\(\int \frac {(c+d x^2)^{3/2}}{x (a+b x^2)} \, dx\) [932]

Optimal result

Integrand size = 24, antiderivative size = 96 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x \left (a+b x^2\right )} \, dx=\frac {d \sqrt {c+d x^2}}{b}-\frac {c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a}+\frac {(b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a b^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.07 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x \left (a+b x^2\right )} \, dx=\frac {a \sqrt {b} d \sqrt {c+d x^2}-(-b c+a d)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )-b^{3/2} c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a b^{3/2}} \] Input:

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

Output:

(a*Sqrt[b]*d*Sqrt[c + d*x^2] - (-(b*c) + a*d)^(3/2)*ArcTan[(Sqrt[b]*Sqrt[c 
 + d*x^2])/Sqrt[-(b*c) + a*d]] - b^(3/2)*c^(3/2)*ArcTanh[Sqrt[c + d*x^2]/S 
qrt[c]])/(a*b^(3/2))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {354, 95, 174, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (c+d x^2\right )^{3/2}}{x \left (a+b x^2\right )} \, dx\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int \frac {\left (d x^2+c\right )^{3/2}}{x^2 \left (b x^2+a\right )}dx^2\)

\(\Big \downarrow \) 95

\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {b c^2+d (2 b c-a d) x^2}{x^2 \left (b x^2+a\right ) \sqrt {d x^2+c}}dx^2}{b}+\frac {2 d \sqrt {c+d x^2}}{b}\right )\)

\(\Big \downarrow \) 174

\(\displaystyle \frac {1}{2} \left (\frac {\frac {b c^2 \int \frac {1}{x^2 \sqrt {d x^2+c}}dx^2}{a}-\frac {(b c-a d)^2 \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx^2}{a}}{b}+\frac {2 d \sqrt {c+d x^2}}{b}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{2} \left (\frac {\frac {2 b c^2 \int \frac {1}{\frac {x^4}{d}-\frac {c}{d}}d\sqrt {d x^2+c}}{a d}-\frac {2 (b c-a d)^2 \int \frac {1}{\frac {b x^4}{d}+a-\frac {b c}{d}}d\sqrt {d x^2+c}}{a d}}{b}+\frac {2 d \sqrt {c+d x^2}}{b}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{2} \left (\frac {\frac {2 (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a \sqrt {b}}-\frac {2 b c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a}}{b}+\frac {2 d \sqrt {c+d x^2}}{b}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 95

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

rule 174

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

rule 221

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 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.07

method	result	size
pseudoelliptic	\(\frac {-\left (a d -b c \right )^{2} \arctan \left (\frac {\sqrt {x^{2} d +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )+\left (-c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}}{\sqrt {c}}\right ) b +a d \sqrt {x^{2} d +c}\right ) \sqrt {\left (a d -b c \right ) b}}{\sqrt {\left (a d -b c \right ) b}\, b a}\)	\(103\)
default	\(\text {Expression too large to display}\)	\(1293\)

Input:

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

Output:

(-(a*d-b*c)^2*arctan((d*x^2+c)^(1/2)*b/((a*d-b*c)*b)^(1/2))+(-c^(3/2)*arct 
anh((d*x^2+c)^(1/2)/c^(1/2))*b+a*d*(d*x^2+c)^(1/2))*((a*d-b*c)*b)^(1/2))/( 
(a*d-b*c)*b)^(1/2)/b/a
 

Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 688, normalized size of antiderivative = 7.17 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x \left (a+b x^2\right )} \, dx=\left [\frac {2 \, b c^{\frac {3}{2}} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 4 \, \sqrt {d x^{2} + c} a d - {\left (b c - a d\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{4 \, a b}, \frac {4 \, b \sqrt {-c} c \arctan \left (\frac {\sqrt {d x^{2} + c} \sqrt {-c}}{c}\right ) + 4 \, \sqrt {d x^{2} + c} a d - {\left (b c - a d\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{4 \, a b}, \frac {b c^{\frac {3}{2}} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, \sqrt {d x^{2} + c} a d + {\left (b c - a d\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right )}{2 \, a b}, \frac {2 \, b \sqrt {-c} c \arctan \left (\frac {\sqrt {d x^{2} + c} \sqrt {-c}}{c}\right ) + 2 \, \sqrt {d x^{2} + c} a d + {\left (b c - a d\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right )}{2 \, a b}\right ] \] Input:

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

Output:

[1/4*(2*b*c^(3/2)*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 4* 
sqrt(d*x^2 + c)*a*d - (b*c - a*d)*sqrt((b*c - a*d)/b)*log((b^2*d^2*x^4 + 8 
*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d - 3*a*b*d^2)*x^2 - 4*(b^2*d* 
x^2 + 2*b^2*c - a*b*d)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/b))/(b^2*x^4 + 2*a 
*b*x^2 + a^2)))/(a*b), 1/4*(4*b*sqrt(-c)*c*arctan(sqrt(d*x^2 + c)*sqrt(-c) 
/c) + 4*sqrt(d*x^2 + c)*a*d - (b*c - a*d)*sqrt((b*c - a*d)/b)*log((b^2*d^2 
*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d - 3*a*b*d^2)*x^2 - 4 
*(b^2*d*x^2 + 2*b^2*c - a*b*d)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/b))/(b^2*x 
^4 + 2*a*b*x^2 + a^2)))/(a*b), 1/2*(b*c^(3/2)*log(-(d*x^2 - 2*sqrt(d*x^2 + 
 c)*sqrt(c) + 2*c)/x^2) + 2*sqrt(d*x^2 + c)*a*d + (b*c - a*d)*sqrt(-(b*c - 
 a*d)/b)*arctan(-1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt(-(b*c - 
a*d)/b)/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2)))/(a*b), 1/2*(2*b*sqrt(-c)*c 
*arctan(sqrt(d*x^2 + c)*sqrt(-c)/c) + 2*sqrt(d*x^2 + c)*a*d + (b*c - a*d)* 
sqrt(-(b*c - a*d)/b)*arctan(-1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*s 
qrt(-(b*c - a*d)/b)/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2)))/(a*b)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (80) = 160\).

Time = 5.52 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.79 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x \left (a+b x^2\right )} \, dx=\begin {cases} \frac {2 \left (\frac {d^{2} \sqrt {c + d x^{2}}}{2 b} + \frac {c^{2} d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{2 a \sqrt {- c}} - \frac {d \left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{2 a b^{2} \sqrt {\frac {a d - b c}{b}}}\right )}{d} & \text {for}\: d \neq 0 \\c^{\frac {3}{2}} \left (- \frac {b \left (\begin {cases} \frac {\frac {a}{2 b} + x^{2}}{a} & \text {for}\: b = 0 \\- \frac {\log {\left (a - 2 b \left (\frac {a}{2 b} + x^{2}\right ) \right )}}{2 b} & \text {otherwise} \end {cases}\right )}{a} - \frac {b \left (\begin {cases} \frac {\frac {a}{2 b} + x^{2}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + 2 b \left (\frac {a}{2 b} + x^{2}\right ) \right )}}{2 b} & \text {otherwise} \end {cases}\right )}{a}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((2*(d**2*sqrt(c + d*x**2)/(2*b) + c**2*d*atan(sqrt(c + d*x**2)/s 
qrt(-c))/(2*a*sqrt(-c)) - d*(a*d - b*c)**2*atan(sqrt(c + d*x**2)/sqrt((a*d 
 - b*c)/b))/(2*a*b**2*sqrt((a*d - b*c)/b)))/d, Ne(d, 0)), (c**(3/2)*(-b*Pi 
ecewise(((a/(2*b) + x**2)/a, Eq(b, 0)), (-log(a - 2*b*(a/(2*b) + x**2))/(2 
*b), True))/a - b*Piecewise(((a/(2*b) + x**2)/a, Eq(b, 0)), (log(a + 2*b*( 
a/(2*b) + x**2))/(2*b), True))/a), True))
 

Maxima [F]

\[ \int \frac {\left (c+d x^2\right )^{3/2}}{x \left (a+b x^2\right )} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}}}{{\left (b x^{2} + a\right )} x} \,d x } \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.15 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x \left (a+b x^2\right )} \, dx=\frac {c^{2} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{a \sqrt {-c}} + \frac {\sqrt {d x^{2} + c} d}{b} - \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a b} \] Input:

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

Output:

c^2*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(a*sqrt(-c)) + sqrt(d*x^2 + c)*d/b - 
(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b 
*d))/(sqrt(-b^2*c + a*b*d)*a*b)
 

Mupad [B] (verification not implemented)

Time = 1.16 (sec) , antiderivative size = 711, normalized size of antiderivative = 7.41 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x \left (a+b x^2\right )} \, dx=\frac {d\,\sqrt {d\,x^2+c}}{b}-\frac {\mathrm {atanh}\left (\frac {2\,a^3\,d^6\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{2\,a^3\,c^2\,d^6-8\,a^2\,b\,c^3\,d^5+12\,a\,b^2\,c^4\,d^4-6\,b^3\,c^5\,d^3}+\frac {8\,a^2\,c\,d^5\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{8\,a^2\,c^3\,d^5+6\,b^2\,c^5\,d^3-\frac {2\,a^3\,c^2\,d^6}{b}-12\,a\,b\,c^4\,d^4}+\frac {6\,b^2\,c^3\,d^3\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{8\,a^2\,c^3\,d^5+6\,b^2\,c^5\,d^3-\frac {2\,a^3\,c^2\,d^6}{b}-12\,a\,b\,c^4\,d^4}-\frac {12\,a\,b\,c^2\,d^4\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{8\,a^2\,c^3\,d^5+6\,b^2\,c^5\,d^3-\frac {2\,a^3\,c^2\,d^6}{b}-12\,a\,b\,c^4\,d^4}\right )\,\sqrt {c^3}}{a}+\frac {\mathrm {atanh}\left (\frac {6\,c^3\,d^3\,\sqrt {d\,x^2+c}\,\sqrt {-a^3\,b^3\,d^3+3\,a^2\,b^4\,c\,d^2-3\,a\,b^5\,c^2\,d+b^6\,c^3}}{6\,b^3\,c^5\,d^3-10\,a^3\,c^2\,d^6-18\,a\,b^2\,c^4\,d^4+20\,a^2\,b\,c^3\,d^5+\frac {2\,a^4\,c\,d^7}{b}}-\frac {6\,a\,c^2\,d^4\,\sqrt {d\,x^2+c}\,\sqrt {-a^3\,b^3\,d^3+3\,a^2\,b^4\,c\,d^2-3\,a\,b^5\,c^2\,d+b^6\,c^3}}{2\,a^4\,c\,d^7-10\,a^3\,b\,c^2\,d^6+20\,a^2\,b^2\,c^3\,d^5-18\,a\,b^3\,c^4\,d^4+6\,b^4\,c^5\,d^3}+\frac {2\,a^2\,c\,d^5\,\sqrt {d\,x^2+c}\,\sqrt {-a^3\,b^3\,d^3+3\,a^2\,b^4\,c\,d^2-3\,a\,b^5\,c^2\,d+b^6\,c^3}}{2\,a^4\,b\,c\,d^7-10\,a^3\,b^2\,c^2\,d^6+20\,a^2\,b^3\,c^3\,d^5-18\,a\,b^4\,c^4\,d^4+6\,b^5\,c^5\,d^3}\right )\,\sqrt {-b^3\,{\left (a\,d-b\,c\right )}^3}}{a\,b^3} \] Input:

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

Output:

(d*(c + d*x^2)^(1/2))/b - (atanh((2*a^3*d^6*(c + d*x^2)^(1/2)*(c^3)^(1/2)) 
/(2*a^3*c^2*d^6 - 6*b^3*c^5*d^3 + 12*a*b^2*c^4*d^4 - 8*a^2*b*c^3*d^5) + (8 
*a^2*c*d^5*(c + d*x^2)^(1/2)*(c^3)^(1/2))/(8*a^2*c^3*d^5 + 6*b^2*c^5*d^3 - 
 (2*a^3*c^2*d^6)/b - 12*a*b*c^4*d^4) + (6*b^2*c^3*d^3*(c + d*x^2)^(1/2)*(c 
^3)^(1/2))/(8*a^2*c^3*d^5 + 6*b^2*c^5*d^3 - (2*a^3*c^2*d^6)/b - 12*a*b*c^4 
*d^4) - (12*a*b*c^2*d^4*(c + d*x^2)^(1/2)*(c^3)^(1/2))/(8*a^2*c^3*d^5 + 6* 
b^2*c^5*d^3 - (2*a^3*c^2*d^6)/b - 12*a*b*c^4*d^4))*(c^3)^(1/2))/a + (atanh 
((6*c^3*d^3*(c + d*x^2)^(1/2)*(b^6*c^3 - a^3*b^3*d^3 + 3*a^2*b^4*c*d^2 - 3 
*a*b^5*c^2*d)^(1/2))/(6*b^3*c^5*d^3 - 10*a^3*c^2*d^6 - 18*a*b^2*c^4*d^4 + 
20*a^2*b*c^3*d^5 + (2*a^4*c*d^7)/b) - (6*a*c^2*d^4*(c + d*x^2)^(1/2)*(b^6* 
c^3 - a^3*b^3*d^3 + 3*a^2*b^4*c*d^2 - 3*a*b^5*c^2*d)^(1/2))/(2*a^4*c*d^7 + 
 6*b^4*c^5*d^3 - 18*a*b^3*c^4*d^4 - 10*a^3*b*c^2*d^6 + 20*a^2*b^2*c^3*d^5) 
 + (2*a^2*c*d^5*(c + d*x^2)^(1/2)*(b^6*c^3 - a^3*b^3*d^3 + 3*a^2*b^4*c*d^2 
 - 3*a*b^5*c^2*d)^(1/2))/(6*b^5*c^5*d^3 - 18*a*b^4*c^4*d^4 + 20*a^2*b^3*c^ 
3*d^5 - 10*a^3*b^2*c^2*d^6 + 2*a^4*b*c*d^7))*(-b^3*(a*d - b*c)^3)^(1/2))/( 
a*b^3)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 1336, normalized size of antiderivative = 13.92 \[ \int \frac {\left (c+d x^2\right )^{3/2}}{x \left (a+b x^2\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 2*sqrt(d)*sqrt(b)*sqrt(a)*sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a 
*d - b*c)*sqrt(a*d - b*c)*atan((sqrt(c + d*x**2)*b + sqrt(d)*b*x)/(sqrt(b) 
*sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)))*a*d + 2*sqrt(d)*s 
qrt(b)*sqrt(a)*sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)*sqrt( 
a*d - b*c)*atan((sqrt(c + d*x**2)*b + sqrt(d)*b*x)/(sqrt(b)*sqrt(2*sqrt(d) 
*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)))*b*c + 2*sqrt(b)*sqrt(2*sqrt(d)*s 
qrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)*atan((sqrt(c + d*x**2)*b + sqrt(d)*b 
*x)/(sqrt(b)*sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)))*a**2* 
d**2 - 4*sqrt(b)*sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)*ata 
n((sqrt(c + d*x**2)*b + sqrt(d)*b*x)/(sqrt(b)*sqrt(2*sqrt(d)*sqrt(a)*sqrt( 
a*d - b*c) + 2*a*d - b*c)))*a*b*c*d + 2*sqrt(b)*sqrt(2*sqrt(d)*sqrt(a)*sqr 
t(a*d - b*c) + 2*a*d - b*c)*atan((sqrt(c + d*x**2)*b + sqrt(d)*b*x)/(sqrt( 
b)*sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*a*d - b*c)))*b**2*c**2 - sqr 
t(d)*sqrt(b)*sqrt(a)*sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) 
*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b 
*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*a*d + sqrt(d)*sqrt(b)* 
sqrt(a)*sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c)*sqrt(a*d - b 
*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(b) 
*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*b*c + sqrt(d)*sqrt(b)*sqrt(a)*sqrt( 
2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c)*sqrt(a*d - b*c)*log(sq...