\(\int \frac {(a+b x^2)^2}{x^2 (c+d x)^{5/2}} \, dx\) [559]

Optimal result

Integrand size = 22, antiderivative size = 126 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 (c+d x)^{5/2}} \, dx=-\frac {2 \left (b c^2+a d^2\right )^2}{3 c^2 d^3 (c+d x)^{3/2}}+\frac {4 \left (\frac {b^2 c}{d^3}-\frac {a^2 d}{c^3}\right )}{\sqrt {c+d x}}+\frac {2 b^2 \sqrt {c+d x}}{d^3}-\frac {a^2 \sqrt {c+d x}}{c^3 x}+\frac {5 a^2 d \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{7/2}} \] Output:

-2/3*(a*d^2+b*c^2)^2/c^2/d^3/(d*x+c)^(3/2)+4*(b^2*c/d^3-a^2*d/c^3)/(d*x+c) 
^(1/2)+2*b^2*(d*x+c)^(1/2)/d^3-a^2*(d*x+c)^(1/2)/c^3/x+5*a^2*d*arctanh((d* 
x+c)^(1/2)/c^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 (c+d x)^{5/2}} \, dx=\frac {-4 a b c^3 d^2 x+2 b^2 c^3 x \left (8 c^2+12 c d x+3 d^2 x^2\right )-a^2 d^3 \left (3 c^2+20 c d x+15 d^2 x^2\right )}{3 c^3 d^3 x (c+d x)^{3/2}}+\frac {5 a^2 d \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{7/2}} \] Input:

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

Output:

(-4*a*b*c^3*d^2*x + 2*b^2*c^3*x*(8*c^2 + 12*c*d*x + 3*d^2*x^2) - a^2*d^3*( 
3*c^2 + 20*c*d*x + 15*d^2*x^2))/(3*c^3*d^3*x*(c + d*x)^(3/2)) + (5*a^2*d*A 
rcTanh[Sqrt[c + d*x]/Sqrt[c]])/c^(7/2)
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {517, 1582, 2333, 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.

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

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 1582

\(\displaystyle \frac {2 \left (\frac {\int -\frac {-2 b^2 c^3 (c+d x)^3+\left (6 b^2 c^4+a^2 d^4\right ) (c+d x)^2-2 c \left (3 b c^2-a d^2\right ) \left (b c^2+a d^2\right ) (c+d x)+2 c^2 \left (b c^2+a d^2\right )^2}{d x (c+d x)^2}d\sqrt {c+d x}}{2 c^3}-\frac {a^2 d^3 \sqrt {c+d x}}{2 c^3 x}\right )}{d^3}\)

\(\Big \downarrow \) 2333

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(2*(-1/2*(a^2*d^3*Sqrt[c + d*x])/(c^3*x) + ((-2*c*(b*c^2 + a*d^2)^2)/(3*(c 
 + d*x)^(3/2)) + (4*(b^2*c^4 - a^2*d^4))/Sqrt[c + d*x] + 2*b^2*c^3*Sqrt[c 
+ d*x] + (5*a^2*d^4*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/Sqrt[c])/(2*c^3)))/d^3
 

Defintions of rubi rules used

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 1582

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^ 
4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d 
+ e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[(-d)^(m/2 - 1)/(2*e^ 
(2*p)*(q + 1))   Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e 
*x^2))*(2*(-d)^(-m/2 + 1)*e^(2*p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - 
 b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x], x] /; Fre 
eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] 
&& ILtQ[m/2, 0]
 

rule 2009

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

rule 2333

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ 
ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] 
&& PolyQ[Pq, x] && IGtQ[p, -2]
 

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00

method	result	size
pseudoelliptic	\(\frac {5 \left (d x +c \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right ) a^{2} c^{3} d^{4} x +\frac {16 c^{\frac {7}{2}} \left (c^{5} b^{2} x +\frac {3 b^{2} c^{4} d \,x^{2}}{2}-\frac {d^{2} x b \left (-\frac {3 b \,x^{2}}{2}+a \right ) c^{3}}{4}-\frac {3 d^{3} a^{2} c^{2}}{16}-\frac {5 a^{2} c \,d^{4} x}{4}-\frac {15 a^{2} x^{2} d^{5}}{16}\right )}{3}}{d^{3} c^{\frac {13}{2}} \left (d x +c \right )^{\frac {3}{2}} x}\)	\(126\)
derivativedivides	\(\frac {2 b^{2} \sqrt {d x +c}+\frac {2 a^{2} d^{4} \left (-\frac {\sqrt {d x +c}}{2 d x}+\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )}{c^{3}}-\frac {2 \left (2 a^{2} d^{4}-2 b^{2} c^{4}\right )}{c^{3} \sqrt {d x +c}}-\frac {2 \left (a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}\right )}{3 c^{2} \left (d x +c \right )^{\frac {3}{2}}}}{d^{3}}\)	\(127\)
default	\(\frac {2 b^{2} \sqrt {d x +c}+\frac {2 a^{2} d^{4} \left (-\frac {\sqrt {d x +c}}{2 d x}+\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )}{c^{3}}-\frac {2 \left (2 a^{2} d^{4}-2 b^{2} c^{4}\right )}{c^{3} \sqrt {d x +c}}-\frac {2 \left (a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}\right )}{3 c^{2} \left (d x +c \right )^{\frac {3}{2}}}}{d^{3}}\)	\(127\)
risch	\(-\frac {a^{2} \sqrt {d x +c}}{c^{3} x}+\frac {2 c^{3} b^{2} \sqrt {d x +c}-\frac {4 a^{2} d^{4}-4 b^{2} c^{4}}{\sqrt {d x +c}}-\frac {2 c \left (a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}+\frac {5 a^{2} d^{4} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\sqrt {c}}}{d^{3} c^{3}}\)	\(127\)

Input:

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

Output:

5*((d*x+c)^(3/2)*arctanh((d*x+c)^(1/2)/c^(1/2))*a^2*c^3*d^4*x+16/15*c^(7/2 
)*(c^5*b^2*x+3/2*b^2*c^4*d*x^2-1/4*d^2*x*b*(-3/2*b*x^2+a)*c^3-3/16*d^3*a^2 
*c^2-5/4*a^2*c*d^4*x-15/16*a^2*x^2*d^5))/(d*x+c)^(3/2)/d^3/c^(13/2)/x
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 373, normalized size of antiderivative = 2.96 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 (c+d x)^{5/2}} \, dx=\left [\frac {15 \, {\left (a^{2} d^{6} x^{3} + 2 \, a^{2} c d^{5} x^{2} + a^{2} c^{2} d^{4} x\right )} \sqrt {c} \log \left (\frac {d x + 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (6 \, b^{2} c^{4} d^{2} x^{3} - 3 \, a^{2} c^{3} d^{3} + 3 \, {\left (8 \, b^{2} c^{5} d - 5 \, a^{2} c d^{5}\right )} x^{2} + 4 \, {\left (4 \, b^{2} c^{6} - a b c^{4} d^{2} - 5 \, a^{2} c^{2} d^{4}\right )} x\right )} \sqrt {d x + c}}{6 \, {\left (c^{4} d^{5} x^{3} + 2 \, c^{5} d^{4} x^{2} + c^{6} d^{3} x\right )}}, -\frac {15 \, {\left (a^{2} d^{6} x^{3} + 2 \, a^{2} c d^{5} x^{2} + a^{2} c^{2} d^{4} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) - {\left (6 \, b^{2} c^{4} d^{2} x^{3} - 3 \, a^{2} c^{3} d^{3} + 3 \, {\left (8 \, b^{2} c^{5} d - 5 \, a^{2} c d^{5}\right )} x^{2} + 4 \, {\left (4 \, b^{2} c^{6} - a b c^{4} d^{2} - 5 \, a^{2} c^{2} d^{4}\right )} x\right )} \sqrt {d x + c}}{3 \, {\left (c^{4} d^{5} x^{3} + 2 \, c^{5} d^{4} x^{2} + c^{6} d^{3} x\right )}}\right ] \] Input:

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

Output:

[1/6*(15*(a^2*d^6*x^3 + 2*a^2*c*d^5*x^2 + a^2*c^2*d^4*x)*sqrt(c)*log((d*x 
+ 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*(6*b^2*c^4*d^2*x^3 - 3*a^2*c^3*d^3 
 + 3*(8*b^2*c^5*d - 5*a^2*c*d^5)*x^2 + 4*(4*b^2*c^6 - a*b*c^4*d^2 - 5*a^2* 
c^2*d^4)*x)*sqrt(d*x + c))/(c^4*d^5*x^3 + 2*c^5*d^4*x^2 + c^6*d^3*x), -1/3 
*(15*(a^2*d^6*x^3 + 2*a^2*c*d^5*x^2 + a^2*c^2*d^4*x)*sqrt(-c)*arctan(sqrt( 
-c)/sqrt(d*x + c)) - (6*b^2*c^4*d^2*x^3 - 3*a^2*c^3*d^3 + 3*(8*b^2*c^5*d - 
 5*a^2*c*d^5)*x^2 + 4*(4*b^2*c^6 - a*b*c^4*d^2 - 5*a^2*c^2*d^4)*x)*sqrt(d* 
x + c))/(c^4*d^5*x^3 + 2*c^5*d^4*x^2 + c^6*d^3*x)]
 

Sympy [F]

\[ \int \frac {\left (a+b x^2\right )^2}{x^2 (c+d x)^{5/2}} \, dx=\int \frac {\left (a + b x^{2}\right )^{2}}{x^{2} \left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.43 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 (c+d x)^{5/2}} \, dx=\frac {1}{6} \, d {\left (\frac {2 \, {\left (2 \, b^{2} c^{6} + 4 \, a b c^{4} d^{2} + 2 \, a^{2} c^{2} d^{4} + 3 \, {\left (4 \, b^{2} c^{4} - 5 \, a^{2} d^{4}\right )} {\left (d x + c\right )}^{2} - 2 \, {\left (7 \, b^{2} c^{5} + 2 \, a b c^{3} d^{2} - 5 \, a^{2} c d^{4}\right )} {\left (d x + c\right )}\right )}}{{\left (d x + c\right )}^{\frac {5}{2}} c^{3} d^{4} - {\left (d x + c\right )}^{\frac {3}{2}} c^{4} d^{4}} - \frac {15 \, a^{2} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{c^{\frac {7}{2}}} + \frac {12 \, \sqrt {d x + c} b^{2}}{d^{4}}\right )} \] Input:

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

Output:

1/6*d*(2*(2*b^2*c^6 + 4*a*b*c^4*d^2 + 2*a^2*c^2*d^4 + 3*(4*b^2*c^4 - 5*a^2 
*d^4)*(d*x + c)^2 - 2*(7*b^2*c^5 + 2*a*b*c^3*d^2 - 5*a^2*c*d^4)*(d*x + c)) 
/((d*x + c)^(5/2)*c^3*d^4 - (d*x + c)^(3/2)*c^4*d^4) - 15*a^2*log((sqrt(d* 
x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c)))/c^(7/2) + 12*sqrt(d*x + c)*b^ 
2/d^4)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 (c+d x)^{5/2}} \, dx=-\frac {5 \, a^{2} d \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{3}} + \frac {2 \, \sqrt {d x + c} b^{2}}{d^{3}} - \frac {\sqrt {d x + c} a^{2}}{c^{3} x} + \frac {2 \, {\left (6 \, {\left (d x + c\right )} b^{2} c^{4} - b^{2} c^{5} - 2 \, a b c^{3} d^{2} - 6 \, {\left (d x + c\right )} a^{2} d^{4} - a^{2} c d^{4}\right )}}{3 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{3} d^{3}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.29 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 (c+d x)^{5/2}} \, dx=\frac {2\,b^2\,\sqrt {c+d\,x}}{d^3}-\frac {\frac {\left (5\,a^2\,d^4-4\,b^2\,c^4\right )\,{\left (c+d\,x\right )}^2}{c^3}-\frac {2\,\left (a^2\,d^4+2\,a\,b\,c^2\,d^2+b^2\,c^4\right )}{3\,c}+\frac {2\,\left (c+d\,x\right )\,\left (-5\,a^2\,d^4+2\,a\,b\,c^2\,d^2+7\,b^2\,c^4\right )}{3\,c^2}}{d^3\,{\left (c+d\,x\right )}^{5/2}-c\,d^3\,{\left (c+d\,x\right )}^{3/2}}+\frac {5\,a^2\,d\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}}{\sqrt {c}}\right )}{c^{7/2}} \] Input:

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

Output:

(2*b^2*(c + d*x)^(1/2))/d^3 - (((5*a^2*d^4 - 4*b^2*c^4)*(c + d*x)^2)/c^3 - 
 (2*(a^2*d^4 + b^2*c^4 + 2*a*b*c^2*d^2))/(3*c) + (2*(c + d*x)*(7*b^2*c^4 - 
 5*a^2*d^4 + 2*a*b*c^2*d^2))/(3*c^2))/(d^3*(c + d*x)^(5/2) - c*d^3*(c + d* 
x)^(3/2)) + (5*a^2*d*atanh((c + d*x)^(1/2)/c^(1/2)))/c^(7/2)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.79 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 (c+d x)^{5/2}} \, dx=\frac {-15 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a^{2} c \,d^{4} x -15 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a^{2} d^{5} x^{2}+15 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a^{2} c \,d^{4} x +15 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a^{2} d^{5} x^{2}-6 a^{2} c^{3} d^{3}-40 a^{2} c^{2} d^{4} x -30 a^{2} c \,d^{5} x^{2}-8 a b \,c^{4} d^{2} x +32 b^{2} c^{6} x +48 b^{2} c^{5} d \,x^{2}+12 b^{2} c^{4} d^{2} x^{3}}{6 \sqrt {d x +c}\, c^{4} d^{3} x \left (d x +c \right )} \] Input:

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

Output:

( - 15*sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*a**2*c*d**4*x - 
15*sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*a**2*d**5*x**2 + 15* 
sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) + sqrt(c))*a**2*c*d**4*x + 15*sqrt 
(c)*sqrt(c + d*x)*log(sqrt(c + d*x) + sqrt(c))*a**2*d**5*x**2 - 6*a**2*c** 
3*d**3 - 40*a**2*c**2*d**4*x - 30*a**2*c*d**5*x**2 - 8*a*b*c**4*d**2*x + 3 
2*b**2*c**6*x + 48*b**2*c**5*d*x**2 + 12*b**2*c**4*d**2*x**3)/(6*sqrt(c + 
d*x)*c**4*d**3*x*(c + d*x))