\(\int \frac {(c+d x)^{3/2}}{x^2 (a+b x^2)} \, dx\) [616]

Optimal result

Integrand size = 22, antiderivative size = 485 \[ \int \frac {(c+d x)^{3/2}}{x^2 \left (a+b x^2\right )} \, dx=-\frac {c \sqrt {c+d x}}{a x}+\frac {d \left (2 \sqrt {b} c-\sqrt {b c^2+a d^2}\right ) \arctan \left (\frac {\sqrt {\sqrt {b} c+\sqrt {b c^2+a d^2}}-\sqrt {2} \sqrt [4]{b} \sqrt {c+d x}}{\sqrt {-\sqrt {b} c+\sqrt {b c^2+a d^2}}}\right )}{\sqrt {2} a \sqrt [4]{b} \sqrt {-\sqrt {b} c+\sqrt {b c^2+a d^2}}}-\frac {d \left (2 \sqrt {b} c-\sqrt {b c^2+a d^2}\right ) \arctan \left (\frac {\sqrt {\sqrt {b} c+\sqrt {b c^2+a d^2}}+\sqrt {2} \sqrt [4]{b} \sqrt {c+d x}}{\sqrt {-\sqrt {b} c+\sqrt {b c^2+a d^2}}}\right )}{\sqrt {2} a \sqrt [4]{b} \sqrt {-\sqrt {b} c+\sqrt {b c^2+a d^2}}}-\frac {3 \sqrt {c} d \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a}+\frac {d \left (2 \sqrt {b} c+\sqrt {b c^2+a d^2}\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {\sqrt {b} c+\sqrt {b c^2+a d^2}} \sqrt {c+d x}}{\sqrt {b c^2+a d^2}+\sqrt {b} (c+d x)}\right )}{\sqrt {2} a \sqrt [4]{b} \sqrt {\sqrt {b} c+\sqrt {b c^2+a d^2}}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.73 (sec) , antiderivative size = 262, normalized size of antiderivative = 0.54 \[ \int \frac {(c+d x)^{3/2}}{x^2 \left (a+b x^2\right )} \, dx=\frac {-\frac {\sqrt {a} c \sqrt {c+d x}}{x}-\frac {i \left (\sqrt {b} c+i \sqrt {a} d\right )^2 \arctan \left (\frac {\sqrt {-b c-i \sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c+i \sqrt {a} d}\right )}{\sqrt {-b c-i \sqrt {a} \sqrt {b} d}}+\frac {i \left (\sqrt {b} c-i \sqrt {a} d\right )^2 \arctan \left (\frac {\sqrt {-b c+i \sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c-i \sqrt {a} d}\right )}{\sqrt {-b c+i \sqrt {a} \sqrt {b} d}}-3 \sqrt {a} \sqrt {c} d \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 2.30 (sec) , antiderivative size = 741, normalized size of antiderivative = 1.53, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {561, 27, 1610, 2009}

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.

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

\(\Big \downarrow \) 561

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1610

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

\(\Big \downarrow \) 2009

\(\displaystyle 2 d \left (\frac {\left (-2 \sqrt {b} c \sqrt {a d^2+b c^2}+a d^2+b c^2\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {a d^2+b c^2}+\sqrt {b} c}-\sqrt {2} \sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a d^2+b c^2}}}\right )}{2 \sqrt {2} a \sqrt [4]{b} \sqrt {a d^2+b c^2} \sqrt {\sqrt {b} c-\sqrt {a d^2+b c^2}}}-\frac {\left (-2 \sqrt {b} c \sqrt {a d^2+b c^2}+a d^2+b c^2\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {a d^2+b c^2}+\sqrt {b} c}+\sqrt {2} \sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a d^2+b c^2}}}\right )}{2 \sqrt {2} a \sqrt [4]{b} \sqrt {a d^2+b c^2} \sqrt {\sqrt {b} c-\sqrt {a d^2+b c^2}}}-\frac {3 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{2 a}-\frac {\left (2 \sqrt {b} c \sqrt {a d^2+b c^2}+a d^2+b c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt {c+d x} \sqrt {\sqrt {a d^2+b c^2}+\sqrt {b} c}+\sqrt {a d^2+b c^2}+\sqrt {b} (c+d x)\right )}{4 \sqrt {2} a \sqrt [4]{b} \sqrt {a d^2+b c^2} \sqrt {\sqrt {a d^2+b c^2}+\sqrt {b} c}}+\frac {\left (2 \sqrt {b} c \sqrt {a d^2+b c^2}+a d^2+b c^2\right ) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt {c+d x} \sqrt {\sqrt {a d^2+b c^2}+\sqrt {b} c}+\sqrt {a d^2+b c^2}+\sqrt {b} (c+d x)\right )}{4 \sqrt {2} a \sqrt [4]{b} \sqrt {a d^2+b c^2} \sqrt {\sqrt {a d^2+b c^2}+\sqrt {b} c}}-\frac {c \sqrt {c+d x}}{2 a d x}\right )\)

Input:

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

Output:

2*d*(-1/2*(c*Sqrt[c + d*x])/(a*d*x) - (3*Sqrt[c]*ArcTanh[Sqrt[c + d*x]/Sqr 
t[c]])/(2*a) + ((b*c^2 + a*d^2 - 2*Sqrt[b]*c*Sqrt[b*c^2 + a*d^2])*ArcTanh[ 
(Sqrt[Sqrt[b]*c + Sqrt[b*c^2 + a*d^2]] - Sqrt[2]*b^(1/4)*Sqrt[c + d*x])/Sq 
rt[Sqrt[b]*c - Sqrt[b*c^2 + a*d^2]]])/(2*Sqrt[2]*a*b^(1/4)*Sqrt[b*c^2 + a* 
d^2]*Sqrt[Sqrt[b]*c - Sqrt[b*c^2 + a*d^2]]) - ((b*c^2 + a*d^2 - 2*Sqrt[b]* 
c*Sqrt[b*c^2 + a*d^2])*ArcTanh[(Sqrt[Sqrt[b]*c + Sqrt[b*c^2 + a*d^2]] + Sq 
rt[2]*b^(1/4)*Sqrt[c + d*x])/Sqrt[Sqrt[b]*c - Sqrt[b*c^2 + a*d^2]]])/(2*Sq 
rt[2]*a*b^(1/4)*Sqrt[b*c^2 + a*d^2]*Sqrt[Sqrt[b]*c - Sqrt[b*c^2 + a*d^2]]) 
 - ((b*c^2 + a*d^2 + 2*Sqrt[b]*c*Sqrt[b*c^2 + a*d^2])*Log[Sqrt[b*c^2 + a*d 
^2] - Sqrt[2]*b^(1/4)*Sqrt[Sqrt[b]*c + Sqrt[b*c^2 + a*d^2]]*Sqrt[c + d*x] 
+ Sqrt[b]*(c + d*x)])/(4*Sqrt[2]*a*b^(1/4)*Sqrt[b*c^2 + a*d^2]*Sqrt[Sqrt[b 
]*c + Sqrt[b*c^2 + a*d^2]]) + ((b*c^2 + a*d^2 + 2*Sqrt[b]*c*Sqrt[b*c^2 + a 
*d^2])*Log[Sqrt[b*c^2 + a*d^2] + Sqrt[2]*b^(1/4)*Sqrt[Sqrt[b]*c + Sqrt[b*c 
^2 + a*d^2]]*Sqrt[c + d*x] + Sqrt[b]*(c + d*x)])/(4*Sqrt[2]*a*b^(1/4)*Sqrt 
[b*c^2 + a*d^2]*Sqrt[Sqrt[b]*c + Sqrt[b*c^2 + a*d^2]]))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 561

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{k = Denominator[n]}, Simp[k/d   Subst[Int[x^(k*(n + 1) - 1)*(-c 
/d + x^k/d)^m*Simp[(b*c^2 + a*d^2)/d^2 - 2*b*c*(x^k/d^2) + b*(x^(2*k)/d^2), 
 x]^p, x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, m, p}, x] && Frac 
tionQ[n] && IntegerQ[p] && IntegerQ[m]
 

rule 1610

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.))/((a_) + (b_.)*(x_)^2 + 
 (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*((d + e*x^2)^q/(a 
+ b*x^2 + c*x^4)), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4 
*a*c, 0] && IntegerQ[q] && IntegerQ[m]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 734, normalized size of antiderivative = 1.51

method	result	size
pseudoelliptic	\(\frac {-\frac {x \left (\left (\sqrt {c}\, \sqrt {a \,d^{2}+b \,c^{2}}\, \sqrt {b}+2 c^{\frac {3}{2}} b \right ) \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}-c^{\frac {3}{2}} \sqrt {a \,d^{2}+b \,c^{2}}\, b^{\frac {3}{2}}-2 b^{2} c^{\frac {5}{2}}\right ) \sqrt {4 \sqrt {a \,d^{2}+b \,c^{2}}\, \sqrt {b}-2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}-2 b c}\, \sqrt {2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}+2 b c}\, \ln \left (\sqrt {b}\, \left (d x +c \right )-\sqrt {d x +c}\, \sqrt {2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}+2 b c}+\sqrt {a \,d^{2}+b \,c^{2}}\right )}{4}+\frac {x \left (\left (\sqrt {c}\, \sqrt {a \,d^{2}+b \,c^{2}}\, \sqrt {b}+2 c^{\frac {3}{2}} b \right ) \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}-c^{\frac {3}{2}} \sqrt {a \,d^{2}+b \,c^{2}}\, b^{\frac {3}{2}}-2 b^{2} c^{\frac {5}{2}}\right ) \sqrt {4 \sqrt {a \,d^{2}+b \,c^{2}}\, \sqrt {b}-2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}-2 b c}\, \sqrt {2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}+2 b c}\, \ln \left (\sqrt {b}\, \left (d x +c \right )+\sqrt {d x +c}\, \sqrt {2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}+2 b c}+\sqrt {a \,d^{2}+b \,c^{2}}\right )}{4}+d \left (-b^{\frac {3}{2}} \left (3 \,\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right ) c d x +c^{\frac {3}{2}} \sqrt {d x +c}\right ) \sqrt {4 \sqrt {a \,d^{2}+b \,c^{2}}\, \sqrt {b}-2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}-2 b c}+d x \left (\sqrt {c}\, \sqrt {a \,d^{2}+b \,c^{2}}\, b^{\frac {3}{2}}-2 c^{\frac {3}{2}} b^{2}\right ) \left (\arctan \left (\frac {2 \sqrt {b}\, \sqrt {d x +c}+\sqrt {2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}+2 b c}}{\sqrt {4 \sqrt {a \,d^{2}+b \,c^{2}}\, \sqrt {b}-2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}-2 b c}}\right )-\arctan \left (\frac {-2 \sqrt {b}\, \sqrt {d x +c}+\sqrt {2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}+2 b c}}{\sqrt {4 \sqrt {a \,d^{2}+b \,c^{2}}\, \sqrt {b}-2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}-2 b c}}\right )\right )\right ) a}{\sqrt {4 \sqrt {a \,d^{2}+b \,c^{2}}\, \sqrt {b}-2 \sqrt {\left (a \,d^{2}+b \,c^{2}\right ) b}-2 b c}\, b^{\frac {3}{2}} \sqrt {c}\, d \,a^{2} x}\)	\(734\)
risch	\(\text {Expression too large to display}\)	\(1265\)
derivativedivides	\(\text {Expression too large to display}\)	\(1280\)
default	\(\text {Expression too large to display}\)	\(1280\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 992 vs. \(2 (387) = 774\).

Time = 0.28 (sec) , antiderivative size = 1993, normalized size of antiderivative = 4.11 \[ \int \frac {(c+d x)^{3/2}}{x^2 \left (a+b x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/2*(a*x*sqrt(-(b*c^3 - 3*a*c*d^2 + a^3*sqrt(-(9*b^2*c^4*d^2 - 6*a*b*c^2 
*d^4 + a^2*d^6)/(a^5*b)))/a^3)*log(-(3*b^2*c^4*d + 2*a*b*c^2*d^3 - a^2*d^5 
)*sqrt(d*x + c) + (3*a^2*b*c^2*d^2 - a^3*d^4 + a^4*b*c*sqrt(-(9*b^2*c^4*d^ 
2 - 6*a*b*c^2*d^4 + a^2*d^6)/(a^5*b)))*sqrt(-(b*c^3 - 3*a*c*d^2 + a^3*sqrt 
(-(9*b^2*c^4*d^2 - 6*a*b*c^2*d^4 + a^2*d^6)/(a^5*b)))/a^3)) - a*x*sqrt(-(b 
*c^3 - 3*a*c*d^2 + a^3*sqrt(-(9*b^2*c^4*d^2 - 6*a*b*c^2*d^4 + a^2*d^6)/(a^ 
5*b)))/a^3)*log(-(3*b^2*c^4*d + 2*a*b*c^2*d^3 - a^2*d^5)*sqrt(d*x + c) - ( 
3*a^2*b*c^2*d^2 - a^3*d^4 + a^4*b*c*sqrt(-(9*b^2*c^4*d^2 - 6*a*b*c^2*d^4 + 
 a^2*d^6)/(a^5*b)))*sqrt(-(b*c^3 - 3*a*c*d^2 + a^3*sqrt(-(9*b^2*c^4*d^2 - 
6*a*b*c^2*d^4 + a^2*d^6)/(a^5*b)))/a^3)) + a*x*sqrt(-(b*c^3 - 3*a*c*d^2 - 
a^3*sqrt(-(9*b^2*c^4*d^2 - 6*a*b*c^2*d^4 + a^2*d^6)/(a^5*b)))/a^3)*log(-(3 
*b^2*c^4*d + 2*a*b*c^2*d^3 - a^2*d^5)*sqrt(d*x + c) + (3*a^2*b*c^2*d^2 - a 
^3*d^4 - a^4*b*c*sqrt(-(9*b^2*c^4*d^2 - 6*a*b*c^2*d^4 + a^2*d^6)/(a^5*b))) 
*sqrt(-(b*c^3 - 3*a*c*d^2 - a^3*sqrt(-(9*b^2*c^4*d^2 - 6*a*b*c^2*d^4 + a^2 
*d^6)/(a^5*b)))/a^3)) - a*x*sqrt(-(b*c^3 - 3*a*c*d^2 - a^3*sqrt(-(9*b^2*c^ 
4*d^2 - 6*a*b*c^2*d^4 + a^2*d^6)/(a^5*b)))/a^3)*log(-(3*b^2*c^4*d + 2*a*b* 
c^2*d^3 - a^2*d^5)*sqrt(d*x + c) - (3*a^2*b*c^2*d^2 - a^3*d^4 - a^4*b*c*sq 
rt(-(9*b^2*c^4*d^2 - 6*a*b*c^2*d^4 + a^2*d^6)/(a^5*b)))*sqrt(-(b*c^3 - 3*a 
*c*d^2 - a^3*sqrt(-(9*b^2*c^4*d^2 - 6*a*b*c^2*d^4 + a^2*d^6)/(a^5*b)))/a^3 
)) - 3*sqrt(c)*d*x*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*sqr...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 390, normalized size of antiderivative = 0.80 \[ \int \frac {(c+d x)^{3/2}}{x^2 \left (a+b x^2\right )} \, dx=\frac {3 \, c d \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{a \sqrt {-c}} - \frac {\sqrt {d x + c} c}{a x} - \frac {{\left (2 \, a^{2} b c d^{3} {\left | b \right |} + {\left (\sqrt {-a b} b c^{2} d + \sqrt {-a b} a d^{3}\right )} {\left | a \right |} {\left | b \right |} {\left | d \right |} + {\left (a b^{2} c^{3} d - a^{2} b c d^{3}\right )} {\left | b \right |}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-\frac {a b c + \sqrt {a^{2} b^{2} c^{2} - {\left (a b c^{2} + a^{2} d^{2}\right )} a b}}{a b}}}\right )}{{\left (a^{2} b d + \sqrt {-a b} a b c\right )} \sqrt {-b^{2} c - \sqrt {-a b} b d} {\left | a \right |} {\left | d \right |}} - \frac {{\left (2 \, \sqrt {-a b} a^{2} c d^{3} {\left | b \right |} + {\left (a b c^{2} d + a^{2} d^{3}\right )} {\left | a \right |} {\left | b \right |} {\left | d \right |} + {\left (\sqrt {-a b} a b c^{3} d - \sqrt {-a b} a^{2} c d^{3}\right )} {\left | b \right |}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-\frac {a b c - \sqrt {a^{2} b^{2} c^{2} - {\left (a b c^{2} + a^{2} d^{2}\right )} a b}}{a b}}}\right )}{{\left (a^{2} b c + \sqrt {-a b} a^{2} d\right )} \sqrt {-b^{2} c + \sqrt {-a b} b d} {\left | a \right |} {\left | d \right |}} \] Input:

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

Output:

3*c*d*arctan(sqrt(d*x + c)/sqrt(-c))/(a*sqrt(-c)) - sqrt(d*x + c)*c/(a*x) 
- (2*a^2*b*c*d^3*abs(b) + (sqrt(-a*b)*b*c^2*d + sqrt(-a*b)*a*d^3)*abs(a)*a 
bs(b)*abs(d) + (a*b^2*c^3*d - a^2*b*c*d^3)*abs(b))*arctan(sqrt(d*x + c)/sq 
rt(-(a*b*c + sqrt(a^2*b^2*c^2 - (a*b*c^2 + a^2*d^2)*a*b))/(a*b)))/((a^2*b* 
d + sqrt(-a*b)*a*b*c)*sqrt(-b^2*c - sqrt(-a*b)*b*d)*abs(a)*abs(d)) - (2*sq 
rt(-a*b)*a^2*c*d^3*abs(b) + (a*b*c^2*d + a^2*d^3)*abs(a)*abs(b)*abs(d) + ( 
sqrt(-a*b)*a*b*c^3*d - sqrt(-a*b)*a^2*c*d^3)*abs(b))*arctan(sqrt(d*x + c)/ 
sqrt(-(a*b*c - sqrt(a^2*b^2*c^2 - (a*b*c^2 + a^2*d^2)*a*b))/(a*b)))/((a^2* 
b*c + sqrt(-a*b)*a^2*d)*sqrt(-b^2*c + sqrt(-a*b)*b*d)*abs(a)*abs(d))
 

Mupad [B] (verification not implemented)

Time = 8.40 (sec) , antiderivative size = 3858, normalized size of antiderivative = 7.95 \[ \int \frac {(c+d x)^{3/2}}{x^2 \left (a+b x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

(c^(1/2)*d*atan((b^4*c^(5/2)*d^15*(c + d*x)^(1/2)*2976i)/(2976*b^4*c^3*d^1 
5 + (4896*b^5*c^5*d^13)/a + (2112*b^6*c^7*d^11)/a^2 + (96*b^7*c^9*d^9)/a^3 
 + 96*a*b^3*c*d^17) + (b^5*c^(9/2)*d^13*(c + d*x)^(1/2)*4896i)/(4896*b^5*c 
^5*d^13 + 2976*a*b^4*c^3*d^15 + 96*a^2*b^3*c*d^17 + (2112*b^6*c^7*d^11)/a 
+ (96*b^7*c^9*d^9)/a^2) + (b^6*c^(13/2)*d^11*(c + d*x)^(1/2)*2112i)/(2112* 
b^6*c^7*d^11 + 4896*a*b^5*c^5*d^13 + 96*a^3*b^3*c*d^17 + 2976*a^2*b^4*c^3* 
d^15 + (96*b^7*c^9*d^9)/a) + (b^7*c^(17/2)*d^9*(c + d*x)^(1/2)*96i)/(96*b^ 
7*c^9*d^9 + 2112*a*b^6*c^7*d^11 + 96*a^4*b^3*c*d^17 + 4896*a^2*b^5*c^5*d^1 
3 + 2976*a^3*b^4*c^3*d^15) + (a*b^3*c^(1/2)*d^17*(c + d*x)^(1/2)*96i)/(297 
6*b^4*c^3*d^15 + (4896*b^5*c^5*d^13)/a + (2112*b^6*c^7*d^11)/a^2 + (96*b^7 
*c^9*d^9)/a^3 + 96*a*b^3*c*d^17))*3i)/a - atan((b^5*c^4*d^12*(c + d*x)^(1/ 
2)*((3*c*d^2)/(4*a^2) - (b*c^3)/(4*a^3) - (3*c^2*d*(-a^7*b)^(1/2))/(4*a^6) 
 + (d^3*(-a^7*b)^(1/2))/(4*a^5*b))^(1/2)*736i)/((128*b^5*c^5*d^13)/a - 16* 
b^4*c^3*d^15 + (176*b^6*c^7*d^11)/a^2 + (48*b^7*c^9*d^9)/a^3 - 16*a*b^3*c* 
d^17 - (80*b^3*c^2*d^16*(-a^7*b)^(1/2))/a^3 + (96*b^4*c^4*d^14*(-a^7*b)^(1 
/2))/a^4 + (368*b^5*c^6*d^12*(-a^7*b)^(1/2))/a^5 + (192*b^6*c^8*d^10*(-a^7 
*b)^(1/2))/a^6) - (b^6*c^6*d^10*(c + d*x)^(1/2)*((3*c*d^2)/(4*a^2) - (b*c^ 
3)/(4*a^3) - (3*c^2*d*(-a^7*b)^(1/2))/(4*a^6) + (d^3*(-a^7*b)^(1/2))/(4*a^ 
5*b))^(1/2)*192i)/(128*b^5*c^5*d^13 - 16*a*b^4*c^3*d^15 - 16*a^2*b^3*c*d^1 
7 + (176*b^6*c^7*d^11)/a + (48*b^7*c^9*d^9)/a^2 - (80*b^3*c^2*d^16*(-a^...
 

Reduce [F]

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

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

Output:

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