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

Optimal. Leaf size=222 \[ -\frac {a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}+\frac {\left (c+d x^2\right )^{5/2} \left (5 a d (3 a d+8 b c)+8 b^2 c^2\right )}{40 c^2}+\frac {\left (c+d x^2\right )^{3/2} \left (5 a d (3 a d+8 b c)+8 b^2 c^2\right )}{24 c}+\frac {1}{8} \sqrt {c+d x^2} \left (5 a d (3 a d+8 b c)+8 b^2 c^2\right )-\frac {1}{8} \sqrt {c} \left (5 a d (3 a d+8 b c)+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )-\frac {a \left (c+d x^2\right )^{7/2} (3 a d+8 b c)}{8 c^2 x^2} \]

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Rubi [A]  time = 0.25, antiderivative size = 219, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 89, 78, 50, 63, 208} \begin {gather*} -\frac {a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}+\frac {1}{40} \left (c+d x^2\right )^{5/2} \left (\frac {5 a d (3 a d+8 b c)}{c^2}+8 b^2\right )+\frac {\left (c+d x^2\right )^{3/2} \left (5 a d (3 a d+8 b c)+8 b^2 c^2\right )}{24 c}+\frac {1}{8} \sqrt {c+d x^2} \left (5 a d (3 a d+8 b c)+8 b^2 c^2\right )-\frac {1}{8} \sqrt {c} \left (5 a d (3 a d+8 b c)+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )-\frac {a \left (c+d x^2\right )^{7/2} (3 a d+8 b c)}{8 c^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((8*b^2*c^2 + 5*a*d*(8*b*c + 3*a*d))*Sqrt[c + d*x^2])/8 + ((8*b^2*c^2 + 5*a*d*(8*b*c + 3*a*d))*(c + d*x^2)^(3/
2))/(24*c) + ((8*b^2 + (5*a*d*(8*b*c + 3*a*d))/c^2)*(c + d*x^2)^(5/2))/40 - (a^2*(c + d*x^2)^(7/2))/(4*c*x^4)
- (a*(8*b*c + 3*a*d)*(c + d*x^2)^(7/2))/(8*c^2*x^2) - (Sqrt[c]*(8*b^2*c^2 + 5*a*d*(8*b*c + 3*a*d))*ArcTanh[Sqr
t[c + d*x^2]/Sqrt[c]])/8

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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 \left (c+d x^2\right )^{5/2}}{x^5} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^2 (c+d x)^{5/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}+\frac {\operatorname {Subst}\left (\int \frac {\left (\frac {1}{2} a (8 b c+3 a d)+2 b^2 c x\right ) (c+d x)^{5/2}}{x^2} \, dx,x,x^2\right )}{4 c}\\ &=-\frac {a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}-\frac {a (8 b c+3 a d) \left (c+d x^2\right )^{7/2}}{8 c^2 x^2}+\frac {1}{16} \left (8 b^2+\frac {5 a d (8 b c+3 a d)}{c^2}\right ) \operatorname {Subst}\left (\int \frac {(c+d x)^{5/2}}{x} \, dx,x,x^2\right )\\ &=\frac {1}{40} \left (8 b^2+\frac {5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}-\frac {a (8 b c+3 a d) \left (c+d x^2\right )^{7/2}}{8 c^2 x^2}+\frac {1}{16} \left (c \left (8 b^2+\frac {5 a d (8 b c+3 a d)}{c^2}\right )\right ) \operatorname {Subst}\left (\int \frac {(c+d x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac {1}{24} c \left (8 b^2+\frac {5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}+\frac {1}{40} \left (8 b^2+\frac {5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}-\frac {a (8 b c+3 a d) \left (c+d x^2\right )^{7/2}}{8 c^2 x^2}+\frac {1}{16} \left (8 b^2 c^2+40 a b c d+15 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{x} \, dx,x,x^2\right )\\ &=\frac {1}{8} \left (8 b^2 c^2+40 a b c d+15 a^2 d^2\right ) \sqrt {c+d x^2}+\frac {1}{24} c \left (8 b^2+\frac {5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}+\frac {1}{40} \left (8 b^2+\frac {5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}-\frac {a (8 b c+3 a d) \left (c+d x^2\right )^{7/2}}{8 c^2 x^2}+\frac {1}{16} \left (c \left (8 b^2 c^2+40 a b c d+15 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )\\ &=\frac {1}{8} \left (8 b^2 c^2+40 a b c d+15 a^2 d^2\right ) \sqrt {c+d x^2}+\frac {1}{24} c \left (8 b^2+\frac {5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}+\frac {1}{40} \left (8 b^2+\frac {5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}-\frac {a (8 b c+3 a d) \left (c+d x^2\right )^{7/2}}{8 c^2 x^2}+\frac {\left (c \left (8 b^2 c^2+40 a b c d+15 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{8 d}\\ &=\frac {1}{8} \left (8 b^2 c^2+40 a b c d+15 a^2 d^2\right ) \sqrt {c+d x^2}+\frac {1}{24} c \left (8 b^2+\frac {5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}+\frac {1}{40} \left (8 b^2+\frac {5 a d (8 b c+3 a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{4 c x^4}-\frac {a (8 b c+3 a d) \left (c+d x^2\right )^{7/2}}{8 c^2 x^2}-\frac {1}{8} \sqrt {c} \left (8 b^2 c^2+40 a b c d+15 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 153, normalized size = 0.69 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-15 a^2 \left (2 c^2+9 c d x^2-8 d^2 x^4\right )+40 a b x^2 \left (-3 c^2+14 c d x^2+2 d^2 x^4\right )+8 b^2 x^4 \left (23 c^2+11 c d x^2+3 d^2 x^4\right )\right )}{120 x^4}-\frac {1}{8} \sqrt {c} \left (15 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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(-15*a^2*(2*c^2 + 9*c*d*x^2 - 8*d^2*x^4) + 40*a*b*x^2*(-3*c^2 + 14*c*d*x^2 + 2*d^2*x^4) + 8*b
^2*x^4*(23*c^2 + 11*c*d*x^2 + 3*d^2*x^4)))/(120*x^4) - (Sqrt[c]*(8*b^2*c^2 + 40*a*b*c*d + 15*a^2*d^2)*ArcTanh[
Sqrt[c + d*x^2]/Sqrt[c]])/8

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IntegrateAlgebraic [A]  time = 0.20, size = 166, normalized size = 0.75 \begin {gather*} \frac {1}{8} \left (-15 a^2 \sqrt {c} d^2-40 a b c^{3/2} d-8 b^2 c^{5/2}\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )+\frac {\sqrt {c+d x^2} \left (-30 a^2 c^2-135 a^2 c d x^2+120 a^2 d^2 x^4-120 a b c^2 x^2+560 a b c d x^4+80 a b d^2 x^6+184 b^2 c^2 x^4+88 b^2 c d x^6+24 b^2 d^2 x^8\right )}{120 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(-30*a^2*c^2 - 120*a*b*c^2*x^2 - 135*a^2*c*d*x^2 + 184*b^2*c^2*x^4 + 560*a*b*c*d*x^4 + 120*a^
2*d^2*x^4 + 88*b^2*c*d*x^6 + 80*a*b*d^2*x^6 + 24*b^2*d^2*x^8))/(120*x^4) + ((-8*b^2*c^(5/2) - 40*a*b*c^(3/2)*d
- 15*a^2*Sqrt[c]*d^2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/8

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fricas [A]  time = 1.50, size = 319, normalized size = 1.44 \begin {gather*} \left [\frac {15 \, {\left (8 \, b^{2} c^{2} + 40 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {c} x^{4} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (24 \, b^{2} d^{2} x^{8} + 8 \, {\left (11 \, b^{2} c d + 10 \, a b d^{2}\right )} x^{6} + 8 \, {\left (23 \, b^{2} c^{2} + 70 \, a b c d + 15 \, a^{2} d^{2}\right )} x^{4} - 30 \, a^{2} c^{2} - 15 \, {\left (8 \, a b c^{2} + 9 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{240 \, x^{4}}, \frac {15 \, {\left (8 \, b^{2} c^{2} + 40 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (24 \, b^{2} d^{2} x^{8} + 8 \, {\left (11 \, b^{2} c d + 10 \, a b d^{2}\right )} x^{6} + 8 \, {\left (23 \, b^{2} c^{2} + 70 \, a b c d + 15 \, a^{2} d^{2}\right )} x^{4} - 30 \, a^{2} c^{2} - 15 \, {\left (8 \, a b c^{2} + 9 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{120 \, x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/240*(15*(8*b^2*c^2 + 40*a*b*c*d + 15*a^2*d^2)*sqrt(c)*x^4*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^
2) + 2*(24*b^2*d^2*x^8 + 8*(11*b^2*c*d + 10*a*b*d^2)*x^6 + 8*(23*b^2*c^2 + 70*a*b*c*d + 15*a^2*d^2)*x^4 - 30*a
^2*c^2 - 15*(8*a*b*c^2 + 9*a^2*c*d)*x^2)*sqrt(d*x^2 + c))/x^4, 1/120*(15*(8*b^2*c^2 + 40*a*b*c*d + 15*a^2*d^2)
*sqrt(-c)*x^4*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (24*b^2*d^2*x^8 + 8*(11*b^2*c*d + 10*a*b*d^2)*x^6 + 8*(23*b^2
*c^2 + 70*a*b*c*d + 15*a^2*d^2)*x^4 - 30*a^2*c^2 - 15*(8*a*b*c^2 + 9*a^2*c*d)*x^2)*sqrt(d*x^2 + c))/x^4]

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giac [A]  time = 0.53, size = 242, normalized size = 1.09 \begin {gather*} \frac {24 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} d + 40 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c d + 120 \, \sqrt {d x^{2} + c} b^{2} c^{2} d + 80 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d^{2} + 480 \, \sqrt {d x^{2} + c} a b c d^{2} + 120 \, \sqrt {d x^{2} + c} a^{2} d^{3} + \frac {15 \, {\left (8 \, b^{2} c^{3} d + 40 \, a b c^{2} d^{2} + 15 \, a^{2} c d^{3}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {15 \, {\left (8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c^{2} d^{2} - 8 \, \sqrt {d x^{2} + c} a b c^{3} d^{2} + 9 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c d^{3} - 7 \, \sqrt {d x^{2} + c} a^{2} c^{2} d^{3}\right )}}{d^{2} x^{4}}}{120 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(24*(d*x^2 + c)^(5/2)*b^2*d + 40*(d*x^2 + c)^(3/2)*b^2*c*d + 120*sqrt(d*x^2 + c)*b^2*c^2*d + 80*(d*x^2 +
c)^(3/2)*a*b*d^2 + 480*sqrt(d*x^2 + c)*a*b*c*d^2 + 120*sqrt(d*x^2 + c)*a^2*d^3 + 15*(8*b^2*c^3*d + 40*a*b*c^2
*d^2 + 15*a^2*c*d^3)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c) - 15*(8*(d*x^2 + c)^(3/2)*a*b*c^2*d^2 - 8*sqrt(
d*x^2 + c)*a*b*c^3*d^2 + 9*(d*x^2 + c)^(3/2)*a^2*c*d^3 - 7*sqrt(d*x^2 + c)*a^2*c^2*d^3)/(d^2*x^4))/d

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maple [A]  time = 0.02, size = 305, normalized size = 1.37 \begin {gather*} -\frac {15 a^{2} \sqrt {c}\, d^{2} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{8}-5 a b \,c^{\frac {3}{2}} d \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )-b^{2} c^{\frac {5}{2}} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )+\frac {15 \sqrt {d \,x^{2}+c}\, a^{2} d^{2}}{8}+5 \sqrt {d \,x^{2}+c}\, a b c d +\sqrt {d \,x^{2}+c}\, b^{2} c^{2}+\frac {5 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2} d^{2}}{8 c}+\frac {5 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a b d}{3}+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} b^{2} c}{3}+\frac {3 \left (d \,x^{2}+c \right )^{\frac {5}{2}} a^{2} d^{2}}{8 c^{2}}+\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} a b d}{c}+\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} b^{2}}{5}-\frac {3 \left (d \,x^{2}+c \right )^{\frac {7}{2}} a^{2} d}{8 c^{2} x^{2}}-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}} a b}{c \,x^{2}}-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}} a^{2}}{4 c \,x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a^2*(d*x^2+c)^(7/2)/c/x^4-3/8*a^2*d/c^2/x^2*(d*x^2+c)^(7/2)+3/8*a^2*d^2/c^2*(d*x^2+c)^(5/2)+5/8*a^2*d^2/c
*(d*x^2+c)^(3/2)-15/8*a^2*d^2*c^(1/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)+15/8*a^2*d^2*(d*x^2+c)^(1/2)-a*b/c
/x^2*(d*x^2+c)^(7/2)+a*b*d/c*(d*x^2+c)^(5/2)+5/3*a*b*d*(d*x^2+c)^(3/2)-5*a*b*d*c^(3/2)*ln((2*c+2*(d*x^2+c)^(1/
2)*c^(1/2))/x)+5*a*b*d*c*(d*x^2+c)^(1/2)+1/5*b^2*(d*x^2+c)^(5/2)+1/3*b^2*c*(d*x^2+c)^(3/2)-b^2*c^(5/2)*ln((2*c
+2*(d*x^2+c)^(1/2)*c^(1/2))/x)+b^2*(d*x^2+c)^(1/2)*c^2

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maxima [A]  time = 0.94, size = 271, normalized size = 1.22 \begin {gather*} -b^{2} c^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - 5 \, a b c^{\frac {3}{2}} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - \frac {15}{8} \, a^{2} \sqrt {c} d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) + \frac {1}{5} \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} + \frac {1}{3} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c + \sqrt {d x^{2} + c} b^{2} c^{2} + \frac {5}{3} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d + \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a b d}{c} + 5 \, \sqrt {d x^{2} + c} a b c d + \frac {15}{8} \, \sqrt {d x^{2} + c} a^{2} d^{2} + \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d^{2}}{8 \, c^{2}} + \frac {5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{2}}{8 \, c} - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a b}{c x^{2}} - \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2} d}{8 \, c^{2} x^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2}}{4 \, c x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^2*c^(5/2)*arcsinh(c/(sqrt(c*d)*abs(x))) - 5*a*b*c^(3/2)*d*arcsinh(c/(sqrt(c*d)*abs(x))) - 15/8*a^2*sqrt(c)*
d^2*arcsinh(c/(sqrt(c*d)*abs(x))) + 1/5*(d*x^2 + c)^(5/2)*b^2 + 1/3*(d*x^2 + c)^(3/2)*b^2*c + sqrt(d*x^2 + c)*
b^2*c^2 + 5/3*(d*x^2 + c)^(3/2)*a*b*d + (d*x^2 + c)^(5/2)*a*b*d/c + 5*sqrt(d*x^2 + c)*a*b*c*d + 15/8*sqrt(d*x^
2 + c)*a^2*d^2 + 3/8*(d*x^2 + c)^(5/2)*a^2*d^2/c^2 + 5/8*(d*x^2 + c)^(3/2)*a^2*d^2/c - (d*x^2 + c)^(7/2)*a*b/(
c*x^2) - 3/8*(d*x^2 + c)^(7/2)*a^2*d/(c^2*x^2) - 1/4*(d*x^2 + c)^(7/2)*a^2/(c*x^4)

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mupad [B]  time = 1.88, size = 262, normalized size = 1.18 \begin {gather*} {\left (d\,x^2+c\right )}^{3/2}\,\left (\frac {c\,b^2}{3}+\frac {2\,a\,d\,b}{3}\right )-\frac {{\left (d\,x^2+c\right )}^{3/2}\,\left (\frac {9\,a^2\,c\,d^2}{8}+b\,a\,c^2\,d\right )-\left (\frac {7\,a^2\,c^2\,d^2}{8}+b\,a\,c^3\,d\right )\,\sqrt {d\,x^2+c}}{{\left (d\,x^2+c\right )}^2-2\,c\,\left (d\,x^2+c\right )+c^2}+\sqrt {d\,x^2+c}\,\left ({\left (a\,d-b\,c\right )}^2+3\,c\,\left (c\,b^2+2\,a\,d\,b\right )-3\,b^2\,c^2\right )+\frac {b^2\,{\left (d\,x^2+c\right )}^{5/2}}{5}+2\,\mathrm {atan}\left (\frac {2\,\sqrt {d\,x^2+c}\,\sqrt {-\frac {c}{256}}\,\left (15\,a^2\,d^2+40\,a\,b\,c\,d+8\,b^2\,c^2\right )}{\frac {15\,a^2\,c\,d^2}{8}+5\,a\,b\,c^2\,d+b^2\,c^3}\right )\,\sqrt {-\frac {c}{256}}\,\left (15\,a^2\,d^2+40\,a\,b\,c\,d+8\,b^2\,c^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c + d*x^2)^(3/2)*((b^2*c)/3 + (2*a*b*d)/3) - ((c + d*x^2)^(3/2)*((9*a^2*c*d^2)/8 + a*b*c^2*d) - ((7*a^2*c^2*d
^2)/8 + a*b*c^3*d)*(c + d*x^2)^(1/2))/((c + d*x^2)^2 - 2*c*(c + d*x^2) + c^2) + (c + d*x^2)^(1/2)*((a*d - b*c)
^2 + 3*c*(b^2*c + 2*a*b*d) - 3*b^2*c^2) + (b^2*(c + d*x^2)^(5/2))/5 + 2*atan((2*(c + d*x^2)^(1/2)*(-c/256)^(1/
2)*(15*a^2*d^2 + 8*b^2*c^2 + 40*a*b*c*d))/(b^2*c^3 + (15*a^2*c*d^2)/8 + 5*a*b*c^2*d))*(-c/256)^(1/2)*(15*a^2*d
^2 + 8*b^2*c^2 + 40*a*b*c*d)

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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)**(5/2)/x**5,x)

[Out]

Timed out

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