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

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

[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]

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Rubi [A]  time = 0.2008, antiderivative size = 192, normalized size of antiderivative = 1., 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} \[ -\frac{\sqrt{a} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{8 c^{5/2}}+\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{\sqrt{d}}-\frac{a \sqrt{a+b x^2} \sqrt{c+d x^2} (7 b c-3 a d)}{8 c^2 x^2}-\frac{a \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{4 c x^4} \]

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 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]]

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 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 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 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 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 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]

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*}

Mathematica [A]  time = 1.09908, size = 206, normalized size = 1.07 \[ -\frac{\sqrt{a} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{8 c^{5/2}}+\frac{a \sqrt{a+b x^2} \sqrt{c+d x^2} \left (-2 a c+3 a d x^2-9 b c x^2\right )}{8 c^2 x^4}+\frac{(b c-a d)^{5/2} \left (\frac{b \left (c+d x^2\right )}{b c-a d}\right )^{5/2} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )}{\sqrt{d} \left (c+d x^2\right )^{5/2}} \]

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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Maple [B]  time = 0.015, size = 464, normalized size = 2.4 \begin{align*}{\frac{1}{16\,{c}^{2}{x}^{4}}\sqrt{b{x}^{2}+a}\sqrt{d{x}^{2}+c} \left ( 8\,\ln \left ( 1/2\,{\frac{2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{4}{b}^{3}{c}^{2}\sqrt{ac}-3\,\ln \left ({\frac{ad{x}^{2}+bc{x}^{2}+2\,\sqrt{ac}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}+2\,ac}{{x}^{2}}} \right ){x}^{4}{a}^{3}{d}^{2}\sqrt{bd}+10\,\ln \left ({\frac{ad{x}^{2}+bc{x}^{2}+2\,\sqrt{ac}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}+2\,ac}{{x}^{2}}} \right ){x}^{4}{a}^{2}bcd\sqrt{bd}-15\,\ln \left ({\frac{ad{x}^{2}+bc{x}^{2}+2\,\sqrt{ac}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}+2\,ac}{{x}^{2}}} \right ){x}^{4}a{b}^{2}{c}^{2}\sqrt{bd}+6\,{x}^{2}{a}^{2}d\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}\sqrt{ac}-18\,{x}^{2}abc\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}\sqrt{ac}-4\,{a}^{2}c\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}\sqrt{ac} \right ){\frac{1}{\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ac}}}} \end{align*}

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*d*x^2*b+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2
)+a*d+b*c)/(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)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a
*c)^(1/2)+2*a*c)/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)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*
c)^(1/2)+2*a*c)/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)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a
*c)^(1/2)+2*a*c)/x^2)*x^4*a*b^2*c^2*(b*d)^(1/2)+6*x^2*a^2*d*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*(a
*c)^(1/2)-18*x^2*a*b*c*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-4*a^2*c*(b*d*x^4+a*d*x^2+b*
c*x^2+a*c)^(1/2)*(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)^(1/2)/(a*c)^(1/2)/x^4

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

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

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Fricas [A]  time = 13.4588, size = 2450, normalized size = 12.76 \begin{align*} \text{result too large to display} \end{align*}

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

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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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

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]

Exception raised: TypeError