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

Optimal. Leaf size=192 \[ -\frac {\sqrt {a} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 c^{5/2}}+\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{\sqrt {d}}-\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2} (7 b c-3 a d)}{8 c^2 x^2}-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 c x^4} \]

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Rubi [A]  time = 0.20, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {446, 98, 149, 157, 63, 217, 206, 93, 208} \begin {gather*} -\frac {\sqrt {a} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 c^{5/2}}+\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{\sqrt {d}}-\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2} (7 b c-3 a d)}{8 c^2 x^2}-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 c x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(7*b*c - 3*a*d)*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(8*c^2*x^2) - (a*(a + b*x^2)^(3/2)*Sqrt[c + d*x^2])/(4*c*
x^4) - (Sqrt[a]*(15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^
2])])/(8*c^(5/2)) + (b^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2])])/Sqrt[d]

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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 98

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

Rule 149

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 )^{5/2}}{x^5 \sqrt {c+d x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^{5/2}}{x^3 \sqrt {c+d x}} \, dx,x,x^2\right )\\ &=-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 c x^4}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a+b x} \left (-\frac {1}{2} a (7 b c-3 a d)-2 b^2 c x\right )}{x^2 \sqrt {c+d x}} \, dx,x,x^2\right )}{4 c}\\ &=-\frac {a (7 b c-3 a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 c^2 x^2}-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 c x^4}-\frac {\operatorname {Subst}\left (\int \frac {-\frac {1}{4} a \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right )-2 b^3 c^2 x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{4 c^2}\\ &=-\frac {a (7 b c-3 a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 c^2 x^2}-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 c x^4}+\frac {1}{2} b^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )+\frac {\left (a \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{16 c^2}\\ &=-\frac {a (7 b c-3 a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 c^2 x^2}-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 c x^4}+b^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x^2}\right )+\frac {\left (a \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x^2}}{\sqrt {c+d x^2}}\right )}{8 c^2}\\ &=-\frac {a (7 b c-3 a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 c^2 x^2}-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 c x^4}-\frac {\sqrt {a} \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 c^{5/2}}+b^2 \operatorname {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x^2}}{\sqrt {c+d x^2}}\right )\\ &=-\frac {a (7 b c-3 a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 c^2 x^2}-\frac {a \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 c x^4}-\frac {\sqrt {a} \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 c^{5/2}}+\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{\sqrt {d}}\\ \end {align*}

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Mathematica [A]  time = 1.25, size = 206, normalized size = 1.07 \begin {gather*} -\frac {\sqrt {a} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 c^{5/2}}+\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2} \left (-2 a c+3 a d x^2-9 b c x^2\right )}{8 c^2 x^4}+\frac {(b c-a d)^{5/2} \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{5/2} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )}{\sqrt {d} \left (c+d x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*(-2*a*c - 9*b*c*x^2 + 3*a*d*x^2))/(8*c^2*x^4) + ((b*c - a*d)^(5/2)*((b*(c +
d*x^2))/(b*c - a*d))^(5/2)*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^2])/Sqrt[b*c - a*d]])/(Sqrt[d]*(c + d*x^2)^(5/2)) -
(Sqrt[a]*(15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])])/(
8*c^(5/2))

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IntegrateAlgebraic [F]  time = 3.04, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a+b x^2\right )^{5/2}}{x^5 \sqrt {c+d x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 6.71, size = 1123, normalized size = 5.85

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^5/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[1/32*(8*b^2*c^2*x^4*sqrt(b/d)*log(8*b^2*d^2*x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x^2 +
4*(2*b*d^2*x^2 + b*c*d + a*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b/d)) + (15*b^2*c^2 - 10*a*b*c*d + 3*a^2
*d^2)*x^4*sqrt(a/c)*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^4 + 8*a^2*c^2 + 8*(a*b*c^2 + a^2*c*d)*x^2 - 4*(2*a*
c^2 + (b*c^2 + a*c*d)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(a/c))/x^4) - 4*(2*a^2*c + 3*(3*a*b*c - a^2*d)*
x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(c^2*x^4), -1/32*(16*b^2*c^2*x^4*sqrt(-b/d)*arctan(1/2*(2*b*d*x^2 + b*c
+ a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-b/d)/(b^2*d*x^4 + a*b*c + (b^2*c + a*b*d)*x^2)) - (15*b^2*c^2 - 1
0*a*b*c*d + 3*a^2*d^2)*x^4*sqrt(a/c)*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^4 + 8*a^2*c^2 + 8*(a*b*c^2 + a^2*c
*d)*x^2 - 4*(2*a*c^2 + (b*c^2 + a*c*d)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(a/c))/x^4) + 4*(2*a^2*c + 3*(
3*a*b*c - a^2*d)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(c^2*x^4), 1/16*(4*b^2*c^2*x^4*sqrt(b/d)*log(8*b^2*d^2*
x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x^2 + 4*(2*b*d^2*x^2 + b*c*d + a*d^2)*sqrt(b*x^2 +
a)*sqrt(d*x^2 + c)*sqrt(b/d)) + (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*x^4*sqrt(-a/c)*arctan(1/2*((b*c + a*d)*
x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-a/c)/(a*b*d*x^4 + a^2*c + (a*b*c + a^2*d)*x^2)) - 2*(2*a^2*
c + 3*(3*a*b*c - a^2*d)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(c^2*x^4), -1/16*(8*b^2*c^2*x^4*sqrt(-b/d)*arcta
n(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-b/d)/(b^2*d*x^4 + a*b*c + (b^2*c + a*b*d)*
x^2)) - (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*x^4*sqrt(-a/c)*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt(b*x^2 +
a)*sqrt(d*x^2 + c)*sqrt(-a/c)/(a*b*d*x^4 + a^2*c + (a*b*c + a^2*d)*x^2)) + 2*(2*a^2*c + 3*(3*a*b*c - a^2*d)*x
^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(c^2*x^4)]

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giac [B]  time = 1.18, size = 1175, normalized size = 6.12 \begin {gather*} -\frac {{\left (\frac {4 \, \sqrt {b d} b^{2} \log \left ({\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}\right )}{d} + \frac {{\left (15 \, \sqrt {b d} a b^{3} c^{2} - 10 \, \sqrt {b d} a^{2} b^{2} c d + 3 \, \sqrt {b d} a^{3} b d^{2}\right )} \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} b c^{2}} + \frac {2 \, {\left (9 \, \sqrt {b d} a b^{9} c^{5} - 39 \, \sqrt {b d} a^{2} b^{8} c^{4} d + 66 \, \sqrt {b d} a^{3} b^{7} c^{3} d^{2} - 54 \, \sqrt {b d} a^{4} b^{6} c^{2} d^{3} + 21 \, \sqrt {b d} a^{5} b^{5} c d^{4} - 3 \, \sqrt {b d} a^{6} b^{4} d^{5} - 27 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a b^{7} c^{4} + 40 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a^{2} b^{6} c^{3} d + 10 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a^{3} b^{5} c^{2} d^{2} - 32 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a^{4} b^{4} c d^{3} + 9 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a^{5} b^{3} d^{4} + 27 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} a b^{5} c^{3} + 9 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} a^{2} b^{4} c^{2} d + 21 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} a^{3} b^{3} c d^{2} - 9 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} a^{4} b^{2} d^{3} - 9 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{6} a b^{3} c^{2} - 10 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{6} a^{2} b^{2} c d + 3 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{6} a^{3} b d^{2}\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4}\right )}^{2} c^{2}}\right )} b}{8 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^5/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

-1/8*(4*sqrt(b*d)*b^2*log((sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2)/d + (15*sqrt(
b*d)*a*b^3*c^2 - 10*sqrt(b*d)*a^2*b^2*c*d + 3*sqrt(b*d)*a^3*b*d^2)*arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*x^2 +
a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*b*c^2) + 2*(9*sqr
t(b*d)*a*b^9*c^5 - 39*sqrt(b*d)*a^2*b^8*c^4*d + 66*sqrt(b*d)*a^3*b^7*c^3*d^2 - 54*sqrt(b*d)*a^4*b^6*c^2*d^3 +
21*sqrt(b*d)*a^5*b^5*c*d^4 - 3*sqrt(b*d)*a^6*b^4*d^5 - 27*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c +
(b*x^2 + a)*b*d - a*b*d))^2*a*b^7*c^4 + 40*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d
- a*b*d))^2*a^2*b^6*c^3*d + 10*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^
2*a^3*b^5*c^2*d^2 - 32*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a^4*b^4
*c*d^3 + 9*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a^5*b^3*d^4 + 27*sq
rt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4*a*b^5*c^3 + 9*sqrt(b*d)*(sqrt(b*
x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4*a^2*b^4*c^2*d + 21*sqrt(b*d)*(sqrt(b*x^2 + a)*sq
rt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4*a^3*b^3*c*d^2 - 9*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sq
rt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4*a^4*b^2*d^3 - 9*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*
x^2 + a)*b*d - a*b*d))^6*a*b^3*c^2 - 10*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d -
a*b*d))^6*a^2*b^2*c*d + 3*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^6*a^3*
b*d^2)/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a
*b*d))^2*b^2*c - 2*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*x^2 +
a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4)^2*c^2))*b/abs(b)

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maple [B]  time = 0.02, size = 464, normalized size = 2.42 \begin {gather*} \frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (-3 \sqrt {b d}\, a^{3} d^{2} x^{4} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+10 \sqrt {b d}\, a^{2} b c d \,x^{4} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )-15 \sqrt {b d}\, a \,b^{2} c^{2} x^{4} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+8 \sqrt {a c}\, b^{3} c^{2} x^{4} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+6 \sqrt {b d}\, \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, a^{2} d \,x^{2}-18 \sqrt {b d}\, \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, a b c \,x^{2}-4 \sqrt {b d}\, \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, a^{2} c \right )}{16 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, \sqrt {a c}\, c^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2)/x^5/(d*x^2+c)^(1/2),x)

[Out]

1/16*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/c^2*(8*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b
*d)^(1/2))/(b*d)^(1/2))*x^4*b^3*c^2*(a*c)^(1/2)-3*ln((a*d*x^2+b*c*x^2+2*a*c+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c
*x^2+a*c)^(1/2))/x^2)*x^4*a^3*d^2*(b*d)^(1/2)+10*ln((a*d*x^2+b*c*x^2+2*a*c+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*
x^2+a*c)^(1/2))/x^2)*x^4*a^2*b*c*d*(b*d)^(1/2)-15*ln((a*d*x^2+b*c*x^2+2*a*c+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c
*x^2+a*c)^(1/2))/x^2)*x^4*a*b^2*c^2*(b*d)^(1/2)+6*x^2*a^2*d*(b*d)^(1/2)*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a
*c)^(1/2)-18*x^2*a*b*c*(b*d)^(1/2)*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)-4*a^2*c*(b*d)^(1/2)*(a*c)^(
1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2))/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)/x^4/(b*d)^(1/2)/(a*c)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^5/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^{5/2}}{x^5\,\sqrt {d\,x^2+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(5/2)/(x^5*(c + d*x^2)^(1/2)),x)

[Out]

int((a + b*x^2)^(5/2)/(x^5*(c + d*x^2)^(1/2)), x)

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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 )^{\frac {5}{2}}}{x^{5} \sqrt {c + d x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/2)/x**5/(d*x**2+c)**(1/2),x)

[Out]

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

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