3.6.86 \(\int \frac {(a+b x^2)^2 \sqrt {c+d x^2}}{x^7} \, dx\)

Optimal. Leaf size=149 \[ -\frac {\sqrt {c+d x^2} \left (a^2 d^2-4 a b c d+8 b^2 c^2\right )}{16 c^2 x^2}-\frac {d \left (a^2 d^2-4 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{5/2}}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{6 c x^6}-\frac {a \left (c+d x^2\right )^{3/2} (4 b c-a d)}{8 c^2 x^4} \]

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Rubi [A]  time = 0.15, antiderivative size = 149, normalized size of antiderivative = 1.00, 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 = {446, 89, 78, 47, 63, 208} \begin {gather*} -\frac {\sqrt {c+d x^2} \left (a^2 d^2-4 a b c d+8 b^2 c^2\right )}{16 c^2 x^2}-\frac {d \left (a^2 d^2-4 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{5/2}}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{6 c x^6}-\frac {a \left (c+d x^2\right )^{3/2} (4 b c-a d)}{8 c^2 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*Sqrt[c + d*x^2])/x^7,x]

[Out]

-((8*b^2*c^2 - 4*a*b*c*d + a^2*d^2)*Sqrt[c + d*x^2])/(16*c^2*x^2) - (a^2*(c + d*x^2)^(3/2))/(6*c*x^6) - (a*(4*
b*c - a*d)*(c + d*x^2)^(3/2))/(8*c^2*x^4) - (d*(8*b^2*c^2 - 4*a*b*c*d + a^2*d^2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[
c]])/(16*c^(5/2))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 78

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

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 208

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

Rule 446

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^7} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^2 \sqrt {c+d x}}{x^4} \, dx,x,x^2\right )\\ &=-\frac {a^2 \left (c+d x^2\right )^{3/2}}{6 c x^6}+\frac {\operatorname {Subst}\left (\int \frac {\left (\frac {3}{2} a (4 b c-a d)+3 b^2 c x\right ) \sqrt {c+d x}}{x^3} \, dx,x,x^2\right )}{6 c}\\ &=-\frac {a^2 \left (c+d x^2\right )^{3/2}}{6 c x^6}-\frac {a (4 b c-a d) \left (c+d x^2\right )^{3/2}}{8 c^2 x^4}+\frac {\left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{x^2} \, dx,x,x^2\right )}{16 c^2}\\ &=-\frac {\left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \sqrt {c+d x^2}}{16 c^2 x^2}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{6 c x^6}-\frac {a (4 b c-a d) \left (c+d x^2\right )^{3/2}}{8 c^2 x^4}+\frac {\left (d \left (8 b^2 c^2-4 a b c d+a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{32 c^2}\\ &=-\frac {\left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \sqrt {c+d x^2}}{16 c^2 x^2}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{6 c x^6}-\frac {a (4 b c-a d) \left (c+d x^2\right )^{3/2}}{8 c^2 x^4}+\frac {\left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{16 c^2}\\ &=-\frac {\left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \sqrt {c+d x^2}}{16 c^2 x^2}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{6 c x^6}-\frac {a (4 b c-a d) \left (c+d x^2\right )^{3/2}}{8 c^2 x^4}-\frac {d \left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 142, normalized size = 0.95 \begin {gather*} \frac {-3 d x^6 \sqrt {\frac {d x^2}{c}+1} \left (a^2 d^2-4 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\sqrt {\frac {d x^2}{c}+1}\right )-\left (c+d x^2\right ) \left (a^2 \left (8 c^2+2 c d x^2-3 d^2 x^4\right )+12 a b c x^2 \left (2 c+d x^2\right )+24 b^2 c^2 x^4\right )}{48 c^2 x^6 \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*Sqrt[c + d*x^2])/x^7,x]

[Out]

(-((c + d*x^2)*(24*b^2*c^2*x^4 + 12*a*b*c*x^2*(2*c + d*x^2) + a^2*(8*c^2 + 2*c*d*x^2 - 3*d^2*x^4))) - 3*d*(8*b
^2*c^2 - 4*a*b*c*d + a^2*d^2)*x^6*Sqrt[1 + (d*x^2)/c]*ArcTanh[Sqrt[1 + (d*x^2)/c]])/(48*c^2*x^6*Sqrt[c + d*x^2
])

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IntegrateAlgebraic [A]  time = 0.23, size = 135, normalized size = 0.91 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-8 a^2 c^2-2 a^2 c d x^2+3 a^2 d^2 x^4-24 a b c^2 x^2-12 a b c d x^4-24 b^2 c^2 x^4\right )}{48 c^2 x^6}+\frac {\left (-a^2 d^3+4 a b c d^2-8 b^2 c^2 d\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x^2)^2*Sqrt[c + d*x^2])/x^7,x]

[Out]

(Sqrt[c + d*x^2]*(-8*a^2*c^2 - 24*a*b*c^2*x^2 - 2*a^2*c*d*x^2 - 24*b^2*c^2*x^4 - 12*a*b*c*d*x^4 + 3*a^2*d^2*x^
4))/(48*c^2*x^6) + ((-8*b^2*c^2*d + 4*a*b*c*d^2 - a^2*d^3)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(16*c^(5/2))

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fricas [A]  time = 1.81, size = 276, normalized size = 1.85 \begin {gather*} \left [\frac {3 \, {\left (8 \, b^{2} c^{2} d - 4 \, a b c d^{2} + a^{2} d^{3}\right )} \sqrt {c} x^{6} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (8 \, a^{2} c^{3} + 3 \, {\left (8 \, b^{2} c^{3} + 4 \, a b c^{2} d - a^{2} c d^{2}\right )} x^{4} + 2 \, {\left (12 \, a b c^{3} + a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{96 \, c^{3} x^{6}}, \frac {3 \, {\left (8 \, b^{2} c^{2} d - 4 \, a b c d^{2} + a^{2} d^{3}\right )} \sqrt {-c} x^{6} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - {\left (8 \, a^{2} c^{3} + 3 \, {\left (8 \, b^{2} c^{3} + 4 \, a b c^{2} d - a^{2} c d^{2}\right )} x^{4} + 2 \, {\left (12 \, a b c^{3} + a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{48 \, c^{3} x^{6}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^7,x, algorithm="fricas")

[Out]

[1/96*(3*(8*b^2*c^2*d - 4*a*b*c*d^2 + a^2*d^3)*sqrt(c)*x^6*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2)
- 2*(8*a^2*c^3 + 3*(8*b^2*c^3 + 4*a*b*c^2*d - a^2*c*d^2)*x^4 + 2*(12*a*b*c^3 + a^2*c^2*d)*x^2)*sqrt(d*x^2 + c
))/(c^3*x^6), 1/48*(3*(8*b^2*c^2*d - 4*a*b*c*d^2 + a^2*d^3)*sqrt(-c)*x^6*arctan(sqrt(-c)/sqrt(d*x^2 + c)) - (8
*a^2*c^3 + 3*(8*b^2*c^3 + 4*a*b*c^2*d - a^2*c*d^2)*x^4 + 2*(12*a*b*c^3 + a^2*c^2*d)*x^2)*sqrt(d*x^2 + c))/(c^3
*x^6)]

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giac [A]  time = 0.48, size = 222, normalized size = 1.49 \begin {gather*} \frac {\frac {3 \, {\left (8 \, b^{2} c^{2} d^{2} - 4 \, a b c d^{3} + a^{2} d^{4}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{2}} - \frac {24 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c^{2} d^{2} - 48 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{3} d^{2} + 24 \, \sqrt {d x^{2} + c} b^{2} c^{4} d^{2} + 12 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b c d^{3} - 12 \, \sqrt {d x^{2} + c} a b c^{3} d^{3} - 3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d^{4} + 8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c d^{4} + 3 \, \sqrt {d x^{2} + c} a^{2} c^{2} d^{4}}{c^{2} d^{3} x^{6}}}{48 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^7,x, algorithm="giac")

[Out]

1/48*(3*(8*b^2*c^2*d^2 - 4*a*b*c*d^3 + a^2*d^4)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(sqrt(-c)*c^2) - (24*(d*x^2 +
c)^(5/2)*b^2*c^2*d^2 - 48*(d*x^2 + c)^(3/2)*b^2*c^3*d^2 + 24*sqrt(d*x^2 + c)*b^2*c^4*d^2 + 12*(d*x^2 + c)^(5/
2)*a*b*c*d^3 - 12*sqrt(d*x^2 + c)*a*b*c^3*d^3 - 3*(d*x^2 + c)^(5/2)*a^2*d^4 + 8*(d*x^2 + c)^(3/2)*a^2*c*d^4 +
3*sqrt(d*x^2 + c)*a^2*c^2*d^4)/(c^2*d^3*x^6))/d

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maple [B]  time = 0.02, size = 281, normalized size = 1.89 \begin {gather*} -\frac {a^{2} d^{3} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{16 c^{\frac {5}{2}}}+\frac {a b \,d^{2} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{4 c^{\frac {3}{2}}}-\frac {b^{2} d \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{2 \sqrt {c}}+\frac {\sqrt {d \,x^{2}+c}\, a^{2} d^{3}}{16 c^{3}}-\frac {\sqrt {d \,x^{2}+c}\, a b \,d^{2}}{4 c^{2}}+\frac {\sqrt {d \,x^{2}+c}\, b^{2} d}{2 c}-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2} d^{2}}{16 c^{3} x^{2}}+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a b d}{4 c^{2} x^{2}}-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} b^{2}}{2 c \,x^{2}}+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2} d}{8 c^{2} x^{4}}-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a b}{2 c \,x^{4}}-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2}}{6 c \,x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^7,x)

[Out]

-1/6*a^2*(d*x^2+c)^(3/2)/c/x^6+1/8*a^2*d/c^2/x^4*(d*x^2+c)^(3/2)-1/16*a^2*d^2/c^3/x^2*(d*x^2+c)^(3/2)-1/16*a^2
*d^3/c^(5/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)+1/16*a^2*d^3/c^3*(d*x^2+c)^(1/2)-1/2*a*b/c/x^4*(d*x^2+c)^(3
/2)+1/4*a*b*d/c^2/x^2*(d*x^2+c)^(3/2)+1/4*a*b*d^2/c^(3/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)-1/4*a*b*d^2/c^
2*(d*x^2+c)^(1/2)-1/2*b^2/c/x^2*(d*x^2+c)^(3/2)-1/2*b^2*d/c^(1/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)+1/2*b^
2*d/c*(d*x^2+c)^(1/2)

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maxima [A]  time = 1.02, size = 247, normalized size = 1.66 \begin {gather*} -\frac {b^{2} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{2 \, \sqrt {c}} + \frac {a b d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{4 \, c^{\frac {3}{2}}} - \frac {a^{2} d^{3} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{16 \, c^{\frac {5}{2}}} + \frac {\sqrt {d x^{2} + c} b^{2} d}{2 \, c} - \frac {\sqrt {d x^{2} + c} a b d^{2}}{4 \, c^{2}} + \frac {\sqrt {d x^{2} + c} a^{2} d^{3}}{16 \, c^{3}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2}}{2 \, c x^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d}{4 \, c^{2} x^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{2}}{16 \, c^{3} x^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b}{2 \, c x^{4}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d}{8 \, c^{2} x^{4}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2}}{6 \, c x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^7,x, algorithm="maxima")

[Out]

-1/2*b^2*d*arcsinh(c/(sqrt(c*d)*abs(x)))/sqrt(c) + 1/4*a*b*d^2*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(3/2) - 1/16*a^
2*d^3*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(5/2) + 1/2*sqrt(d*x^2 + c)*b^2*d/c - 1/4*sqrt(d*x^2 + c)*a*b*d^2/c^2 +
1/16*sqrt(d*x^2 + c)*a^2*d^3/c^3 - 1/2*(d*x^2 + c)^(3/2)*b^2/(c*x^2) + 1/4*(d*x^2 + c)^(3/2)*a*b*d/(c^2*x^2) -
1/16*(d*x^2 + c)^(3/2)*a^2*d^2/(c^3*x^2) - 1/2*(d*x^2 + c)^(3/2)*a*b/(c*x^4) + 1/8*(d*x^2 + c)^(3/2)*a^2*d/(c
^2*x^4) - 1/6*(d*x^2 + c)^(3/2)*a^2/(c*x^6)

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mupad [B]  time = 1.84, size = 193, normalized size = 1.30 \begin {gather*} \frac {\sqrt {d\,x^2+c}\,\left (\frac {a^2\,d^3}{16}-\frac {a\,b\,c\,d^2}{4}+\frac {b^2\,c^2\,d}{2}\right )+\frac {{\left (d\,x^2+c\right )}^{3/2}\,\left (a^2\,d^3-6\,b^2\,c^2\,d\right )}{6\,c}+\frac {{\left (d\,x^2+c\right )}^{5/2}\,\left (-a^2\,d^3+4\,a\,b\,c\,d^2+8\,b^2\,c^2\,d\right )}{16\,c^2}}{3\,c\,{\left (d\,x^2+c\right )}^2-3\,c^2\,\left (d\,x^2+c\right )-{\left (d\,x^2+c\right )}^3+c^3}-\frac {d\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )\,\left (a^2\,d^2-4\,a\,b\,c\,d+8\,b^2\,c^2\right )}{16\,c^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/x^7,x)

[Out]

((c + d*x^2)^(1/2)*((a^2*d^3)/16 + (b^2*c^2*d)/2 - (a*b*c*d^2)/4) + ((c + d*x^2)^(3/2)*(a^2*d^3 - 6*b^2*c^2*d)
)/(6*c) + ((c + d*x^2)^(5/2)*(8*b^2*c^2*d - a^2*d^3 + 4*a*b*c*d^2))/(16*c^2))/(3*c*(c + d*x^2)^2 - 3*c^2*(c +
d*x^2) - (c + d*x^2)^3 + c^3) - (d*atanh((c + d*x^2)^(1/2)/c^(1/2))*(a^2*d^2 + 8*b^2*c^2 - 4*a*b*c*d))/(16*c^(
5/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(1/2)/x**7,x)

[Out]

Timed out

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