3.7.41 \(\int \frac {(a+b x^2)^2}{x^7 (c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=190 \[ -\frac {a^2}{6 c x^6 \sqrt {c+d x^2}}+\frac {d \left (24 b^2 c^2-5 a d (12 b c-7 a d)\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{9/2}}-\frac {d \left (24 b^2 c^2-5 a d (12 b c-7 a d)\right )}{16 c^4 \sqrt {c+d x^2}}-\frac {24 b^2 c^2-5 a d (12 b c-7 a d)}{48 c^3 x^2 \sqrt {c+d x^2}}-\frac {a (12 b c-7 a d)}{24 c^2 x^4 \sqrt {c+d x^2}} \]

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Rubi [A]  time = 0.22, antiderivative size = 193, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 89, 78, 51, 63, 208} \begin {gather*} \frac {35 a^2 d^2-60 a b c d+24 b^2 c^2}{24 c^3 x^2 \sqrt {c+d x^2}}-\frac {a^2}{6 c x^6 \sqrt {c+d x^2}}-\frac {\sqrt {c+d x^2} \left (24 b^2 c^2-5 a d (12 b c-7 a d)\right )}{16 c^4 x^2}+\frac {d \left (24 b^2 c^2-5 a d (12 b c-7 a d)\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{9/2}}-\frac {a (12 b c-7 a d)}{24 c^2 x^4 \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(x^7*(c + d*x^2)^(3/2)),x]

[Out]

-a^2/(6*c*x^6*Sqrt[c + d*x^2]) - (a*(12*b*c - 7*a*d))/(24*c^2*x^4*Sqrt[c + d*x^2]) + (24*b^2*c^2 - 60*a*b*c*d
+ 35*a^2*d^2)/(24*c^3*x^2*Sqrt[c + d*x^2]) - ((24*b^2*c^2 - 5*a*d*(12*b*c - 7*a*d))*Sqrt[c + d*x^2])/(16*c^4*x
^2) + (d*(24*b^2*c^2 - 5*a*d*(12*b*c - 7*a*d))*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(16*c^(9/2))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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}{x^7 \left (c+d x^2\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^2}{x^4 (c+d x)^{3/2}} \, dx,x,x^2\right )\\ &=-\frac {a^2}{6 c x^6 \sqrt {c+d x^2}}+\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} a (12 b c-7 a d)+3 b^2 c x}{x^3 (c+d x)^{3/2}} \, dx,x,x^2\right )}{6 c}\\ &=-\frac {a^2}{6 c x^6 \sqrt {c+d x^2}}-\frac {a (12 b c-7 a d)}{24 c^2 x^4 \sqrt {c+d x^2}}+\frac {1}{48} \left (24 b^2-\frac {5 a d (12 b c-7 a d)}{c^2}\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 (c+d x)^{3/2}} \, dx,x,x^2\right )\\ &=-\frac {a^2}{6 c x^6 \sqrt {c+d x^2}}-\frac {a (12 b c-7 a d)}{24 c^2 x^4 \sqrt {c+d x^2}}+\frac {24 b^2 c^2-60 a b c d+35 a^2 d^2}{24 c^3 x^2 \sqrt {c+d x^2}}+\frac {\left (24 b^2 c^2-60 a b c d+35 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {c+d x}} \, dx,x,x^2\right )}{16 c^3}\\ &=-\frac {a^2}{6 c x^6 \sqrt {c+d x^2}}-\frac {a (12 b c-7 a d)}{24 c^2 x^4 \sqrt {c+d x^2}}+\frac {24 b^2 c^2-60 a b c d+35 a^2 d^2}{24 c^3 x^2 \sqrt {c+d x^2}}-\frac {\left (24 b^2 c^2-60 a b c d+35 a^2 d^2\right ) \sqrt {c+d x^2}}{16 c^4 x^2}-\frac {\left (d \left (24 b^2 c^2-60 a b c d+35 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{32 c^4}\\ &=-\frac {a^2}{6 c x^6 \sqrt {c+d x^2}}-\frac {a (12 b c-7 a d)}{24 c^2 x^4 \sqrt {c+d x^2}}+\frac {24 b^2 c^2-60 a b c d+35 a^2 d^2}{24 c^3 x^2 \sqrt {c+d x^2}}-\frac {\left (24 b^2 c^2-60 a b c d+35 a^2 d^2\right ) \sqrt {c+d x^2}}{16 c^4 x^2}-\frac {\left (24 b^2 c^2-60 a b c d+35 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^4}\\ &=-\frac {a^2}{6 c x^6 \sqrt {c+d x^2}}-\frac {a (12 b c-7 a d)}{24 c^2 x^4 \sqrt {c+d x^2}}+\frac {24 b^2 c^2-60 a b c d+35 a^2 d^2}{24 c^3 x^2 \sqrt {c+d x^2}}-\frac {\left (24 b^2 c^2-60 a b c d+35 a^2 d^2\right ) \sqrt {c+d x^2}}{16 c^4 x^2}+\frac {d \left (24 b^2 c^2-60 a b c d+35 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 92, normalized size = 0.48 \begin {gather*} \frac {d x^6 \left (-35 a^2 d^2+60 a b c d-24 b^2 c^2\right ) \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};\frac {d x^2}{c}+1\right )+a c^2 \left (-4 a c+7 a d x^2-12 b c x^2\right )}{24 c^4 x^6 \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/(x^7*(c + d*x^2)^(3/2)),x]

[Out]

(a*c^2*(-4*a*c - 12*b*c*x^2 + 7*a*d*x^2) + d*(-24*b^2*c^2 + 60*a*b*c*d - 35*a^2*d^2)*x^6*Hypergeometric2F1[-1/
2, 2, 1/2, 1 + (d*x^2)/c])/(24*c^4*x^6*Sqrt[c + d*x^2])

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IntegrateAlgebraic [A]  time = 0.28, size = 174, normalized size = 0.92 \begin {gather*} \frac {\left (35 a^2 d^3-60 a b c d^2+24 b^2 c^2 d\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{9/2}}+\frac {-8 a^2 c^3+14 a^2 c^2 d x^2-35 a^2 c d^2 x^4-105 a^2 d^3 x^6-24 a b c^3 x^2+60 a b c^2 d x^4+180 a b c d^2 x^6-24 b^2 c^3 x^4-72 b^2 c^2 d x^6}{48 c^4 x^6 \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x^2)^2/(x^7*(c + d*x^2)^(3/2)),x]

[Out]

(-8*a^2*c^3 - 24*a*b*c^3*x^2 + 14*a^2*c^2*d*x^2 - 24*b^2*c^3*x^4 + 60*a*b*c^2*d*x^4 - 35*a^2*c*d^2*x^4 - 72*b^
2*c^2*d*x^6 + 180*a*b*c*d^2*x^6 - 105*a^2*d^3*x^6)/(48*c^4*x^6*Sqrt[c + d*x^2]) + ((24*b^2*c^2*d - 60*a*b*c*d^
2 + 35*a^2*d^3)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(16*c^(9/2))

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fricas [A]  time = 1.64, size = 447, normalized size = 2.35 \begin {gather*} \left [\frac {3 \, {\left ({\left (24 \, b^{2} c^{2} d^{2} - 60 \, a b c d^{3} + 35 \, a^{2} d^{4}\right )} x^{8} + {\left (24 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 35 \, a^{2} c d^{3}\right )} x^{6}\right )} \sqrt {c} \log \left (-\frac {d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (3 \, {\left (24 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 35 \, a^{2} c d^{3}\right )} x^{6} + 8 \, a^{2} c^{4} + {\left (24 \, b^{2} c^{4} - 60 \, a b c^{3} d + 35 \, a^{2} c^{2} d^{2}\right )} x^{4} + 2 \, {\left (12 \, a b c^{4} - 7 \, a^{2} c^{3} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{96 \, {\left (c^{5} d x^{8} + c^{6} x^{6}\right )}}, -\frac {3 \, {\left ({\left (24 \, b^{2} c^{2} d^{2} - 60 \, a b c d^{3} + 35 \, a^{2} d^{4}\right )} x^{8} + {\left (24 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 35 \, a^{2} c d^{3}\right )} x^{6}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (3 \, {\left (24 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 35 \, a^{2} c d^{3}\right )} x^{6} + 8 \, a^{2} c^{4} + {\left (24 \, b^{2} c^{4} - 60 \, a b c^{3} d + 35 \, a^{2} c^{2} d^{2}\right )} x^{4} + 2 \, {\left (12 \, a b c^{4} - 7 \, a^{2} c^{3} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{48 \, {\left (c^{5} d x^{8} + c^{6} x^{6}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^7/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

[1/96*(3*((24*b^2*c^2*d^2 - 60*a*b*c*d^3 + 35*a^2*d^4)*x^8 + (24*b^2*c^3*d - 60*a*b*c^2*d^2 + 35*a^2*c*d^3)*x^
6)*sqrt(c)*log(-(d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) - 2*(3*(24*b^2*c^3*d - 60*a*b*c^2*d^2 + 35*a^2*
c*d^3)*x^6 + 8*a^2*c^4 + (24*b^2*c^4 - 60*a*b*c^3*d + 35*a^2*c^2*d^2)*x^4 + 2*(12*a*b*c^4 - 7*a^2*c^3*d)*x^2)*
sqrt(d*x^2 + c))/(c^5*d*x^8 + c^6*x^6), -1/48*(3*((24*b^2*c^2*d^2 - 60*a*b*c*d^3 + 35*a^2*d^4)*x^8 + (24*b^2*c
^3*d - 60*a*b*c^2*d^2 + 35*a^2*c*d^3)*x^6)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (3*(24*b^2*c^3*d - 60*a
*b*c^2*d^2 + 35*a^2*c*d^3)*x^6 + 8*a^2*c^4 + (24*b^2*c^4 - 60*a*b*c^3*d + 35*a^2*c^2*d^2)*x^4 + 2*(12*a*b*c^4
- 7*a^2*c^3*d)*x^2)*sqrt(d*x^2 + c))/(c^5*d*x^8 + c^6*x^6)]

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giac [A]  time = 0.40, size = 267, normalized size = 1.41 \begin {gather*} -\frac {{\left (24 \, b^{2} c^{2} d - 60 \, a b c d^{2} + 35 \, a^{2} d^{3}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{16 \, \sqrt {-c} c^{4}} - \frac {b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}}{\sqrt {d x^{2} + c} c^{4}} - \frac {24 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c^{2} d - 48 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{3} d + 24 \, \sqrt {d x^{2} + c} b^{2} c^{4} d - 84 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b c d^{2} + 192 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c^{2} d^{2} - 108 \, \sqrt {d x^{2} + c} a b c^{3} d^{2} + 57 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d^{3} - 136 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c d^{3} + 87 \, \sqrt {d x^{2} + c} a^{2} c^{2} d^{3}}{48 \, c^{4} d^{3} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^7/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

-1/16*(24*b^2*c^2*d - 60*a*b*c*d^2 + 35*a^2*d^3)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(sqrt(-c)*c^4) - (b^2*c^2*d
- 2*a*b*c*d^2 + a^2*d^3)/(sqrt(d*x^2 + c)*c^4) - 1/48*(24*(d*x^2 + c)^(5/2)*b^2*c^2*d - 48*(d*x^2 + c)^(3/2)*b
^2*c^3*d + 24*sqrt(d*x^2 + c)*b^2*c^4*d - 84*(d*x^2 + c)^(5/2)*a*b*c*d^2 + 192*(d*x^2 + c)^(3/2)*a*b*c^2*d^2 -
108*sqrt(d*x^2 + c)*a*b*c^3*d^2 + 57*(d*x^2 + c)^(5/2)*a^2*d^3 - 136*(d*x^2 + c)^(3/2)*a^2*c*d^3 + 87*sqrt(d*
x^2 + c)*a^2*c^2*d^3)/(c^4*d^3*x^6)

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maple [A]  time = 0.02, size = 281, normalized size = 1.48 \begin {gather*} \frac {35 a^{2} d^{3} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{16 c^{\frac {9}{2}}}-\frac {15 a b \,d^{2} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{4 c^{\frac {7}{2}}}+\frac {3 b^{2} d \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{2 c^{\frac {5}{2}}}-\frac {35 a^{2} d^{3}}{16 \sqrt {d \,x^{2}+c}\, c^{4}}+\frac {15 a b \,d^{2}}{4 \sqrt {d \,x^{2}+c}\, c^{3}}-\frac {3 b^{2} d}{2 \sqrt {d \,x^{2}+c}\, c^{2}}-\frac {35 a^{2} d^{2}}{48 \sqrt {d \,x^{2}+c}\, c^{3} x^{2}}+\frac {5 a b d}{4 \sqrt {d \,x^{2}+c}\, c^{2} x^{2}}-\frac {b^{2}}{2 \sqrt {d \,x^{2}+c}\, c \,x^{2}}+\frac {7 a^{2} d}{24 \sqrt {d \,x^{2}+c}\, c^{2} x^{4}}-\frac {a b}{2 \sqrt {d \,x^{2}+c}\, c \,x^{4}}-\frac {a^{2}}{6 \sqrt {d \,x^{2}+c}\, c \,x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/x^7/(d*x^2+c)^(3/2),x)

[Out]

-1/6*a^2/c/x^6/(d*x^2+c)^(1/2)+7/24*a^2*d/c^2/x^4/(d*x^2+c)^(1/2)-35/48*a^2*d^2/c^3/x^2/(d*x^2+c)^(1/2)-35/16*
a^2*d^3/c^4/(d*x^2+c)^(1/2)+35/16*a^2*d^3/c^(9/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)-1/2*b^2/c/x^2/(d*x^2+c
)^(1/2)-3/2*b^2*d/c^2/(d*x^2+c)^(1/2)+3/2*b^2*d/c^(5/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)-1/2*a*b/c/x^4/(d
*x^2+c)^(1/2)+5/4*a*b*d/c^2/x^2/(d*x^2+c)^(1/2)+15/4*a*b*d^2/c^3/(d*x^2+c)^(1/2)-15/4*a*b*d^2/c^(7/2)*ln((2*c+
2*(d*x^2+c)^(1/2)*c^(1/2))/x)

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maxima [A]  time = 0.95, size = 247, normalized size = 1.30 \begin {gather*} \frac {3 \, b^{2} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{2 \, c^{\frac {5}{2}}} - \frac {15 \, a b d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{4 \, c^{\frac {7}{2}}} + \frac {35 \, a^{2} d^{3} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{16 \, c^{\frac {9}{2}}} - \frac {3 \, b^{2} d}{2 \, \sqrt {d x^{2} + c} c^{2}} + \frac {15 \, a b d^{2}}{4 \, \sqrt {d x^{2} + c} c^{3}} - \frac {35 \, a^{2} d^{3}}{16 \, \sqrt {d x^{2} + c} c^{4}} - \frac {b^{2}}{2 \, \sqrt {d x^{2} + c} c x^{2}} + \frac {5 \, a b d}{4 \, \sqrt {d x^{2} + c} c^{2} x^{2}} - \frac {35 \, a^{2} d^{2}}{48 \, \sqrt {d x^{2} + c} c^{3} x^{2}} - \frac {a b}{2 \, \sqrt {d x^{2} + c} c x^{4}} + \frac {7 \, a^{2} d}{24 \, \sqrt {d x^{2} + c} c^{2} x^{4}} - \frac {a^{2}}{6 \, \sqrt {d x^{2} + c} c x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^7/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

3/2*b^2*d*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(5/2) - 15/4*a*b*d^2*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(7/2) + 35/16*a
^2*d^3*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(9/2) - 3/2*b^2*d/(sqrt(d*x^2 + c)*c^2) + 15/4*a*b*d^2/(sqrt(d*x^2 + c)
*c^3) - 35/16*a^2*d^3/(sqrt(d*x^2 + c)*c^4) - 1/2*b^2/(sqrt(d*x^2 + c)*c*x^2) + 5/4*a*b*d/(sqrt(d*x^2 + c)*c^2
*x^2) - 35/48*a^2*d^2/(sqrt(d*x^2 + c)*c^3*x^2) - 1/2*a*b/(sqrt(d*x^2 + c)*c*x^4) + 7/24*a^2*d/(sqrt(d*x^2 + c
)*c^2*x^4) - 1/6*a^2/(sqrt(d*x^2 + c)*c*x^6)

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mupad [B]  time = 1.34, size = 246, normalized size = 1.29 \begin {gather*} \frac {d\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )\,\left (35\,a^2\,d^2-60\,a\,b\,c\,d+24\,b^2\,c^2\right )}{16\,c^{9/2}}-\frac {\frac {a^2\,d^3-2\,a\,b\,c\,d^2+b^2\,c^2\,d}{c}-\frac {\left (d\,x^2+c\right )\,\left (77\,a^2\,d^3-132\,a\,b\,c\,d^2+56\,b^2\,c^2\,d\right )}{16\,c^2}+\frac {{\left (d\,x^2+c\right )}^2\,\left (35\,a^2\,d^3-60\,a\,b\,c\,d^2+24\,b^2\,c^2\,d\right )}{6\,c^3}-\frac {{\left (d\,x^2+c\right )}^3\,\left (35\,a^2\,d^3-60\,a\,b\,c\,d^2+24\,b^2\,c^2\,d\right )}{16\,c^4}}{3\,c\,{\left (d\,x^2+c\right )}^{5/2}-{\left (d\,x^2+c\right )}^{7/2}+c^3\,\sqrt {d\,x^2+c}-3\,c^2\,{\left (d\,x^2+c\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^2/(x^7*(c + d*x^2)^(3/2)),x)

[Out]

(d*atanh((c + d*x^2)^(1/2)/c^(1/2))*(35*a^2*d^2 + 24*b^2*c^2 - 60*a*b*c*d))/(16*c^(9/2)) - ((a^2*d^3 + b^2*c^2
*d - 2*a*b*c*d^2)/c - ((c + d*x^2)*(77*a^2*d^3 + 56*b^2*c^2*d - 132*a*b*c*d^2))/(16*c^2) + ((c + d*x^2)^2*(35*
a^2*d^3 + 24*b^2*c^2*d - 60*a*b*c*d^2))/(6*c^3) - ((c + d*x^2)^3*(35*a^2*d^3 + 24*b^2*c^2*d - 60*a*b*c*d^2))/(
16*c^4))/(3*c*(c + d*x^2)^(5/2) - (c + d*x^2)^(7/2) + c^3*(c + d*x^2)^(1/2) - 3*c^2*(c + d*x^2)^(3/2))

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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/x**7/(d*x**2+c)**(3/2),x)

[Out]

Timed out

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