3.7.22 \(\int \frac {(a+b x^2)^2 (c+d x^2)^{3/2}}{x^5} \, dx\) [622]

Optimal. Leaf size=181 \[ \frac {\left (8 b^2 c^2+3 a d (8 b c+a d)\right ) \sqrt {c+d x^2}}{8 c}+\frac {\left (8 b^2 c^2+3 a d (8 b c+a d)\right ) \left (c+d x^2\right )^{3/2}}{24 c^2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac {a (8 b c+a d) \left (c+d x^2\right )^{5/2}}{8 c^2 x^2}-\frac {\left (8 b^2 c^2+3 a d (8 b c+a d)\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 \sqrt {c}} \]

[Out]

1/24*(8*b^2*c^2+3*a*d*(a*d+8*b*c))*(d*x^2+c)^(3/2)/c^2-1/4*a^2*(d*x^2+c)^(5/2)/c/x^4-1/8*a*(a*d+8*b*c)*(d*x^2+
c)^(5/2)/c^2/x^2-1/8*(8*b^2*c^2+3*a*d*(a*d+8*b*c))*arctanh((d*x^2+c)^(1/2)/c^(1/2))/c^(1/2)+1/8*(8*b^2*c^2+3*a
*d*(a*d+8*b*c))*(d*x^2+c)^(1/2)/c

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Rubi [A]
time = 0.14, antiderivative size = 178, normalized size of antiderivative = 0.98, 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 = {457, 91, 79, 52, 65, 214} \begin {gather*} -\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}+\frac {1}{24} \left (c+d x^2\right )^{3/2} \left (\frac {3 a d (a d+8 b c)}{c^2}+8 b^2\right )+\frac {\sqrt {c+d x^2} \left (3 a d (a d+8 b c)+8 b^2 c^2\right )}{8 c}-\frac {\left (3 a d (a d+8 b c)+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 \sqrt {c}}-\frac {a \left (c+d x^2\right )^{5/2} (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)^(3/2))/x^5,x]

[Out]

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

Rule 52

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 65

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 79

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] || L
tQ[p, n]))))

Rule 91

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 214

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

Rule 457

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

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Mathematica [A]
time = 0.23, size = 116, normalized size = 0.64 \begin {gather*} \frac {1}{24} \left (\frac {\sqrt {c+d x^2} \left (-24 a b x^2 \left (c-2 d x^2\right )+8 b^2 x^4 \left (4 c+d x^2\right )-3 a^2 \left (2 c+5 d x^2\right )\right )}{x^4}-\frac {3 \left (8 b^2 c^2+24 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{\sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 242, normalized size = 1.34

method	result	size
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, a \left (5 a d \,x^{2}+8 c \,x^{2} b +2 a c \right )}{8 x^{4}}+\frac {b^{2} d \,x^{2} \sqrt {d \,x^{2}+c}}{3}+\frac {4 b^{2} c \sqrt {d \,x^{2}+c}}{3}+2 a b d \sqrt {d \,x^{2}+c}-\frac {3 \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) a^{2} d^{2}}{8 \sqrt {c}}-3 \sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) a b d -c^{\frac {3}{2}} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) b^{2}\)	\(179\)
default	\(a^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}}}{4 c \,x^{4}}+\frac {d \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}}}{2 c \,x^{2}}+\frac {3 d \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )}{2 c}\right )}{4 c}\right )+2 a b \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}}}{2 c \,x^{2}}+\frac {3 d \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )}{2 c}\right )+b^{2} \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )\)	\(242\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)^(3/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 222, normalized size = 1.23 \begin {gather*} -b^{2} c^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - 3 \, a b \sqrt {c} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - \frac {3 \, a^{2} d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{8 \, \sqrt {c}} + \frac {1}{3} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} + \sqrt {d x^{2} + c} b^{2} c + 3 \, \sqrt {d x^{2} + c} a b d + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d}{c} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{2}}{8 \, c^{2}} + \frac {3 \, \sqrt {d x^{2} + c} a^{2} d^{2}}{8 \, c} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a b}{c x^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d}{8 \, c^{2} x^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{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)^(3/2)/x^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.74, size = 267, normalized size = 1.48 \begin {gather*} \left [\frac {3 \, {\left (8 \, b^{2} c^{2} + 24 \, a b c d + 3 \, 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 (8 \, b^{2} c d x^{6} + 16 \, {\left (2 \, b^{2} c^{2} + 3 \, a b c d\right )} x^{4} - 6 \, a^{2} c^{2} - 3 \, {\left (8 \, a b c^{2} + 5 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{48 \, c x^{4}}, \frac {3 \, {\left (8 \, b^{2} c^{2} + 24 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (8 \, b^{2} c d x^{6} + 16 \, {\left (2 \, b^{2} c^{2} + 3 \, a b c d\right )} x^{4} - 6 \, a^{2} c^{2} - 3 \, {\left (8 \, a b c^{2} + 5 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{24 \, c 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)^(3/2)/x^5,x, algorithm="fricas")

[Out]

[1/48*(3*(8*b^2*c^2 + 24*a*b*c*d + 3*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*(8*b^2*c*d*x^6 + 16*(2*b^2*c^2 + 3*a*b*c*d)*x^4 - 6*a^2*c^2 - 3*(8*a*b*c^2 + 5*a^2*c*d)*x^2)*sqrt(d*x^2 +
c))/(c*x^4), 1/24*(3*(8*b^2*c^2 + 24*a*b*c*d + 3*a^2*d^2)*sqrt(-c)*x^4*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (8*b
^2*c*d*x^6 + 16*(2*b^2*c^2 + 3*a*b*c*d)*x^4 - 6*a^2*c^2 - 3*(8*a*b*c^2 + 5*a^2*c*d)*x^2)*sqrt(d*x^2 + c))/(c*x
^4)]

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Sympy [A]
time = 77.68, size = 332, normalized size = 1.83 \begin {gather*} - \frac {a^{2} c^{2}}{4 \sqrt {d} x^{5} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {3 a^{2} c \sqrt {d}}{8 x^{3} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {a^{2} d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{2 x} - \frac {a^{2} d^{\frac {3}{2}}}{8 x \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {3 a^{2} d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{8 \sqrt {c}} - 3 a b \sqrt {c} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )} - \frac {a b c \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{x} + \frac {2 a b c \sqrt {d}}{x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {2 a b d^{\frac {3}{2}} x}{\sqrt {\frac {c}{d x^{2}} + 1}} - b^{2} c^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )} + \frac {b^{2} c^{2}}{\sqrt {d} x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {b^{2} c \sqrt {d} x}{\sqrt {\frac {c}{d x^{2}} + 1}} + b^{2} 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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(3/2)/x**5,x)

[Out]

-a**2*c**2/(4*sqrt(d)*x**5*sqrt(c/(d*x**2) + 1)) - 3*a**2*c*sqrt(d)/(8*x**3*sqrt(c/(d*x**2) + 1)) - a**2*d**(3
/2)*sqrt(c/(d*x**2) + 1)/(2*x) - a**2*d**(3/2)/(8*x*sqrt(c/(d*x**2) + 1)) - 3*a**2*d**2*asinh(sqrt(c)/(sqrt(d)
*x))/(8*sqrt(c)) - 3*a*b*sqrt(c)*d*asinh(sqrt(c)/(sqrt(d)*x)) - a*b*c*sqrt(d)*sqrt(c/(d*x**2) + 1)/x + 2*a*b*c
*sqrt(d)/(x*sqrt(c/(d*x**2) + 1)) + 2*a*b*d**(3/2)*x/sqrt(c/(d*x**2) + 1) - b**2*c**(3/2)*asinh(sqrt(c)/(sqrt(
d)*x)) + b**2*c**2/(sqrt(d)*x*sqrt(c/(d*x**2) + 1)) + b**2*c*sqrt(d)*x/sqrt(c/(d*x**2) + 1) + b**2*d*Piecewise
((sqrt(c)*x**2/2, Eq(d, 0)), ((c + d*x**2)**(3/2)/(3*d), True))

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Giac [A]
time = 1.07, size = 182, normalized size = 1.01 \begin {gather*} \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d + 24 \, \sqrt {d x^{2} + c} b^{2} c d + 48 \, \sqrt {d x^{2} + c} a b d^{2} + \frac {3 \, {\left (8 \, b^{2} c^{2} d + 24 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {3 \, {\left (8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c d^{2} - 8 \, \sqrt {d x^{2} + c} a b c^{2} d^{2} + 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{3} - 3 \, \sqrt {d x^{2} + c} a^{2} c d^{3}\right )}}{d^{2} x^{4}}}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/x^5,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.76, size = 208, normalized size = 1.15 \begin {gather*} \frac {\sqrt {d\,x^2+c}\,\left (\frac {3\,a^2\,c\,d^2}{8}+b\,a\,c^2\,d\right )-\left (\frac {5\,a^2\,d^2}{8}+b\,c\,a\,d\right )\,{\left (d\,x^2+c\right )}^{3/2}}{{\left (d\,x^2+c\right )}^2-2\,c\,\left (d\,x^2+c\right )+c^2}+\sqrt {d\,x^2+c}\,\left (c\,b^2+2\,a\,d\,b\right )+\frac {b^2\,{\left (d\,x^2+c\right )}^{3/2}}{3}+\frac {\mathrm {atan}\left (\frac {\sqrt {d\,x^2+c}\,\left (3\,a^2\,d^2+24\,a\,b\,c\,d+8\,b^2\,c^2\right )\,1{}\mathrm {i}}{4\,\sqrt {c}\,\left (\frac {3\,a^2\,d^2}{4}+6\,a\,b\,c\,d+2\,b^2\,c^2\right )}\right )\,\left (3\,a^2\,d^2+24\,a\,b\,c\,d+8\,b^2\,c^2\right )\,1{}\mathrm {i}}{8\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^2*(c + d*x^2)^(3/2))/x^5,x)

[Out]

((c + d*x^2)^(1/2)*((3*a^2*c*d^2)/8 + a*b*c^2*d) - ((5*a^2*d^2)/8 + a*b*c*d)*(c + d*x^2)^(3/2))/((c + d*x^2)^2
- 2*c*(c + d*x^2) + c^2) + (c + d*x^2)^(1/2)*(b^2*c + 2*a*b*d) + (b^2*(c + d*x^2)^(3/2))/3 + (atan(((c + d*x^
2)^(1/2)*(3*a^2*d^2 + 8*b^2*c^2 + 24*a*b*c*d)*1i)/(4*c^(1/2)*((3*a^2*d^2)/4 + 2*b^2*c^2 + 6*a*b*c*d)))*(3*a^2*
d^2 + 8*b^2*c^2 + 24*a*b*c*d)*1i)/(8*c^(1/2))

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