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

Optimal. Leaf size=187 \[ \frac{\sqrt{b} \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{8 d^{5/2}}-\frac{a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{c}}-\frac{b \sqrt{a+b x^2} \sqrt{c+d x^2} (3 b c-7 a d)}{8 d^2}+\frac{b \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{4 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.225928, antiderivative size = 187, 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, 102, 154, 157, 63, 217, 206, 93, 208} \[ \frac{\sqrt{b} \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{8 d^{5/2}}-\frac{a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{c}}-\frac{b \sqrt{a+b x^2} \sqrt{c+d x^2} (3 b c-7 a d)}{8 d^2}+\frac{b \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 102

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

Rule 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

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

Mathematica [A]  time = 0.637364, size = 213, normalized size = 1.14 \[ \frac{1}{8} \left (\frac{\left (25 a^2 b c d^2-15 a^3 d^3-13 a b^2 c^2 d+3 b^3 c^3\right ) \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )}{d^{5/2} \sqrt{c+d x^2} \sqrt{b c-a d}}-\frac{8 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{c}}+\frac{b \sqrt{a+b x^2} \sqrt{c+d x^2} \left (9 a d-3 b c+2 b d x^2\right )}{d^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*(-3*b*c + 9*a*d + 2*b*d*x^2))/d^2 + ((3*b^3*c^3 - 13*a*b^2*c^2*d + 25*a^2*
b*c*d^2 - 15*a^3*d^3)*Sqrt[(b*(c + d*x^2))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^2])/Sqrt[b*c - a*d]])/(d
^(5/2)*Sqrt[b*c - a*d]*Sqrt[c + d*x^2]) - (8*a^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2
])])/Sqrt[c])/8

________________________________________________________________________________________

Maple [B]  time = 0.017, size = 446, normalized size = 2.4 \begin{align*}{\frac{1}{16\,{d}^{2}}\sqrt{b{x}^{2}+a}\sqrt{d{x}^{2}+c} \left ( 4\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}\sqrt{ac}{x}^{2}{b}^{2}d+15\,\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 ) \sqrt{ac}{a}^{2}b{d}^{2}-10\,\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 ) \sqrt{ac}a{b}^{2}cd+3\,\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 ) \sqrt{ac}{b}^{3}{c}^{2}-8\,\sqrt{bd}\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 ){a}^{3}{d}^{2}+18\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}\sqrt{ac}abd-6\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}\sqrt{ac}{b}^{2}c \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/(d*x^2+c)^(1/2),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 18.0371, size = 2361, normalized size = 12.63 \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/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[1/32*(8*a^2*d^2*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) + (3*b^2*c^2 - 10*a*b*c*d +
15*a^2*d^2)*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)) + 4*(2*b^2*d*x^2 - 3*b^2*c + 9*a*b*d)*sqrt(
b*x^2 + a)*sqrt(d*x^2 + c))/d^2, 1/16*(4*a^2*d^2*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) - (3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*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)) + 2*(2*b^2*d*x^2 - 3*b^2*c + 9*a*b*d)*sqrt(b
*x^2 + a)*sqrt(d*x^2 + c))/d^2, 1/32*(16*a^2*d^2*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)) + (3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^
2)*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)) + 4*(2*b^2*d*x^2 - 3*b^2*c + 9*a*b*d)*sqrt(b*x^2 + a
)*sqrt(d*x^2 + c))/d^2, 1/16*(8*a^2*d^2*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)) - (3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*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)) + 2*(2*b^2*d*x^2 - 3*b^2*c + 9*a*b*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/d^2]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2}\right )^{\frac{5}{2}}}{x \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/(d*x**2+c)**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.23519, size = 354, normalized size = 1.89 \begin{align*} -\frac{{\left (\frac{16 \, \sqrt{b d} a^{3} \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} - 2 \, \sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d} \sqrt{b x^{2} + a}{\left (\frac{2 \,{\left (b x^{2} + a\right )}}{b d} - \frac{3 \, b^{2} c d - 7 \, a b d^{2}}{b^{2} d^{3}}\right )} + \frac{{\left (3 \, \sqrt{b d} b^{2} c^{2} - 10 \, \sqrt{b d} a b c d + 15 \, \sqrt{b d} a^{2} d^{2}\right )} \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 )}{b d^{3}}\right )} b^{2}}{16 \,{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(16*sqrt(b*d)*a^3*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) - 2*sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)*sqrt(b*x^2 + a)
*(2*(b*x^2 + a)/(b*d) - (3*b^2*c*d - 7*a*b*d^2)/(b^2*d^3)) + (3*sqrt(b*d)*b^2*c^2 - 10*sqrt(b*d)*a*b*c*d + 15*
sqrt(b*d)*a^2*d^2)*log((sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2)/(b*d^3))*b^2/abs
(b)