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

Optimal. Leaf size=185 \[ -\frac {a^2}{4 c x^4 \left (c+d x^2\right )^{3/2}}-\frac {\left (8 b^2 c^2-5 a d (8 b c-7 a d)\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{9/2}}+\frac {8 b^2 c^2-5 a d (8 b c-7 a d)}{8 c^4 \sqrt {c+d x^2}}+\frac {8 b^2 c^2-5 a d (8 b c-7 a d)}{24 c^3 \left (c+d x^2\right )^{3/2}}-\frac {a (8 b c-7 a d)}{8 c^2 x^2 \left (c+d x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.22, antiderivative size = 185, normalized size of antiderivative = 1.00, 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 {a^2}{4 c x^4 \left (c+d x^2\right )^{3/2}}+\frac {8 b^2 c^2-5 a d (8 b c-7 a d)}{8 c^4 \sqrt {c+d x^2}}+\frac {8 b^2-\frac {5 a d (8 b c-7 a d)}{c^2}}{24 c \left (c+d x^2\right )^{3/2}}-\frac {\left (8 b^2 c^2-5 a d (8 b c-7 a d)\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{9/2}}-\frac {a (8 b c-7 a d)}{8 c^2 x^2 \left (c+d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b^2 - (5*a*d*(8*b*c - 7*a*d))/c^2)/(24*c*(c + d*x^2)^(3/2)) - a^2/(4*c*x^4*(c + d*x^2)^(3/2)) - (a*(8*b*c -
7*a*d))/(8*c^2*x^2*(c + d*x^2)^(3/2)) + (8*b^2*c^2 - 5*a*d*(8*b*c - 7*a*d))/(8*c^4*Sqrt[c + d*x^2]) - ((8*b^2
*c^2 - 5*a*d*(8*b*c - 7*a*d))*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(8*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^5 \left (c+d x^2\right )^{5/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^2}{x^3 (c+d x)^{5/2}} \, dx,x,x^2\right )\\ &=-\frac {a^2}{4 c x^4 \left (c+d x^2\right )^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} a (8 b c-7 a d)+2 b^2 c x}{x^2 (c+d x)^{5/2}} \, dx,x,x^2\right )}{4 c}\\ &=-\frac {a^2}{4 c x^4 \left (c+d x^2\right )^{3/2}}-\frac {a (8 b c-7 a d)}{8 c^2 x^2 \left (c+d x^2\right )^{3/2}}+\frac {1}{16} \left (8 b^2-\frac {5 a d (8 b c-7 a d)}{c^2}\right ) \operatorname {Subst}\left (\int \frac {1}{x (c+d x)^{5/2}} \, dx,x,x^2\right )\\ &=\frac {8 b^2-\frac {5 a d (8 b c-7 a d)}{c^2}}{24 c \left (c+d x^2\right )^{3/2}}-\frac {a^2}{4 c x^4 \left (c+d x^2\right )^{3/2}}-\frac {a (8 b c-7 a d)}{8 c^2 x^2 \left (c+d x^2\right )^{3/2}}+\frac {\left (8 b^2-\frac {5 a d (8 b c-7 a d)}{c^2}\right ) \operatorname {Subst}\left (\int \frac {1}{x (c+d x)^{3/2}} \, dx,x,x^2\right )}{16 c}\\ &=\frac {8 b^2-\frac {5 a d (8 b c-7 a d)}{c^2}}{24 c \left (c+d x^2\right )^{3/2}}-\frac {a^2}{4 c x^4 \left (c+d x^2\right )^{3/2}}-\frac {a (8 b c-7 a d)}{8 c^2 x^2 \left (c+d x^2\right )^{3/2}}+\frac {8 b^2-\frac {5 a d (8 b c-7 a d)}{c^2}}{8 c^2 \sqrt {c+d x^2}}+\frac {\left (8 b^2-\frac {5 a d (8 b c-7 a d)}{c^2}\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{16 c^2}\\ &=\frac {8 b^2-\frac {5 a d (8 b c-7 a d)}{c^2}}{24 c \left (c+d x^2\right )^{3/2}}-\frac {a^2}{4 c x^4 \left (c+d x^2\right )^{3/2}}-\frac {a (8 b c-7 a d)}{8 c^2 x^2 \left (c+d x^2\right )^{3/2}}+\frac {8 b^2-\frac {5 a d (8 b c-7 a d)}{c^2}}{8 c^2 \sqrt {c+d x^2}}+\frac {\left (8 b^2-\frac {5 a d (8 b c-7 a d)}{c^2}\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{8 c^2 d}\\ &=\frac {8 b^2-\frac {5 a d (8 b c-7 a d)}{c^2}}{24 c \left (c+d x^2\right )^{3/2}}-\frac {a^2}{4 c x^4 \left (c+d x^2\right )^{3/2}}-\frac {a (8 b c-7 a d)}{8 c^2 x^2 \left (c+d x^2\right )^{3/2}}+\frac {8 b^2-\frac {5 a d (8 b c-7 a d)}{c^2}}{8 c^2 \sqrt {c+d x^2}}-\frac {\left (8 b^2-\frac {5 a d (8 b c-7 a d)}{c^2}\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 90, normalized size = 0.49 \begin {gather*} \frac {x^4 \left (35 a^2 d^2-40 a b c d+8 b^2 c^2\right ) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {d x^2}{c}+1\right )-3 a c \left (2 a c-7 a d x^2+8 b c x^2\right )}{24 c^3 x^4 \left (c+d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*c*(2*a*c + 8*b*c*x^2 - 7*a*d*x^2) + (8*b^2*c^2 - 40*a*b*c*d + 35*a^2*d^2)*x^4*Hypergeometric2F1[-3/2, 1,
-1/2, 1 + (d*x^2)/c])/(24*c^3*x^4*(c + d*x^2)^(3/2))

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IntegrateAlgebraic [A]  time = 0.23, size = 171, normalized size = 0.92 \begin {gather*} \frac {\left (-35 a^2 d^2+40 a b c d-8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{9/2}}+\frac {-6 a^2 c^3+21 a^2 c^2 d x^2+140 a^2 c d^2 x^4+105 a^2 d^3 x^6-24 a b c^3 x^2-160 a b c^2 d x^4-120 a b c d^2 x^6+32 b^2 c^3 x^4+24 b^2 c^2 d x^6}{24 c^4 x^4 \left (c+d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a^2*c^3 - 24*a*b*c^3*x^2 + 21*a^2*c^2*d*x^2 + 32*b^2*c^3*x^4 - 160*a*b*c^2*d*x^4 + 140*a^2*c*d^2*x^4 + 24*
b^2*c^2*d*x^6 - 120*a*b*c*d^2*x^6 + 105*a^2*d^3*x^6)/(24*c^4*x^4*(c + d*x^2)^(3/2)) + ((-8*b^2*c^2 + 40*a*b*c*
d - 35*a^2*d^2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(8*c^(9/2))

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fricas [A]  time = 1.33, size = 537, normalized size = 2.90 \begin {gather*} \left [\frac {3 \, {\left ({\left (8 \, b^{2} c^{2} d^{2} - 40 \, a b c d^{3} + 35 \, a^{2} d^{4}\right )} x^{8} + 2 \, {\left (8 \, b^{2} c^{3} d - 40 \, a b c^{2} d^{2} + 35 \, a^{2} c d^{3}\right )} x^{6} + {\left (8 \, b^{2} c^{4} - 40 \, a b c^{3} d + 35 \, a^{2} c^{2} d^{2}\right )} x^{4}\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 (8 \, b^{2} c^{3} d - 40 \, a b c^{2} d^{2} + 35 \, a^{2} c d^{3}\right )} x^{6} - 6 \, a^{2} c^{4} + 4 \, {\left (8 \, b^{2} c^{4} - 40 \, a b c^{3} d + 35 \, a^{2} c^{2} d^{2}\right )} x^{4} - 3 \, {\left (8 \, a b c^{4} - 7 \, a^{2} c^{3} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{48 \, {\left (c^{5} d^{2} x^{8} + 2 \, c^{6} d x^{6} + c^{7} x^{4}\right )}}, \frac {3 \, {\left ({\left (8 \, b^{2} c^{2} d^{2} - 40 \, a b c d^{3} + 35 \, a^{2} d^{4}\right )} x^{8} + 2 \, {\left (8 \, b^{2} c^{3} d - 40 \, a b c^{2} d^{2} + 35 \, a^{2} c d^{3}\right )} x^{6} + {\left (8 \, b^{2} c^{4} - 40 \, a b c^{3} d + 35 \, a^{2} c^{2} d^{2}\right )} x^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (3 \, {\left (8 \, b^{2} c^{3} d - 40 \, a b c^{2} d^{2} + 35 \, a^{2} c d^{3}\right )} x^{6} - 6 \, a^{2} c^{4} + 4 \, {\left (8 \, b^{2} c^{4} - 40 \, a b c^{3} d + 35 \, a^{2} c^{2} d^{2}\right )} x^{4} - 3 \, {\left (8 \, a b c^{4} - 7 \, a^{2} c^{3} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{24 \, {\left (c^{5} d^{2} x^{8} + 2 \, c^{6} d x^{6} + c^{7} x^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^5/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[1/48*(3*((8*b^2*c^2*d^2 - 40*a*b*c*d^3 + 35*a^2*d^4)*x^8 + 2*(8*b^2*c^3*d - 40*a*b*c^2*d^2 + 35*a^2*c*d^3)*x^
6 + (8*b^2*c^4 - 40*a*b*c^3*d + 35*a^2*c^2*d^2)*x^4)*sqrt(c)*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^
2) + 2*(3*(8*b^2*c^3*d - 40*a*b*c^2*d^2 + 35*a^2*c*d^3)*x^6 - 6*a^2*c^4 + 4*(8*b^2*c^4 - 40*a*b*c^3*d + 35*a^2
*c^2*d^2)*x^4 - 3*(8*a*b*c^4 - 7*a^2*c^3*d)*x^2)*sqrt(d*x^2 + c))/(c^5*d^2*x^8 + 2*c^6*d*x^6 + c^7*x^4), 1/24*
(3*((8*b^2*c^2*d^2 - 40*a*b*c*d^3 + 35*a^2*d^4)*x^8 + 2*(8*b^2*c^3*d - 40*a*b*c^2*d^2 + 35*a^2*c*d^3)*x^6 + (8
*b^2*c^4 - 40*a*b*c^3*d + 35*a^2*c^2*d^2)*x^4)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (3*(8*b^2*c^3*d - 4
0*a*b*c^2*d^2 + 35*a^2*c*d^3)*x^6 - 6*a^2*c^4 + 4*(8*b^2*c^4 - 40*a*b*c^3*d + 35*a^2*c^2*d^2)*x^4 - 3*(8*a*b*c
^4 - 7*a^2*c^3*d)*x^2)*sqrt(d*x^2 + c))/(c^5*d^2*x^8 + 2*c^6*d*x^6 + c^7*x^4)]

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giac [A]  time = 0.40, size = 210, normalized size = 1.14 \begin {gather*} \frac {{\left (8 \, b^{2} c^{2} - 40 \, a b c d + 35 \, a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{8 \, \sqrt {-c} c^{4}} + \frac {3 \, {\left (d x^{2} + c\right )} b^{2} c^{2} + b^{2} c^{3} - 12 \, {\left (d x^{2} + c\right )} a b c d - 2 \, a b c^{2} d + 9 \, {\left (d x^{2} + c\right )} a^{2} d^{2} + a^{2} c d^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{4}} - \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c d - 8 \, \sqrt {d x^{2} + c} a b c^{2} d - 11 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{2} + 13 \, \sqrt {d x^{2} + c} a^{2} c d^{2}}{8 \, c^{4} d^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^5/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

1/8*(8*b^2*c^2 - 40*a*b*c*d + 35*a^2*d^2)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(sqrt(-c)*c^4) + 1/3*(3*(d*x^2 + c)
*b^2*c^2 + b^2*c^3 - 12*(d*x^2 + c)*a*b*c*d - 2*a*b*c^2*d + 9*(d*x^2 + c)*a^2*d^2 + a^2*c*d^2)/((d*x^2 + c)^(3
/2)*c^4) - 1/8*(8*(d*x^2 + c)^(3/2)*a*b*c*d - 8*sqrt(d*x^2 + c)*a*b*c^2*d - 11*(d*x^2 + c)^(3/2)*a^2*d^2 + 13*
sqrt(d*x^2 + c)*a^2*c*d^2)/(c^4*d^2*x^4)

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maple [A]  time = 0.02, size = 265, normalized size = 1.43 \begin {gather*} \frac {35 a^{2} d^{2}}{24 \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{3}}-\frac {5 a b d}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{2}}+\frac {b^{2}}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} c}-\frac {35 a^{2} d^{2} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{8 c^{\frac {9}{2}}}+\frac {5 a b d \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{c^{\frac {7}{2}}}-\frac {b^{2} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{c^{\frac {5}{2}}}+\frac {35 a^{2} d^{2}}{8 \sqrt {d \,x^{2}+c}\, c^{4}}-\frac {5 a b d}{\sqrt {d \,x^{2}+c}\, c^{3}}+\frac {b^{2}}{\sqrt {d \,x^{2}+c}\, c^{2}}+\frac {7 a^{2} d}{8 \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{2} x^{2}}-\frac {a b}{\left (d \,x^{2}+c \right )^{\frac {3}{2}} c \,x^{2}}-\frac {a^{2}}{4 \left (d \,x^{2}+c \right )^{\frac {3}{2}} c \,x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*b/c/x^2/(d*x^2+c)^(3/2)-5/3*a*b*d/c^2/(d*x^2+c)^(3/2)-5*a*b*d/c^3/(d*x^2+c)^(1/2)+5*a*b*d/c^(7/2)*ln((2*c+2
*(d*x^2+c)^(1/2)*c^(1/2))/x)-1/4*a^2/c/x^4/(d*x^2+c)^(3/2)+7/8*a^2*d/c^2/x^2/(d*x^2+c)^(3/2)+35/24*a^2*d^2/c^3
/(d*x^2+c)^(3/2)+35/8*a^2*d^2/c^4/(d*x^2+c)^(1/2)-35/8*a^2*d^2/c^(9/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)+1
/3*b^2/c/(d*x^2+c)^(3/2)+b^2/c^2/(d*x^2+c)^(1/2)-b^2/c^(5/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)

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maxima [A]  time = 0.94, size = 231, normalized size = 1.25 \begin {gather*} -\frac {b^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{c^{\frac {5}{2}}} + \frac {5 \, a b d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{c^{\frac {7}{2}}} - \frac {35 \, a^{2} d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{8 \, c^{\frac {9}{2}}} + \frac {b^{2}}{\sqrt {d x^{2} + c} c^{2}} + \frac {b^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c} - \frac {5 \, a b d}{\sqrt {d x^{2} + c} c^{3}} - \frac {5 \, a b d}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2}} + \frac {35 \, a^{2} d^{2}}{8 \, \sqrt {d x^{2} + c} c^{4}} + \frac {35 \, a^{2} d^{2}}{24 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{3}} - \frac {a b}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} c x^{2}} + \frac {7 \, a^{2} d}{8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2} x^{2}} - \frac {a^{2}}{4 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^5/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

-b^2*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(5/2) + 5*a*b*d*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(7/2) - 35/8*a^2*d^2*arcs
inh(c/(sqrt(c*d)*abs(x)))/c^(9/2) + b^2/(sqrt(d*x^2 + c)*c^2) + 1/3*b^2/((d*x^2 + c)^(3/2)*c) - 5*a*b*d/(sqrt(
d*x^2 + c)*c^3) - 5/3*a*b*d/((d*x^2 + c)^(3/2)*c^2) + 35/8*a^2*d^2/(sqrt(d*x^2 + c)*c^4) + 35/24*a^2*d^2/((d*x
^2 + c)^(3/2)*c^3) - a*b/((d*x^2 + c)^(3/2)*c*x^2) + 7/8*a^2*d/((d*x^2 + c)^(3/2)*c^2*x^2) - 1/4*a^2/((d*x^2 +
c)^(3/2)*c*x^4)

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mupad [B]  time = 1.11, size = 216, normalized size = 1.17 \begin {gather*} \frac {\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{3\,c}+\frac {\left (d\,x^2+c\right )\,\left (7\,a^2\,d^2-8\,a\,b\,c\,d+b^2\,c^2\right )}{3\,c^2}-\frac {5\,{\left (d\,x^2+c\right )}^2\,\left (35\,a^2\,d^2-40\,a\,b\,c\,d+8\,b^2\,c^2\right )}{24\,c^3}+\frac {{\left (d\,x^2+c\right )}^3\,\left (35\,a^2\,d^2-40\,a\,b\,c\,d+8\,b^2\,c^2\right )}{8\,c^4}}{{\left (d\,x^2+c\right )}^{7/2}-2\,c\,{\left (d\,x^2+c\right )}^{5/2}+c^2\,{\left (d\,x^2+c\right )}^{3/2}}-\frac {\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )\,\left (35\,a^2\,d^2-40\,a\,b\,c\,d+8\,b^2\,c^2\right )}{8\,c^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^2*d^2 + b^2*c^2 - 2*a*b*c*d)/(3*c) + ((c + d*x^2)*(7*a^2*d^2 + b^2*c^2 - 8*a*b*c*d))/(3*c^2) - (5*(c + d*x
^2)^2*(35*a^2*d^2 + 8*b^2*c^2 - 40*a*b*c*d))/(24*c^3) + ((c + d*x^2)^3*(35*a^2*d^2 + 8*b^2*c^2 - 40*a*b*c*d))/
(8*c^4))/((c + d*x^2)^(7/2) - 2*c*(c + d*x^2)^(5/2) + c^2*(c + d*x^2)^(3/2)) - (atanh((c + d*x^2)^(1/2)/c^(1/2
))*(35*a^2*d^2 + 8*b^2*c^2 - 40*a*b*c*d))/(8*c^(9/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x^{2}\right )^{2}}{x^{5} \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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