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

Optimal. Leaf size=191 \[ -\frac {c^2 \left (16 a^2 d^2+b c (5 b c-16 a d)\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{128 d^{7/2}}+\frac {x^3 \sqrt {c+d x^2} \left (16 a^2 d^2+b c (5 b c-16 a d)\right )}{64 d^2}+\frac {c x \sqrt {c+d x^2} \left (16 a^2 d^2+b c (5 b c-16 a d)\right )}{128 d^3}-\frac {b x^3 \left (c+d x^2\right )^{3/2} (5 b c-16 a d)}{48 d^2}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d} \]

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Rubi [A]  time = 0.19, antiderivative size = 188, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {464, 459, 279, 321, 217, 206} \begin {gather*} -\frac {c^2 \left (16 a^2 d^2+b c (5 b c-16 a d)\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{128 d^{7/2}}+\frac {1}{64} x^3 \sqrt {c+d x^2} \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right )+\frac {c x \sqrt {c+d x^2} \left (16 a^2 d^2+b c (5 b c-16 a d)\right )}{128 d^3}-\frac {b x^3 \left (c+d x^2\right )^{3/2} (5 b c-16 a d)}{48 d^2}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(16*a^2*d^2 + b*c*(5*b*c - 16*a*d))*x*Sqrt[c + d*x^2])/(128*d^3) + ((16*a^2 + (b*c*(5*b*c - 16*a*d))/d^2)*x
^3*Sqrt[c + d*x^2])/64 - (b*(5*b*c - 16*a*d)*x^3*(c + d*x^2)^(3/2))/(48*d^2) + (b^2*x^5*(c + d*x^2)^(3/2))/(8*
d) - (c^2*(16*a^2*d^2 + b*c*(5*b*c - 16*a*d))*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(128*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 279

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

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 x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx &=\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}+\frac {\int x^2 \sqrt {c+d x^2} \left (8 a^2 d-b (5 b c-16 a d) x^2\right ) \, dx}{8 d}\\ &=-\frac {b (5 b c-16 a d) x^3 \left (c+d x^2\right )^{3/2}}{48 d^2}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}+\frac {1}{16} \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) \int x^2 \sqrt {c+d x^2} \, dx\\ &=\frac {1}{64} \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) x^3 \sqrt {c+d x^2}-\frac {b (5 b c-16 a d) x^3 \left (c+d x^2\right )^{3/2}}{48 d^2}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}+\frac {1}{64} \left (c \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right )\right ) \int \frac {x^2}{\sqrt {c+d x^2}} \, dx\\ &=\frac {c \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) x \sqrt {c+d x^2}}{128 d}+\frac {1}{64} \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) x^3 \sqrt {c+d x^2}-\frac {b (5 b c-16 a d) x^3 \left (c+d x^2\right )^{3/2}}{48 d^2}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}-\frac {\left (c^2 \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{128 d}\\ &=\frac {c \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) x \sqrt {c+d x^2}}{128 d}+\frac {1}{64} \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) x^3 \sqrt {c+d x^2}-\frac {b (5 b c-16 a d) x^3 \left (c+d x^2\right )^{3/2}}{48 d^2}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}-\frac {\left (c^2 \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{128 d}\\ &=\frac {c \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) x \sqrt {c+d x^2}}{128 d}+\frac {1}{64} \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) x^3 \sqrt {c+d x^2}-\frac {b (5 b c-16 a d) x^3 \left (c+d x^2\right )^{3/2}}{48 d^2}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}-\frac {c^2 \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{128 d^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 157, normalized size = 0.82 \begin {gather*} \frac {\sqrt {d} x \sqrt {c+d x^2} \left (48 a^2 d^2 \left (c+2 d x^2\right )+16 a b d \left (-3 c^2+2 c d x^2+8 d^2 x^4\right )+b^2 \left (15 c^3-10 c^2 d x^2+8 c d^2 x^4+48 d^3 x^6\right )\right )-3 c^2 \left (16 a^2 d^2-16 a b c d+5 b^2 c^2\right ) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{384 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]*(48*a^2*d^2*(c + 2*d*x^2) + 16*a*b*d*(-3*c^2 + 2*c*d*x^2 + 8*d^2*x^4) + b^2*(15*c^3
- 10*c^2*d*x^2 + 8*c*d^2*x^4 + 48*d^3*x^6)) - 3*c^2*(5*b^2*c^2 - 16*a*b*c*d + 16*a^2*d^2)*Log[d*x + Sqrt[d]*S
qrt[c + d*x^2]])/(384*d^(7/2))

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IntegrateAlgebraic [A]  time = 0.22, size = 173, normalized size = 0.91 \begin {gather*} \frac {\sqrt {c+d x^2} \left (48 a^2 c d^2 x+96 a^2 d^3 x^3-48 a b c^2 d x+32 a b c d^2 x^3+128 a b d^3 x^5+15 b^2 c^3 x-10 b^2 c^2 d x^3+8 b^2 c d^2 x^5+48 b^2 d^3 x^7\right )}{384 d^3}+\frac {\left (16 a^2 c^2 d^2-16 a b c^3 d+5 b^2 c^4\right ) \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )}{128 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^3*x - 48*a*b*c^2*d*x + 48*a^2*c*d^2*x - 10*b^2*c^2*d*x^3 + 32*a*b*c*d^2*x^3 + 96*a^
2*d^3*x^3 + 8*b^2*c*d^2*x^5 + 128*a*b*d^3*x^5 + 48*b^2*d^3*x^7))/(384*d^3) + ((5*b^2*c^4 - 16*a*b*c^3*d + 16*a
^2*c^2*d^2)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/(128*d^(7/2))

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fricas [A]  time = 1.66, size = 341, normalized size = 1.79 \begin {gather*} \left [\frac {3 \, {\left (5 \, b^{2} c^{4} - 16 \, a b c^{3} d + 16 \, a^{2} c^{2} d^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (48 \, b^{2} d^{4} x^{7} + 8 \, {\left (b^{2} c d^{3} + 16 \, a b d^{4}\right )} x^{5} - 2 \, {\left (5 \, b^{2} c^{2} d^{2} - 16 \, a b c d^{3} - 48 \, a^{2} d^{4}\right )} x^{3} + 3 \, {\left (5 \, b^{2} c^{3} d - 16 \, a b c^{2} d^{2} + 16 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{768 \, d^{4}}, \frac {3 \, {\left (5 \, b^{2} c^{4} - 16 \, a b c^{3} d + 16 \, a^{2} c^{2} d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (48 \, b^{2} d^{4} x^{7} + 8 \, {\left (b^{2} c d^{3} + 16 \, a b d^{4}\right )} x^{5} - 2 \, {\left (5 \, b^{2} c^{2} d^{2} - 16 \, a b c d^{3} - 48 \, a^{2} d^{4}\right )} x^{3} + 3 \, {\left (5 \, b^{2} c^{3} d - 16 \, a b c^{2} d^{2} + 16 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{384 \, 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]

[1/768*(3*(5*b^2*c^4 - 16*a*b*c^3*d + 16*a^2*c^2*d^2)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c)
+ 2*(48*b^2*d^4*x^7 + 8*(b^2*c*d^3 + 16*a*b*d^4)*x^5 - 2*(5*b^2*c^2*d^2 - 16*a*b*c*d^3 - 48*a^2*d^4)*x^3 + 3*(
5*b^2*c^3*d - 16*a*b*c^2*d^2 + 16*a^2*c*d^3)*x)*sqrt(d*x^2 + c))/d^4, 1/384*(3*(5*b^2*c^4 - 16*a*b*c^3*d + 16*
a^2*c^2*d^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (48*b^2*d^4*x^7 + 8*(b^2*c*d^3 + 16*a*b*d^4)*x^5 -
2*(5*b^2*c^2*d^2 - 16*a*b*c*d^3 - 48*a^2*d^4)*x^3 + 3*(5*b^2*c^3*d - 16*a*b*c^2*d^2 + 16*a^2*c*d^3)*x)*sqrt(d*
x^2 + c))/d^4]

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giac [A]  time = 0.49, size = 174, normalized size = 0.91 \begin {gather*} \frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, b^{2} x^{2} + \frac {b^{2} c d^{5} + 16 \, a b d^{6}}{d^{6}}\right )} x^{2} - \frac {5 \, b^{2} c^{2} d^{4} - 16 \, a b c d^{5} - 48 \, a^{2} d^{6}}{d^{6}}\right )} x^{2} + \frac {3 \, {\left (5 \, b^{2} c^{3} d^{3} - 16 \, a b c^{2} d^{4} + 16 \, a^{2} c d^{5}\right )}}{d^{6}}\right )} \sqrt {d x^{2} + c} x + \frac {{\left (5 \, b^{2} c^{4} - 16 \, a b c^{3} d + 16 \, a^{2} c^{2} d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{128 \, 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]

1/384*(2*(4*(6*b^2*x^2 + (b^2*c*d^5 + 16*a*b*d^6)/d^6)*x^2 - (5*b^2*c^2*d^4 - 16*a*b*c*d^5 - 48*a^2*d^6)/d^6)*
x^2 + 3*(5*b^2*c^3*d^3 - 16*a*b*c^2*d^4 + 16*a^2*c*d^5)/d^6)*sqrt(d*x^2 + c)*x + 1/128*(5*b^2*c^4 - 16*a*b*c^3
*d + 16*a^2*c^2*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(7/2)

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maple [A]  time = 0.02, size = 259, normalized size = 1.36 \begin {gather*} \frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} b^{2} x^{5}}{8 d}+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a b \,x^{3}}{3 d}-\frac {5 \left (d \,x^{2}+c \right )^{\frac {3}{2}} b^{2} c \,x^{3}}{48 d^{2}}-\frac {a^{2} c^{2} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{8 d^{\frac {3}{2}}}+\frac {a b \,c^{3} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{8 d^{\frac {5}{2}}}-\frac {5 b^{2} c^{4} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{128 d^{\frac {7}{2}}}-\frac {\sqrt {d \,x^{2}+c}\, a^{2} c x}{8 d}+\frac {\sqrt {d \,x^{2}+c}\, a b \,c^{2} x}{8 d^{2}}-\frac {5 \sqrt {d \,x^{2}+c}\, b^{2} c^{3} x}{128 d^{3}}+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2} x}{4 d}-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a b c x}{4 d^{2}}+\frac {5 \left (d \,x^{2}+c \right )^{\frac {3}{2}} b^{2} c^{2} x}{64 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]

1/8*b^2*x^5*(d*x^2+c)^(3/2)/d-5/48*b^2*c/d^2*x^3*(d*x^2+c)^(3/2)+5/64*b^2*c^2/d^3*x*(d*x^2+c)^(3/2)-5/128*b^2*
c^3/d^3*x*(d*x^2+c)^(1/2)-5/128*b^2*c^4/d^(7/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))+1/3*a*b*x^3*(d*x^2+c)^(3/2)/d-1/
4*a*b*c/d^2*x*(d*x^2+c)^(3/2)+1/8*a*b*c^2/d^2*x*(d*x^2+c)^(1/2)+1/8*a*b*c^3/d^(5/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/
2))+1/4*a^2*x*(d*x^2+c)^(3/2)/d-1/8*a^2*c/d*x*(d*x^2+c)^(1/2)-1/8*a^2*c^2/d^(3/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2)
)

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maxima [A]  time = 1.11, size = 237, normalized size = 1.24 \begin {gather*} \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} x^{5}}{8 \, d} - \frac {5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c x^{3}}{48 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b x^{3}}{3 \, d} + \frac {5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{2} x}{64 \, d^{3}} - \frac {5 \, \sqrt {d x^{2} + c} b^{2} c^{3} x}{128 \, d^{3}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c x}{4 \, d^{2}} + \frac {\sqrt {d x^{2} + c} a b c^{2} x}{8 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} x}{4 \, d} - \frac {\sqrt {d x^{2} + c} a^{2} c x}{8 \, d} - \frac {5 \, b^{2} c^{4} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{128 \, d^{\frac {7}{2}}} + \frac {a b c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {5}{2}}} - \frac {a^{2} c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, 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]

1/8*(d*x^2 + c)^(3/2)*b^2*x^5/d - 5/48*(d*x^2 + c)^(3/2)*b^2*c*x^3/d^2 + 1/3*(d*x^2 + c)^(3/2)*a*b*x^3/d + 5/6
4*(d*x^2 + c)^(3/2)*b^2*c^2*x/d^3 - 5/128*sqrt(d*x^2 + c)*b^2*c^3*x/d^3 - 1/4*(d*x^2 + c)^(3/2)*a*b*c*x/d^2 +
1/8*sqrt(d*x^2 + c)*a*b*c^2*x/d^2 + 1/4*(d*x^2 + c)^(3/2)*a^2*x/d - 1/8*sqrt(d*x^2 + c)*a^2*c*x/d - 5/128*b^2*
c^4*arcsinh(d*x/sqrt(c*d))/d^(7/2) + 1/8*a*b*c^3*arcsinh(d*x/sqrt(c*d))/d^(5/2) - 1/8*a^2*c^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 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 = 21.91, size = 411, normalized size = 2.15 \begin {gather*} \frac {a^{2} c^{\frac {3}{2}} x}{8 d \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {3 a^{2} \sqrt {c} x^{3}}{8 \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {a^{2} c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{8 d^{\frac {3}{2}}} + \frac {a^{2} d x^{5}}{4 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {a b c^{\frac {5}{2}} x}{8 d^{2} \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {a b c^{\frac {3}{2}} x^{3}}{24 d \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {5 a b \sqrt {c} x^{5}}{12 \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {a b c^{3} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{8 d^{\frac {5}{2}}} + \frac {a b d x^{7}}{3 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {5 b^{2} c^{\frac {7}{2}} x}{128 d^{3} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {5 b^{2} c^{\frac {5}{2}} x^{3}}{384 d^{2} \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {b^{2} c^{\frac {3}{2}} x^{5}}{192 d \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {7 b^{2} \sqrt {c} x^{7}}{48 \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {5 b^{2} c^{4} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{128 d^{\frac {7}{2}}} + \frac {b^{2} d x^{9}}{8 \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*c**(3/2)*x/(8*d*sqrt(1 + d*x**2/c)) + 3*a**2*sqrt(c)*x**3/(8*sqrt(1 + d*x**2/c)) - a**2*c**2*asinh(sqrt(d
)*x/sqrt(c))/(8*d**(3/2)) + a**2*d*x**5/(4*sqrt(c)*sqrt(1 + d*x**2/c)) - a*b*c**(5/2)*x/(8*d**2*sqrt(1 + d*x**
2/c)) - a*b*c**(3/2)*x**3/(24*d*sqrt(1 + d*x**2/c)) + 5*a*b*sqrt(c)*x**5/(12*sqrt(1 + d*x**2/c)) + a*b*c**3*as
inh(sqrt(d)*x/sqrt(c))/(8*d**(5/2)) + a*b*d*x**7/(3*sqrt(c)*sqrt(1 + d*x**2/c)) + 5*b**2*c**(7/2)*x/(128*d**3*
sqrt(1 + d*x**2/c)) + 5*b**2*c**(5/2)*x**3/(384*d**2*sqrt(1 + d*x**2/c)) - b**2*c**(3/2)*x**5/(192*d*sqrt(1 +
d*x**2/c)) + 7*b**2*sqrt(c)*x**7/(48*sqrt(1 + d*x**2/c)) - 5*b**2*c**4*asinh(sqrt(d)*x/sqrt(c))/(128*d**(7/2))
+ b**2*d*x**9/(8*sqrt(c)*sqrt(1 + d*x**2/c))

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