )}{16 d^3}-\frac {b x^3 \sqrt {c+d x^2} (5 b c-12 a d)}{24 d^2}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d} \]

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Rubi [A]  time = 0.14, antiderivative size = 146, 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 = {464, 459, 321, 217, 206} \begin {gather*} \frac {x \sqrt {c+d x^2} \left (8 a^2+\frac {b c (5 b c-12 a d)}{d^2}\right )}{16 d}-\frac {c \left (8 a^2 d^2+b c (5 b c-12 a d)\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{7/2}}-\frac {b x^3 \sqrt {c+d x^2} (5 b c-12 a d)}{24 d^2}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^2 + (b*c*(5*b*c - 12*a*d))/d^2)*x*Sqrt[c + d*x^2])/(16*d) - (b*(5*b*c - 12*a*d)*x^3*Sqrt[c + d*x^2])/(24
*d^2) + (b^2*x^5*Sqrt[c + d*x^2])/(6*d) - (c*(8*a^2*d^2 + b*c*(5*b*c - 12*a*d))*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d
*x^2]])/(16*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 464

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

Rubi steps

\begin {align*} \int \frac {x^2 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx &=\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d}+\frac {\int \frac {x^2 \left (6 a^2 d-b (5 b c-12 a d) x^2\right )}{\sqrt {c+d x^2}} \, dx}{6 d}\\ &=-\frac {b (5 b c-12 a d) x^3 \sqrt {c+d x^2}}{24 d^2}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d}+\frac {1}{8} \left (8 a^2+\frac {b c (5 b c-12 a d)}{d^2}\right ) \int \frac {x^2}{\sqrt {c+d x^2}} \, dx\\ &=\frac {\left (8 a^2+\frac {b c (5 b c-12 a d)}{d^2}\right ) x \sqrt {c+d x^2}}{16 d}-\frac {b (5 b c-12 a d) x^3 \sqrt {c+d x^2}}{24 d^2}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d}-\frac {\left (c \left (5 b^2 c^2-12 a b c d+8 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{16 d^3}\\ &=\frac {\left (8 a^2+\frac {b c (5 b c-12 a d)}{d^2}\right ) x \sqrt {c+d x^2}}{16 d}-\frac {b (5 b c-12 a d) x^3 \sqrt {c+d x^2}}{24 d^2}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d}-\frac {\left (c \left (5 b^2 c^2-12 a b c d+8 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{16 d^3}\\ &=\frac {\left (8 a^2+\frac {b c (5 b c-12 a d)}{d^2}\right ) x \sqrt {c+d x^2}}{16 d}-\frac {b (5 b c-12 a d) x^3 \sqrt {c+d x^2}}{24 d^2}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d}-\frac {c \left (5 b^2 c^2-12 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 125, normalized size = 0.86 \begin {gather*} \frac {\sqrt {d} x \sqrt {c+d x^2} \left (24 a^2 d^2+12 a b d \left (2 d x^2-3 c\right )+b^2 \left (15 c^2-10 c d x^2+8 d^2 x^4\right )\right )-3 c \left (8 a^2 d^2-12 a b c d+5 b^2 c^2\right ) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{48 d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d]*x*Sqrt[c + d*x^2]*(24*a^2*d^2 + 12*a*b*d*(-3*c + 2*d*x^2) + b^2*(15*c^2 - 10*c*d*x^2 + 8*d^2*x^4)) -
3*c*(5*b^2*c^2 - 12*a*b*c*d + 8*a^2*d^2)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/(48*d^(7/2))

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IntegrateAlgebraic [A]  time = 0.22, size = 132, normalized size = 0.90 \begin {gather*} \frac {\sqrt {c+d x^2} \left (24 a^2 d^2 x-36 a b c d x+24 a b d^2 x^3+15 b^2 c^2 x-10 b^2 c d x^3+8 b^2 d^2 x^5\right )}{48 d^3}+\frac {\left (8 a^2 c d^2-12 a b c^2 d+5 b^2 c^3\right ) \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )}{16 d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(15*b^2*c^2*x - 36*a*b*c*d*x + 24*a^2*d^2*x - 10*b^2*c*d*x^3 + 24*a*b*d^2*x^3 + 8*b^2*d^2*x^5
))/(48*d^3) + ((5*b^2*c^3 - 12*a*b*c^2*d + 8*a^2*c*d^2)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/(16*d^(7/2))

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fricas [A]  time = 0.92, size = 267, normalized size = 1.83 \begin {gather*} \left [\frac {3 \, {\left (5 \, b^{2} c^{3} - 12 \, a b c^{2} d + 8 \, a^{2} c d^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (8 \, b^{2} d^{3} x^{5} - 2 \, {\left (5 \, b^{2} c d^{2} - 12 \, a b d^{3}\right )} x^{3} + 3 \, {\left (5 \, b^{2} c^{2} d - 12 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{96 \, d^{4}}, \frac {3 \, {\left (5 \, b^{2} c^{3} - 12 \, a b c^{2} d + 8 \, a^{2} c d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (8 \, b^{2} d^{3} x^{5} - 2 \, {\left (5 \, b^{2} c d^{2} - 12 \, a b d^{3}\right )} x^{3} + 3 \, {\left (5 \, b^{2} c^{2} d - 12 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{48 \, d^{4}}\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)^(1/2),x, algorithm="fricas")

[Out]

[1/96*(3*(5*b^2*c^3 - 12*a*b*c^2*d + 8*a^2*c*d^2)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*
(8*b^2*d^3*x^5 - 2*(5*b^2*c*d^2 - 12*a*b*d^3)*x^3 + 3*(5*b^2*c^2*d - 12*a*b*c*d^2 + 8*a^2*d^3)*x)*sqrt(d*x^2 +
c))/d^4, 1/48*(3*(5*b^2*c^3 - 12*a*b*c^2*d + 8*a^2*c*d^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (8*b^
2*d^3*x^5 - 2*(5*b^2*c*d^2 - 12*a*b*d^3)*x^3 + 3*(5*b^2*c^2*d - 12*a*b*c*d^2 + 8*a^2*d^3)*x)*sqrt(d*x^2 + c))/
d^4]

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giac [A]  time = 0.37, size = 135, normalized size = 0.92 \begin {gather*} \frac {1}{48} \, {\left (2 \, {\left (\frac {4 \, b^{2} x^{2}}{d} - \frac {5 \, b^{2} c d^{3} - 12 \, a b d^{4}}{d^{5}}\right )} x^{2} + \frac {3 \, {\left (5 \, b^{2} c^{2} d^{2} - 12 \, a b c d^{3} + 8 \, a^{2} d^{4}\right )}}{d^{5}}\right )} \sqrt {d x^{2} + c} x + \frac {{\left (5 \, b^{2} c^{3} - 12 \, a b c^{2} d + 8 \, a^{2} c d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{16 \, 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)^(1/2),x, algorithm="giac")

[Out]

1/48*(2*(4*b^2*x^2/d - (5*b^2*c*d^3 - 12*a*b*d^4)/d^5)*x^2 + 3*(5*b^2*c^2*d^2 - 12*a*b*c*d^3 + 8*a^2*d^4)/d^5)
*sqrt(d*x^2 + c)*x + 1/16*(5*b^2*c^3 - 12*a*b*c^2*d + 8*a^2*c*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 = 197, normalized size = 1.35 \begin {gather*} \frac {\sqrt {d \,x^{2}+c}\, b^{2} x^{5}}{6 d}+\frac {\sqrt {d \,x^{2}+c}\, a b \,x^{3}}{2 d}-\frac {5 \sqrt {d \,x^{2}+c}\, b^{2} c \,x^{3}}{24 d^{2}}-\frac {a^{2} c \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {3}{2}}}+\frac {3 a b \,c^{2} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{4 d^{\frac {5}{2}}}-\frac {5 b^{2} c^{3} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{16 d^{\frac {7}{2}}}+\frac {\sqrt {d \,x^{2}+c}\, a^{2} x}{2 d}-\frac {3 \sqrt {d \,x^{2}+c}\, a b c x}{4 d^{2}}+\frac {5 \sqrt {d \,x^{2}+c}\, b^{2} c^{2} x}{16 d^{3}} \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)^(1/2),x)

[Out]

1/6*b^2*x^5*(d*x^2+c)^(1/2)/d-5/24*b^2*c/d^2*x^3*(d*x^2+c)^(1/2)+5/16*b^2*c^2/d^3*x*(d*x^2+c)^(1/2)-5/16*b^2*c
^3/d^(7/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))+1/2*a*b*x^3/d*(d*x^2+c)^(1/2)-3/4*a*b*c/d^2*x*(d*x^2+c)^(1/2)+3/4*a*b
*c^2/d^(5/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))+1/2*a^2*x/d*(d*x^2+c)^(1/2)-1/2*a^2*c/d^(3/2)*ln(d^(1/2)*x+(d*x^2+c
)^(1/2))

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maxima [A]  time = 0.92, size = 175, normalized size = 1.20 \begin {gather*} \frac {\sqrt {d x^{2} + c} b^{2} x^{5}}{6 \, d} - \frac {5 \, \sqrt {d x^{2} + c} b^{2} c x^{3}}{24 \, d^{2}} + \frac {\sqrt {d x^{2} + c} a b x^{3}}{2 \, d} + \frac {5 \, \sqrt {d x^{2} + c} b^{2} c^{2} x}{16 \, d^{3}} - \frac {3 \, \sqrt {d x^{2} + c} a b c x}{4 \, d^{2}} + \frac {\sqrt {d x^{2} + c} a^{2} x}{2 \, d} - \frac {5 \, b^{2} c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{16 \, d^{\frac {7}{2}}} + \frac {3 \, a b c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{4 \, d^{\frac {5}{2}}} - \frac {a^{2} c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, 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)^(1/2),x, algorithm="maxima")

[Out]

1/6*sqrt(d*x^2 + c)*b^2*x^5/d - 5/24*sqrt(d*x^2 + c)*b^2*c*x^3/d^2 + 1/2*sqrt(d*x^2 + c)*a*b*x^3/d + 5/16*sqrt
(d*x^2 + c)*b^2*c^2*x/d^3 - 3/4*sqrt(d*x^2 + c)*a*b*c*x/d^2 + 1/2*sqrt(d*x^2 + c)*a^2*x/d - 5/16*b^2*c^3*arcsi
nh(d*x/sqrt(c*d))/d^(7/2) + 3/4*a*b*c^2*arcsinh(d*x/sqrt(c*d))/d^(5/2) - 1/2*a^2*c*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}{\sqrt {d\,x^2+c}} \,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)^(1/2),x)

[Out]

int((x^2*(a + b*x^2)^2)/(c + d*x^2)^(1/2), x)

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sympy [B]  time = 16.35, size = 301, normalized size = 2.06 \begin {gather*} \frac {a^{2} \sqrt {c} x \sqrt {1 + \frac {d x^{2}}{c}}}{2 d} - \frac {a^{2} c \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{2 d^{\frac {3}{2}}} - \frac {3 a b c^{\frac {3}{2}} x}{4 d^{2} \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {a b \sqrt {c} x^{3}}{4 d \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {3 a b c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{4 d^{\frac {5}{2}}} + \frac {a b x^{5}}{2 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {5 b^{2} c^{\frac {5}{2}} x}{16 d^{3} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {5 b^{2} c^{\frac {3}{2}} x^{3}}{48 d^{2} \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {b^{2} \sqrt {c} x^{5}}{24 d \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {5 b^{2} c^{3} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{16 d^{\frac {7}{2}}} + \frac {b^{2} x^{7}}{6 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} \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)**(1/2),x)

[Out]

a**2*sqrt(c)*x*sqrt(1 + d*x**2/c)/(2*d) - a**2*c*asinh(sqrt(d)*x/sqrt(c))/(2*d**(3/2)) - 3*a*b*c**(3/2)*x/(4*d
**2*sqrt(1 + d*x**2/c)) - a*b*sqrt(c)*x**3/(4*d*sqrt(1 + d*x**2/c)) + 3*a*b*c**2*asinh(sqrt(d)*x/sqrt(c))/(4*d
**(5/2)) + a*b*x**5/(2*sqrt(c)*sqrt(1 + d*x**2/c)) + 5*b**2*c**(5/2)*x/(16*d**3*sqrt(1 + d*x**2/c)) + 5*b**2*c
**(3/2)*x**3/(48*d**2*sqrt(1 + d*x**2/c)) - b**2*sqrt(c)*x**5/(24*d*sqrt(1 + d*x**2/c)) - 5*b**2*c**3*asinh(sq
rt(d)*x/sqrt(c))/(16*d**(7/2)) + b**2*x**7/(6*sqrt(c)*sqrt(1 + d*x**2/c))

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