\(\int \frac {(a+b x^2)^2}{x (c+d x)^{5/2}} \, dx\) [558]

Optimal result

Integrand size = 22, antiderivative size = 137 \[ \int \frac {\left (a+b x^2\right )^2}{x (c+d x)^{5/2}} \, dx=\frac {2 \left (b c^2+a d^2\right )^2}{3 c d^4 (c+d x)^{3/2}}-\frac {2 \left (3 b c^2-a d^2\right ) \left (b c^2+a d^2\right )}{c^2 d^4 \sqrt {c+d x}}-\frac {6 b^2 c \sqrt {c+d x}}{d^4}+\frac {2 b^2 (c+d x)^{3/2}}{3 d^4}-\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{5/2}} \] Output:

2/3*(a*d^2+b*c^2)^2/c/d^4/(d*x+c)^(3/2)-2*(-a*d^2+3*b*c^2)*(a*d^2+b*c^2)/c 
^2/d^4/(d*x+c)^(1/2)-6*b^2*c*(d*x+c)^(1/2)/d^4+2/3*b^2*(d*x+c)^(3/2)/d^4-2 
*a^2*arctanh((d*x+c)^(1/2)/c^(1/2))/c^(5/2)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^2\right )^2}{x (c+d x)^{5/2}} \, dx=\frac {2 \left (-2 a b c^2 d^2 (2 c+3 d x)+a^2 d^4 (4 c+3 d x)+b^2 c^2 \left (-16 c^3-24 c^2 d x-6 c d^2 x^2+d^3 x^3\right )\right )}{3 c^2 d^4 (c+d x)^{3/2}}-\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{5/2}} \] Input:

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

Output:

(2*(-2*a*b*c^2*d^2*(2*c + 3*d*x) + a^2*d^4*(4*c + 3*d*x) + b^2*c^2*(-16*c^ 
3 - 24*c^2*d*x - 6*c*d^2*x^2 + d^3*x^3)))/(3*c^2*d^4*(c + d*x)^(3/2)) - (2 
*a^2*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/c^(5/2)
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {517, 25, 1584, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^2}{x (c+d x)^{5/2}} \, dx\)

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1584

\(\displaystyle -\frac {2 \int \left (-\frac {a^2 d^3}{c^2 x}+3 b^2 c-b^2 (c+d x)+\frac {-3 b^2 c^4-2 a b d^2 c^2+a^2 d^4}{c^2 (c+d x)}+\frac {\left (b c^2+a d^2\right )^2}{c (c+d x)^2}\right )d\sqrt {c+d x}}{d^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (-\frac {a^2 d^4 \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{5/2}}-\frac {\left (3 b c^2-a d^2\right ) \left (a d^2+b c^2\right )}{c^2 \sqrt {c+d x}}+\frac {\left (a d^2+b c^2\right )^2}{3 c (c+d x)^{3/2}}+\frac {1}{3} b^2 (c+d x)^{3/2}-3 b^2 c \sqrt {c+d x}\right )}{d^4}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 517

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> Simp[2*(e^m/d^(m + 2*p + 1))   Subst[Int[x^(2*n + 1)*(-c + x^ 
2)^m*(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4)^p, x], x, Sqrt[c + d*x]], x] /; Fr 
eeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[m, 0] && IntegerQ[n + 1/2]
 

rule 1584

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

rule 2009

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

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.86

method	result	size
pseudoelliptic	\(\frac {-2 a^{2} d^{4} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right ) c^{2} \left (d x +c \right )^{\frac {3}{2}}-\frac {8 d^{2} \left (\frac {3 b \,x^{2}}{2}+a \right ) b \,c^{\frac {11}{2}}}{3}-4 d^{3} x b \left (-\frac {b \,x^{2}}{6}+a \right ) c^{\frac {9}{2}}-16 c^{\frac {13}{2}} b^{2} d x -\frac {32 c^{\frac {15}{2}} b^{2}}{3}+\frac {8 \left (c^{\frac {7}{2}}+\frac {3 d x \,c^{\frac {5}{2}}}{4}\right ) d^{4} a^{2}}{3}}{d^{4} c^{\frac {9}{2}} \left (d x +c \right )^{\frac {3}{2}}}\)	\(118\)
derivativedivides	\(\frac {\frac {2 b^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}-6 b^{2} c \sqrt {d x +c}-\frac {2 \left (-a^{2} d^{4}-2 b \,c^{2} d^{2} a -b^{2} c^{4}\right )}{3 c \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (-a^{2} d^{4}+2 b \,c^{2} d^{2} a +3 b^{2} c^{4}\right )}{c^{2} \sqrt {d x +c}}-\frac {2 a^{2} d^{4} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{c^{\frac {5}{2}}}}{d^{4}}\)	\(133\)
default	\(\frac {\frac {2 b^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}-6 b^{2} c \sqrt {d x +c}-\frac {2 \left (-a^{2} d^{4}-2 b \,c^{2} d^{2} a -b^{2} c^{4}\right )}{3 c \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (-a^{2} d^{4}+2 b \,c^{2} d^{2} a +3 b^{2} c^{4}\right )}{c^{2} \sqrt {d x +c}}-\frac {2 a^{2} d^{4} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{c^{\frac {5}{2}}}}{d^{4}}\)	\(133\)

Input:

int((b*x^2+a)^2/x/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

8/3*(-3/4*a^2*d^4*arctanh((d*x+c)^(1/2)/c^(1/2))*c^2*(d*x+c)^(3/2)-d^2*(3/ 
2*b*x^2+a)*b*c^(11/2)-3/2*d^3*x*b*(-1/6*b*x^2+a)*c^(9/2)-6*c^(13/2)*b^2*d* 
x-4*c^(15/2)*b^2+(c^(7/2)+3/4*d*x*c^(5/2))*d^4*a^2)/c^(9/2)/(d*x+c)^(3/2)/ 
d^4
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 372, normalized size of antiderivative = 2.72 \[ \int \frac {\left (a+b x^2\right )^2}{x (c+d x)^{5/2}} \, dx=\left [\frac {3 \, {\left (a^{2} d^{6} x^{2} + 2 \, a^{2} c d^{5} x + a^{2} c^{2} d^{4}\right )} \sqrt {c} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (b^{2} c^{3} d^{3} x^{3} - 6 \, b^{2} c^{4} d^{2} x^{2} - 16 \, b^{2} c^{6} - 4 \, a b c^{4} d^{2} + 4 \, a^{2} c^{2} d^{4} - 3 \, {\left (8 \, b^{2} c^{5} d + 2 \, a b c^{3} d^{3} - a^{2} c d^{5}\right )} x\right )} \sqrt {d x + c}}{3 \, {\left (c^{3} d^{6} x^{2} + 2 \, c^{4} d^{5} x + c^{5} d^{4}\right )}}, \frac {2 \, {\left (3 \, {\left (a^{2} d^{6} x^{2} + 2 \, a^{2} c d^{5} x + a^{2} c^{2} d^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) + {\left (b^{2} c^{3} d^{3} x^{3} - 6 \, b^{2} c^{4} d^{2} x^{2} - 16 \, b^{2} c^{6} - 4 \, a b c^{4} d^{2} + 4 \, a^{2} c^{2} d^{4} - 3 \, {\left (8 \, b^{2} c^{5} d + 2 \, a b c^{3} d^{3} - a^{2} c d^{5}\right )} x\right )} \sqrt {d x + c}\right )}}{3 \, {\left (c^{3} d^{6} x^{2} + 2 \, c^{4} d^{5} x + c^{5} d^{4}\right )}}\right ] \] Input:

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

Output:

[1/3*(3*(a^2*d^6*x^2 + 2*a^2*c*d^5*x + a^2*c^2*d^4)*sqrt(c)*log((d*x - 2*s 
qrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*(b^2*c^3*d^3*x^3 - 6*b^2*c^4*d^2*x^2 - 
16*b^2*c^6 - 4*a*b*c^4*d^2 + 4*a^2*c^2*d^4 - 3*(8*b^2*c^5*d + 2*a*b*c^3*d^ 
3 - a^2*c*d^5)*x)*sqrt(d*x + c))/(c^3*d^6*x^2 + 2*c^4*d^5*x + c^5*d^4), 2/ 
3*(3*(a^2*d^6*x^2 + 2*a^2*c*d^5*x + a^2*c^2*d^4)*sqrt(-c)*arctan(sqrt(-c)/ 
sqrt(d*x + c)) + (b^2*c^3*d^3*x^3 - 6*b^2*c^4*d^2*x^2 - 16*b^2*c^6 - 4*a*b 
*c^4*d^2 + 4*a^2*c^2*d^4 - 3*(8*b^2*c^5*d + 2*a*b*c^3*d^3 - a^2*c*d^5)*x)* 
sqrt(d*x + c))/(c^3*d^6*x^2 + 2*c^4*d^5*x + c^5*d^4)]
 

Sympy [A] (verification not implemented)

Time = 6.74 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.22 \[ \int \frac {\left (a+b x^2\right )^2}{x (c+d x)^{5/2}} \, dx=\begin {cases} \frac {2 a^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{c^{2} \sqrt {- c}} - \frac {6 b^{2} c \sqrt {c + d x}}{d^{4}} + \frac {2 b^{2} \left (c + d x\right )^{\frac {3}{2}}}{3 d^{4}} + \frac {2 \left (a d^{2} + b c^{2}\right )^{2}}{3 c d^{4} \left (c + d x\right )^{\frac {3}{2}}} + \frac {2 \left (a d^{2} - 3 b c^{2}\right ) \left (a d^{2} + b c^{2}\right )}{c^{2} d^{4} \sqrt {c + d x}} & \text {for}\: d \neq 0 \\\frac {\frac {a^{2} \log {\left (x^{2} \right )}}{2} + a b x^{2} + \frac {b^{2} x^{4}}{4}}{c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((2*a**2*atan(sqrt(c + d*x)/sqrt(-c))/(c**2*sqrt(-c)) - 6*b**2*c* 
sqrt(c + d*x)/d**4 + 2*b**2*(c + d*x)**(3/2)/(3*d**4) + 2*(a*d**2 + b*c**2 
)**2/(3*c*d**4*(c + d*x)**(3/2)) + 2*(a*d**2 - 3*b*c**2)*(a*d**2 + b*c**2) 
/(c**2*d**4*sqrt(c + d*x)), Ne(d, 0)), ((a**2*log(x**2)/2 + a*b*x**2 + b** 
2*x**4/4)/c**(5/2), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^2\right )^2}{x (c+d x)^{5/2}} \, dx=\frac {a^{2} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{c^{\frac {5}{2}}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} b^{2} - 9 \, \sqrt {d x + c} b^{2} c\right )}}{3 \, d^{4}} + \frac {2 \, {\left (b^{2} c^{5} + 2 \, a b c^{3} d^{2} + a^{2} c d^{4} - 3 \, {\left (3 \, b^{2} c^{4} + 2 \, a b c^{2} d^{2} - a^{2} d^{4}\right )} {\left (d x + c\right )}\right )}}{3 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{2} d^{4}} \] Input:

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

Output:

a^2*log((sqrt(d*x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c)))/c^(5/2) + 2/3 
*((d*x + c)^(3/2)*b^2 - 9*sqrt(d*x + c)*b^2*c)/d^4 + 2/3*(b^2*c^5 + 2*a*b* 
c^3*d^2 + a^2*c*d^4 - 3*(3*b^2*c^4 + 2*a*b*c^2*d^2 - a^2*d^4)*(d*x + c))/( 
(d*x + c)^(3/2)*c^2*d^4)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right )^2}{x (c+d x)^{5/2}} \, dx=\frac {2 \, a^{2} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{2}} - \frac {2 \, {\left (9 \, {\left (d x + c\right )} b^{2} c^{4} - b^{2} c^{5} + 6 \, {\left (d x + c\right )} a b c^{2} d^{2} - 2 \, a b c^{3} d^{2} - 3 \, {\left (d x + c\right )} a^{2} d^{4} - a^{2} c d^{4}\right )}}{3 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{2} d^{4}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} b^{2} d^{8} - 9 \, \sqrt {d x + c} b^{2} c d^{8}\right )}}{3 \, d^{12}} \] Input:

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

Output:

2*a^2*arctan(sqrt(d*x + c)/sqrt(-c))/(sqrt(-c)*c^2) - 2/3*(9*(d*x + c)*b^2 
*c^4 - b^2*c^5 + 6*(d*x + c)*a*b*c^2*d^2 - 2*a*b*c^3*d^2 - 3*(d*x + c)*a^2 
*d^4 - a^2*c*d^4)/((d*x + c)^(3/2)*c^2*d^4) + 2/3*((d*x + c)^(3/2)*b^2*d^8 
 - 9*sqrt(d*x + c)*b^2*c*d^8)/d^12
 

Mupad [B] (verification not implemented)

Time = 8.44 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2\right )^2}{x (c+d x)^{5/2}} \, dx=\frac {2\,b^2\,{\left (c+d\,x\right )}^{3/2}}{3\,d^4}+\frac {\frac {2\,\left (a^2\,d^4+2\,a\,b\,c^2\,d^2+b^2\,c^4\right )}{3\,c}-\frac {2\,\left (c+d\,x\right )\,\left (-a^2\,d^4+2\,a\,b\,c^2\,d^2+3\,b^2\,c^4\right )}{c^2}}{d^4\,{\left (c+d\,x\right )}^{3/2}}-\frac {6\,b^2\,c\,\sqrt {c+d\,x}}{d^4}+\frac {a^2\,\mathrm {atan}\left (\frac {\sqrt {c+d\,x}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,2{}\mathrm {i}}{c^{5/2}} \] Input:

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

Output:

(2*b^2*(c + d*x)^(3/2))/(3*d^4) + ((2*(a^2*d^4 + b^2*c^4 + 2*a*b*c^2*d^2)) 
/(3*c) - (2*(c + d*x)*(3*b^2*c^4 - a^2*d^4 + 2*a*b*c^2*d^2))/c^2)/(d^4*(c 
+ d*x)^(3/2)) + (a^2*atan(((c + d*x)^(1/2)*1i)/c^(1/2))*2i)/c^(5/2) - (6*b 
^2*c*(c + d*x)^(1/2))/d^4
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.64 \[ \int \frac {\left (a+b x^2\right )^2}{x (c+d x)^{5/2}} \, dx=\frac {3 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a^{2} c \,d^{4}+3 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a^{2} d^{5} x -3 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a^{2} c \,d^{4}-3 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a^{2} d^{5} x +8 a^{2} c^{2} d^{4}+6 a^{2} c \,d^{5} x -8 a b \,c^{4} d^{2}-12 a b \,c^{3} d^{3} x -32 b^{2} c^{6}-48 b^{2} c^{5} d x -12 b^{2} c^{4} d^{2} x^{2}+2 b^{2} c^{3} d^{3} x^{3}}{3 \sqrt {d x +c}\, c^{3} d^{4} \left (d x +c \right )} \] Input:

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

Output:

(3*sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*a**2*c*d**4 + 3*sqrt 
(c)*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*a**2*d**5*x - 3*sqrt(c)*sqr 
t(c + d*x)*log(sqrt(c + d*x) + sqrt(c))*a**2*c*d**4 - 3*sqrt(c)*sqrt(c + d 
*x)*log(sqrt(c + d*x) + sqrt(c))*a**2*d**5*x + 8*a**2*c**2*d**4 + 6*a**2*c 
*d**5*x - 8*a*b*c**4*d**2 - 12*a*b*c**3*d**3*x - 32*b**2*c**6 - 48*b**2*c* 
*5*d*x - 12*b**2*c**4*d**2*x**2 + 2*b**2*c**3*d**3*x**3)/(3*sqrt(c + d*x)* 
c**3*d**4*(c + d*x))