3.7.13 \(\int \frac {(a+b x^2)^2 (c+d x^2)^{5/2}}{x^3} \, dx\)

Optimal. Leaf size=162 \[ -\frac {a^2 \left (c+d x^2\right )^{7/2}}{2 c x^2}-\frac {1}{2} a c^{3/2} (5 a d+4 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )+\frac {a \left (c+d x^2\right )^{5/2} (5 a d+4 b c)}{10 c}+\frac {1}{6} a \left (c+d x^2\right )^{3/2} (5 a d+4 b c)+\frac {1}{2} a c \sqrt {c+d x^2} (5 a d+4 b c)+\frac {b^2 \left (c+d x^2\right )^{7/2}}{7 d} \]

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Rubi [A]  time = 0.13, antiderivative size = 162, normalized size of antiderivative = 1.00, 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, 80, 50, 63, 208} \begin {gather*} -\frac {a^2 \left (c+d x^2\right )^{7/2}}{2 c x^2}-\frac {1}{2} a c^{3/2} (5 a d+4 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )+\frac {a \left (c+d x^2\right )^{5/2} (5 a d+4 b c)}{10 c}+\frac {1}{6} a \left (c+d x^2\right )^{3/2} (5 a d+4 b c)+\frac {1}{2} a c \sqrt {c+d x^2} (5 a d+4 b c)+\frac {b^2 \left (c+d x^2\right )^{7/2}}{7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^3,x]

[Out]

(a*c*(4*b*c + 5*a*d)*Sqrt[c + d*x^2])/2 + (a*(4*b*c + 5*a*d)*(c + d*x^2)^(3/2))/6 + (a*(4*b*c + 5*a*d)*(c + d*
x^2)^(5/2))/(10*c) + (b^2*(c + d*x^2)^(7/2))/(7*d) - (a^2*(c + d*x^2)^(7/2))/(2*c*x^2) - (a*c^(3/2)*(4*b*c + 5
*a*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/2

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 80

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

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^3} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^2 (c+d x)^{5/2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {a^2 \left (c+d x^2\right )^{7/2}}{2 c x^2}+\frac {\operatorname {Subst}\left (\int \frac {\left (\frac {1}{2} a (4 b c+5 a d)+b^2 c x\right ) (c+d x)^{5/2}}{x} \, dx,x,x^2\right )}{2 c}\\ &=\frac {b^2 \left (c+d x^2\right )^{7/2}}{7 d}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{2 c x^2}+\frac {(a (4 b c+5 a d)) \operatorname {Subst}\left (\int \frac {(c+d x)^{5/2}}{x} \, dx,x,x^2\right )}{4 c}\\ &=\frac {a (4 b c+5 a d) \left (c+d x^2\right )^{5/2}}{10 c}+\frac {b^2 \left (c+d x^2\right )^{7/2}}{7 d}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{2 c x^2}+\frac {1}{4} (a (4 b c+5 a d)) \operatorname {Subst}\left (\int \frac {(c+d x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac {1}{6} a (4 b c+5 a d) \left (c+d x^2\right )^{3/2}+\frac {a (4 b c+5 a d) \left (c+d x^2\right )^{5/2}}{10 c}+\frac {b^2 \left (c+d x^2\right )^{7/2}}{7 d}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{2 c x^2}+\frac {1}{4} (a c (4 b c+5 a d)) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{x} \, dx,x,x^2\right )\\ &=\frac {1}{2} a c (4 b c+5 a d) \sqrt {c+d x^2}+\frac {1}{6} a (4 b c+5 a d) \left (c+d x^2\right )^{3/2}+\frac {a (4 b c+5 a d) \left (c+d x^2\right )^{5/2}}{10 c}+\frac {b^2 \left (c+d x^2\right )^{7/2}}{7 d}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{2 c x^2}+\frac {1}{4} \left (a c^2 (4 b c+5 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )\\ &=\frac {1}{2} a c (4 b c+5 a d) \sqrt {c+d x^2}+\frac {1}{6} a (4 b c+5 a d) \left (c+d x^2\right )^{3/2}+\frac {a (4 b c+5 a d) \left (c+d x^2\right )^{5/2}}{10 c}+\frac {b^2 \left (c+d x^2\right )^{7/2}}{7 d}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{2 c x^2}+\frac {\left (a c^2 (4 b c+5 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 d}\\ &=\frac {1}{2} a c (4 b c+5 a d) \sqrt {c+d x^2}+\frac {1}{6} a (4 b c+5 a d) \left (c+d x^2\right )^{3/2}+\frac {a (4 b c+5 a d) \left (c+d x^2\right )^{5/2}}{10 c}+\frac {b^2 \left (c+d x^2\right )^{7/2}}{7 d}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{2 c x^2}-\frac {1}{2} a c^{3/2} (4 b c+5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 122, normalized size = 0.75 \begin {gather*} -\frac {\frac {a^2 \left (c+d x^2\right )^{7/2}}{x^2}+\frac {1}{15} a (5 a d+4 b c) \left (15 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )-\sqrt {c+d x^2} \left (23 c^2+11 c d x^2+3 d^2 x^4\right )\right )-\frac {2 b^2 c \left (c+d x^2\right )^{7/2}}{7 d}}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^3,x]

[Out]

-1/2*((-2*b^2*c*(c + d*x^2)^(7/2))/(7*d) + (a^2*(c + d*x^2)^(7/2))/x^2 + (a*(4*b*c + 5*a*d)*(-(Sqrt[c + d*x^2]
*(23*c^2 + 11*c*d*x^2 + 3*d^2*x^4)) + 15*c^(5/2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]]))/15)/c

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IntegrateAlgebraic [A]  time = 0.19, size = 176, normalized size = 1.09 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-105 a^2 c^2 d+490 a^2 c d^2 x^2+70 a^2 d^3 x^4+644 a b c^2 d x^2+308 a b c d^2 x^4+84 a b d^3 x^6+30 b^2 c^3 x^2+90 b^2 c^2 d x^4+90 b^2 c d^2 x^6+30 b^2 d^3 x^8\right )}{210 d x^2}+\frac {1}{2} \left (-5 a^2 c^{3/2} d-4 a b c^{5/2}\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^3,x]

[Out]

(Sqrt[c + d*x^2]*(-105*a^2*c^2*d + 30*b^2*c^3*x^2 + 644*a*b*c^2*d*x^2 + 490*a^2*c*d^2*x^2 + 90*b^2*c^2*d*x^4 +
308*a*b*c*d^2*x^4 + 70*a^2*d^3*x^4 + 90*b^2*c*d^2*x^6 + 84*a*b*d^3*x^6 + 30*b^2*d^3*x^8))/(210*d*x^2) + ((-4*
a*b*c^(5/2) - 5*a^2*c^(3/2)*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/2

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fricas [A]  time = 1.58, size = 349, normalized size = 2.15 \begin {gather*} \left [\frac {105 \, {\left (4 \, a b c^{2} d + 5 \, a^{2} c d^{2}\right )} \sqrt {c} x^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (30 \, b^{2} d^{3} x^{8} + 6 \, {\left (15 \, b^{2} c d^{2} + 14 \, a b d^{3}\right )} x^{6} - 105 \, a^{2} c^{2} d + 2 \, {\left (45 \, b^{2} c^{2} d + 154 \, a b c d^{2} + 35 \, a^{2} d^{3}\right )} x^{4} + 2 \, {\left (15 \, b^{2} c^{3} + 322 \, a b c^{2} d + 245 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{420 \, d x^{2}}, \frac {105 \, {\left (4 \, a b c^{2} d + 5 \, a^{2} c d^{2}\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (30 \, b^{2} d^{3} x^{8} + 6 \, {\left (15 \, b^{2} c d^{2} + 14 \, a b d^{3}\right )} x^{6} - 105 \, a^{2} c^{2} d + 2 \, {\left (45 \, b^{2} c^{2} d + 154 \, a b c d^{2} + 35 \, a^{2} d^{3}\right )} x^{4} + 2 \, {\left (15 \, b^{2} c^{3} + 322 \, a b c^{2} d + 245 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{210 \, d x^{2}}\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^3,x, algorithm="fricas")

[Out]

[1/420*(105*(4*a*b*c^2*d + 5*a^2*c*d^2)*sqrt(c)*x^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 2*(3
0*b^2*d^3*x^8 + 6*(15*b^2*c*d^2 + 14*a*b*d^3)*x^6 - 105*a^2*c^2*d + 2*(45*b^2*c^2*d + 154*a*b*c*d^2 + 35*a^2*d
^3)*x^4 + 2*(15*b^2*c^3 + 322*a*b*c^2*d + 245*a^2*c*d^2)*x^2)*sqrt(d*x^2 + c))/(d*x^2), 1/210*(105*(4*a*b*c^2*
d + 5*a^2*c*d^2)*sqrt(-c)*x^2*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (30*b^2*d^3*x^8 + 6*(15*b^2*c*d^2 + 14*a*b*d^
3)*x^6 - 105*a^2*c^2*d + 2*(45*b^2*c^2*d + 154*a*b*c*d^2 + 35*a^2*d^3)*x^4 + 2*(15*b^2*c^3 + 322*a*b*c^2*d + 2
45*a^2*c*d^2)*x^2)*sqrt(d*x^2 + c))/(d*x^2)]

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giac [A]  time = 0.50, size = 165, normalized size = 1.02 \begin {gather*} \frac {30 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} + 84 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b d + 140 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c d + 420 \, \sqrt {d x^{2} + c} a b c^{2} d + 70 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{2} + 420 \, \sqrt {d x^{2} + c} a^{2} c d^{2} - \frac {105 \, \sqrt {d x^{2} + c} a^{2} c^{2} d}{x^{2}} + \frac {105 \, {\left (4 \, a b c^{3} d + 5 \, a^{2} c^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}}}{210 \, 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^3,x, algorithm="giac")

[Out]

1/210*(30*(d*x^2 + c)^(7/2)*b^2 + 84*(d*x^2 + c)^(5/2)*a*b*d + 140*(d*x^2 + c)^(3/2)*a*b*c*d + 420*sqrt(d*x^2
+ c)*a*b*c^2*d + 70*(d*x^2 + c)^(3/2)*a^2*d^2 + 420*sqrt(d*x^2 + c)*a^2*c*d^2 - 105*sqrt(d*x^2 + c)*a^2*c^2*d/
x^2 + 105*(4*a*b*c^3*d + 5*a^2*c^2*d^2)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c))/d

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maple [A]  time = 0.01, size = 193, normalized size = 1.19 \begin {gather*} -\frac {5 a^{2} c^{\frac {3}{2}} d \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{2}-2 a b \,c^{\frac {5}{2}} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )+\frac {5 \sqrt {d \,x^{2}+c}\, a^{2} c d}{2}+2 \sqrt {d \,x^{2}+c}\, a b \,c^{2}+\frac {5 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2} d}{6}+\frac {2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} a b c}{3}+\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} a^{2} d}{2 c}+\frac {2 \left (d \,x^{2}+c \right )^{\frac {5}{2}} a b}{5}+\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}} b^{2}}{7 d}-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}} a^{2}}{2 c \,x^{2}} \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^3,x)

[Out]

1/7*b^2*(d*x^2+c)^(7/2)/d-1/2*a^2*(d*x^2+c)^(7/2)/c/x^2+1/2*a^2*d/c*(d*x^2+c)^(5/2)+5/6*a^2*d*(d*x^2+c)^(3/2)-
5/2*a^2*d*c^(3/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)+5/2*a^2*d*c*(d*x^2+c)^(1/2)+2/5*a*b*(d*x^2+c)^(5/2)+2/
3*a*b*c*(d*x^2+c)^(3/2)-2*a*b*c^(5/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)+2*a*b*(d*x^2+c)^(1/2)*c^2

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maxima [A]  time = 0.95, size = 170, normalized size = 1.05 \begin {gather*} -2 \, a b c^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - \frac {5}{2} \, a^{2} c^{\frac {3}{2}} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) + \frac {2}{5} \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b + \frac {2}{3} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c + 2 \, \sqrt {d x^{2} + c} a b c^{2} + \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2}}{7 \, d} + \frac {5}{6} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d + \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d}{2 \, c} + \frac {5}{2} \, \sqrt {d x^{2} + c} a^{2} c d - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2}}{2 \, c x^{2}} \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^3,x, algorithm="maxima")

[Out]

-2*a*b*c^(5/2)*arcsinh(c/(sqrt(c*d)*abs(x))) - 5/2*a^2*c^(3/2)*d*arcsinh(c/(sqrt(c*d)*abs(x))) + 2/5*(d*x^2 +
c)^(5/2)*a*b + 2/3*(d*x^2 + c)^(3/2)*a*b*c + 2*sqrt(d*x^2 + c)*a*b*c^2 + 1/7*(d*x^2 + c)^(7/2)*b^2/d + 5/6*(d*
x^2 + c)^(3/2)*a^2*d + 1/2*(d*x^2 + c)^(5/2)*a^2*d/c + 5/2*sqrt(d*x^2 + c)*a^2*c*d - 1/2*(d*x^2 + c)^(7/2)*a^2
/(c*x^2)

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mupad [B]  time = 1.45, size = 274, normalized size = 1.69 \begin {gather*} \sqrt {d\,x^2+c}\,\left (c^2\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d}-\frac {2\,b^2\,c}{d}\right )-2\,c\,\left (2\,c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d}-\frac {2\,b^2\,c}{d}\right )-\frac {{\left (a\,d-b\,c\right )}^2}{d}+\frac {b^2\,c^2}{d}\right )\right )-\left (\frac {2\,b^2\,c-2\,a\,b\,d}{5\,d}-\frac {2\,b^2\,c}{5\,d}\right )\,{\left (d\,x^2+c\right )}^{5/2}-{\left (d\,x^2+c\right )}^{3/2}\,\left (\frac {2\,c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d}-\frac {2\,b^2\,c}{d}\right )}{3}-\frac {{\left (a\,d-b\,c\right )}^2}{3\,d}+\frac {b^2\,c^2}{3\,d}\right )+\frac {b^2\,{\left (d\,x^2+c\right )}^{7/2}}{7\,d}-\frac {a^2\,c^2\,\sqrt {d\,x^2+c}}{2\,x^2}+\frac {a\,c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {d\,x^2+c}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,\left (5\,a\,d+4\,b\,c\right )\,1{}\mathrm {i}}{2} \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^3,x)

[Out]

(c + d*x^2)^(1/2)*(c^2*((2*b^2*c - 2*a*b*d)/d - (2*b^2*c)/d) - 2*c*(2*c*((2*b^2*c - 2*a*b*d)/d - (2*b^2*c)/d)
- (a*d - b*c)^2/d + (b^2*c^2)/d)) - ((2*b^2*c - 2*a*b*d)/(5*d) - (2*b^2*c)/(5*d))*(c + d*x^2)^(5/2) - (c + d*x
^2)^(3/2)*((2*c*((2*b^2*c - 2*a*b*d)/d - (2*b^2*c)/d))/3 - (a*d - b*c)^2/(3*d) + (b^2*c^2)/(3*d)) + (b^2*(c +
d*x^2)^(7/2))/(7*d) + (a*c^(3/2)*atan(((c + d*x^2)^(1/2)*1i)/c^(1/2))*(5*a*d + 4*b*c)*1i)/2 - (a^2*c^2*(c + d*
x^2)^(1/2))/(2*x^2)

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sympy [A]  time = 91.81, size = 518, normalized size = 3.20 \begin {gather*} - \frac {5 a^{2} c^{\frac {3}{2}} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{2} - \frac {a^{2} c^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{2 x} + \frac {2 a^{2} c^{2} \sqrt {d}}{x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {2 a^{2} c d^{\frac {3}{2}} x}{\sqrt {\frac {c}{d x^{2}} + 1}} + a^{2} d^{2} \left (\begin {cases} \frac {\sqrt {c} x^{2}}{2} & \text {for}\: d = 0 \\\frac {\left (c + d x^{2}\right )^{\frac {3}{2}}}{3 d} & \text {otherwise} \end {cases}\right ) - 2 a b c^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )} + \frac {2 a b c^{3}}{\sqrt {d} x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {2 a b c^{2} \sqrt {d} x}{\sqrt {\frac {c}{d x^{2}} + 1}} + 4 a b c d \left (\begin {cases} \frac {\sqrt {c} x^{2}}{2} & \text {for}\: d = 0 \\\frac {\left (c + d x^{2}\right )^{\frac {3}{2}}}{3 d} & \text {otherwise} \end {cases}\right ) + 2 a b d^{2} \left (\begin {cases} - \frac {2 c^{2} \sqrt {c + d x^{2}}}{15 d^{2}} + \frac {c x^{2} \sqrt {c + d x^{2}}}{15 d} + \frac {x^{4} \sqrt {c + d x^{2}}}{5} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + b^{2} c^{2} \left (\begin {cases} \frac {\sqrt {c} x^{2}}{2} & \text {for}\: d = 0 \\\frac {\left (c + d x^{2}\right )^{\frac {3}{2}}}{3 d} & \text {otherwise} \end {cases}\right ) + 2 b^{2} c d \left (\begin {cases} - \frac {2 c^{2} \sqrt {c + d x^{2}}}{15 d^{2}} + \frac {c x^{2} \sqrt {c + d x^{2}}}{15 d} + \frac {x^{4} \sqrt {c + d x^{2}}}{5} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + b^{2} d^{2} \left (\begin {cases} \frac {8 c^{3} \sqrt {c + d x^{2}}}{105 d^{3}} - \frac {4 c^{2} x^{2} \sqrt {c + d x^{2}}}{105 d^{2}} + \frac {c x^{4} \sqrt {c + d x^{2}}}{35 d} + \frac {x^{6} \sqrt {c + d x^{2}}}{7} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{6}}{6} & \text {otherwise} \end {cases}\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**3,x)

[Out]

-5*a**2*c**(3/2)*d*asinh(sqrt(c)/(sqrt(d)*x))/2 - a**2*c**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(2*x) + 2*a**2*c**2*s
qrt(d)/(x*sqrt(c/(d*x**2) + 1)) + 2*a**2*c*d**(3/2)*x/sqrt(c/(d*x**2) + 1) + a**2*d**2*Piecewise((sqrt(c)*x**2
/2, Eq(d, 0)), ((c + d*x**2)**(3/2)/(3*d), True)) - 2*a*b*c**(5/2)*asinh(sqrt(c)/(sqrt(d)*x)) + 2*a*b*c**3/(sq
rt(d)*x*sqrt(c/(d*x**2) + 1)) + 2*a*b*c**2*sqrt(d)*x/sqrt(c/(d*x**2) + 1) + 4*a*b*c*d*Piecewise((sqrt(c)*x**2/
2, Eq(d, 0)), ((c + d*x**2)**(3/2)/(3*d), True)) + 2*a*b*d**2*Piecewise((-2*c**2*sqrt(c + d*x**2)/(15*d**2) +
c*x**2*sqrt(c + d*x**2)/(15*d) + x**4*sqrt(c + d*x**2)/5, Ne(d, 0)), (sqrt(c)*x**4/4, True)) + b**2*c**2*Piece
wise((sqrt(c)*x**2/2, Eq(d, 0)), ((c + d*x**2)**(3/2)/(3*d), True)) + 2*b**2*c*d*Piecewise((-2*c**2*sqrt(c + d
*x**2)/(15*d**2) + c*x**2*sqrt(c + d*x**2)/(15*d) + x**4*sqrt(c + d*x**2)/5, Ne(d, 0)), (sqrt(c)*x**4/4, True)
) + b**2*d**2*Piecewise((8*c**3*sqrt(c + d*x**2)/(105*d**3) - 4*c**2*x**2*sqrt(c + d*x**2)/(105*d**2) + c*x**4
*sqrt(c + d*x**2)/(35*d) + x**6*sqrt(c + d*x**2)/7, Ne(d, 0)), (sqrt(c)*x**6/6, True))

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