\(\int \frac {\sqrt {c+d x} (a+b x^2)^2}{x^3} \, dx\) [520]

Optimal result

Integrand size = 22, antiderivative size = 134 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )^2}{x^3} \, dx=4 a b \sqrt {c+d x}-\frac {a^2 \sqrt {c+d x}}{2 x^2}-\frac {a^2 d \sqrt {c+d x}}{4 c x}-\frac {2 b^2 c (c+d x)^{3/2}}{3 d^2}+\frac {2 b^2 (c+d x)^{5/2}}{5 d^2}-\frac {a \left (16 b c^2-a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 c^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )^2}{x^3} \, dx=\frac {\sqrt {c+d x} \left (240 a b c d^2 x^2-15 a^2 d^2 (2 c+d x)+8 b^2 c x^2 \left (-2 c^2+c d x+3 d^2 x^2\right )\right )}{60 c d^2 x^2}+\frac {a \left (-16 b c^2+a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 c^{3/2}} \] Input:

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

Output:

(Sqrt[c + d*x]*(240*a*b*c*d^2*x^2 - 15*a^2*d^2*(2*c + d*x) + 8*b^2*c*x^2*( 
-2*c^2 + c*d*x + 3*d^2*x^2)))/(60*c*d^2*x^2) + (a*(-16*b*c^2 + a*d^2)*ArcT 
anh[Sqrt[c + d*x]/Sqrt[c]])/(4*c^(3/2))
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.19, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {517, 25, 1580, 25, 2345, 25, 2341, 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.

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

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1580

\(\displaystyle \frac {2 \left (-\frac {1}{4} \int -\frac {a^2 d^4+4 b^2 (c+d x)^4-12 b^2 c (c+d x)^3+4 b \left (3 b c^2+2 a d^2\right ) (c+d x)^2-4 b c \left (b c^2+2 a d^2\right ) (c+d x)}{d^2 x^2}d\sqrt {c+d x}-\frac {a^2 d^2 \sqrt {c+d x}}{4 x^2}\right )}{d^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \left (\frac {1}{4} \int \frac {a^2 d^4+4 b^2 (c+d x)^4-12 b^2 c (c+d x)^3+4 b \left (3 b c^2+2 a d^2\right ) (c+d x)^2-4 b c \left (b c^2+2 a d^2\right ) (c+d x)}{d^2 x^2}d\sqrt {c+d x}-\frac {a^2 d^2 \sqrt {c+d x}}{4 x^2}\right )}{d^2}\)

\(\Big \downarrow \) 2345

\(\displaystyle \frac {2 \left (\frac {1}{4} \left (-\frac {\int \frac {a^2 d^4-8 b^2 c (c+d x)^3+16 b^2 c^2 (c+d x)^2-8 b c \left (b c^2+2 a d^2\right ) (c+d x)}{d x}d\sqrt {c+d x}}{2 c}-\frac {a^2 d^3 \sqrt {c+d x}}{2 c x}\right )-\frac {a^2 d^2 \sqrt {c+d x}}{4 x^2}\right )}{d^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \left (\frac {1}{4} \left (\frac {\int -\frac {a^2 d^4-8 b^2 c (c+d x)^3+16 b^2 c^2 (c+d x)^2-8 b c \left (b c^2+2 a d^2\right ) (c+d x)}{d x}d\sqrt {c+d x}}{2 c}-\frac {a^2 d^3 \sqrt {c+d x}}{2 c x}\right )-\frac {a^2 d^2 \sqrt {c+d x}}{4 x^2}\right )}{d^2}\)

\(\Big \downarrow \) 2341

\(\displaystyle \frac {2 \left (\frac {1}{4} \left (\frac {\int \left (-8 b^2 (c+d x) c^2+16 a b d^2 c+8 b^2 (c+d x)^2 c-\frac {a^2 d^4-16 a b c^2 d^2}{d x}\right )d\sqrt {c+d x}}{2 c}-\frac {a^2 d^3 \sqrt {c+d x}}{2 c x}\right )-\frac {a^2 d^2 \sqrt {c+d x}}{4 x^2}\right )}{d^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (\frac {1}{4} \left (\frac {-\frac {a d^2 \left (16 b c^2-a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\sqrt {c}}+16 a b c d^2 \sqrt {c+d x}-\frac {8}{3} b^2 c^2 (c+d x)^{3/2}+\frac {8}{5} b^2 c (c+d x)^{5/2}}{2 c}-\frac {a^2 d^3 \sqrt {c+d x}}{2 c x}\right )-\frac {a^2 d^2 \sqrt {c+d x}}{4 x^2}\right )}{d^2}\)

Input:

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

Output:

(2*(-1/4*(a^2*d^2*Sqrt[c + d*x])/x^2 + (-1/2*(a^2*d^3*Sqrt[c + d*x])/(c*x) 
 + (16*a*b*c*d^2*Sqrt[c + d*x] - (8*b^2*c^2*(c + d*x)^(3/2))/3 + (8*b^2*c* 
(c + d*x)^(5/2))/5 - (a*d^2*(16*b*c^2 - a*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[ 
c]])/Sqrt[c])/(2*c))/4))/d^2
 

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 1580

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[1/(2*e^(2*p + m/2)* 
(q + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2* 
e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4)^p - (-d)^(m/2 - 1)*(c*d^2 - b 
*d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2))], x], x], x] /; FreeQ[{a, b, c, d, e 
}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]
 

rule 2009

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

rule 2341

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

rule 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b 
*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.83

method	result	size
risch	\(-\frac {a^{2} \sqrt {d x +c}\, \left (d x +2 c \right )}{4 x^{2} c}+\frac {\frac {8 b^{2} c \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {8 b^{2} c^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+16 a b c \,d^{2} \sqrt {d x +c}+\frac {a \,d^{2} \left (a \,d^{2}-16 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\sqrt {c}}}{4 c \,d^{2}}\)	\(111\)
pseudoelliptic	\(-\frac {-a \,d^{2} x^{2} \left (a \,d^{2}-16 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )+\left (2 d^{2} \left (-\frac {4}{5} b^{2} x^{4}-8 a b \,x^{2}+a^{2}\right ) c^{\frac {3}{2}}+x \left (a^{2} d^{3} \sqrt {c}-\frac {8 c^{\frac {5}{2}} b^{2} d \,x^{2}}{15}+\frac {16 c^{\frac {7}{2}} b^{2} x}{15}\right )\right ) \sqrt {d x +c}}{4 c^{\frac {3}{2}} d^{2} x^{2}}\)	\(116\)
derivativedivides	\(\frac {\frac {2 b^{2} \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {2 b^{2} c \left (d x +c \right )^{\frac {3}{2}}}{3}+4 a b \,d^{2} \sqrt {d x +c}-2 a \,d^{2} \left (\frac {\frac {a \,d^{2} \left (d x +c \right )^{\frac {3}{2}}}{8 c}+\frac {a \,d^{2} \sqrt {d x +c}}{8}}{d^{2} x^{2}}-\frac {\left (a \,d^{2}-16 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{8 c^{\frac {3}{2}}}\right )}{d^{2}}\)	\(119\)
default	\(\frac {\frac {2 b^{2} \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {2 b^{2} c \left (d x +c \right )^{\frac {3}{2}}}{3}+4 a b \,d^{2} \sqrt {d x +c}-2 a \,d^{2} \left (\frac {\frac {a \,d^{2} \left (d x +c \right )^{\frac {3}{2}}}{8 c}+\frac {a \,d^{2} \sqrt {d x +c}}{8}}{d^{2} x^{2}}-\frac {\left (a \,d^{2}-16 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{8 c^{\frac {3}{2}}}\right )}{d^{2}}\)	\(119\)

Input:

int((d*x+c)^(1/2)*(b*x^2+a)^2/x^3,x,method=_RETURNVERBOSE)
 

Output:

-1/4*a^2*(d*x+c)^(1/2)*(d*x+2*c)/x^2/c+1/4/c/d^2*(8/5*b^2*c*(d*x+c)^(5/2)- 
8/3*b^2*c^2*(d*x+c)^(3/2)+16*a*b*c*d^2*(d*x+c)^(1/2)+a*d^2*(a*d^2-16*b*c^2 
)/c^(1/2)*arctanh((d*x+c)^(1/2)/c^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.07 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )^2}{x^3} \, dx=\left [-\frac {15 \, {\left (16 \, a b c^{2} d^{2} - a^{2} d^{4}\right )} \sqrt {c} x^{2} \log \left (\frac {d x + 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (24 \, b^{2} c^{2} d^{2} x^{4} + 8 \, b^{2} c^{3} d x^{3} - 15 \, a^{2} c d^{3} x - 30 \, a^{2} c^{2} d^{2} - 16 \, {\left (b^{2} c^{4} - 15 \, a b c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{120 \, c^{2} d^{2} x^{2}}, \frac {15 \, {\left (16 \, a b c^{2} d^{2} - a^{2} d^{4}\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) + {\left (24 \, b^{2} c^{2} d^{2} x^{4} + 8 \, b^{2} c^{3} d x^{3} - 15 \, a^{2} c d^{3} x - 30 \, a^{2} c^{2} d^{2} - 16 \, {\left (b^{2} c^{4} - 15 \, a b c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{60 \, c^{2} d^{2} x^{2}}\right ] \] Input:

integrate((d*x+c)^(1/2)*(b*x^2+a)^2/x^3,x, algorithm="fricas")
 

Output:

[-1/120*(15*(16*a*b*c^2*d^2 - a^2*d^4)*sqrt(c)*x^2*log((d*x + 2*sqrt(d*x + 
 c)*sqrt(c) + 2*c)/x) - 2*(24*b^2*c^2*d^2*x^4 + 8*b^2*c^3*d*x^3 - 15*a^2*c 
*d^3*x - 30*a^2*c^2*d^2 - 16*(b^2*c^4 - 15*a*b*c^2*d^2)*x^2)*sqrt(d*x + c) 
)/(c^2*d^2*x^2), 1/60*(15*(16*a*b*c^2*d^2 - a^2*d^4)*sqrt(-c)*x^2*arctan(s 
qrt(-c)/sqrt(d*x + c)) + (24*b^2*c^2*d^2*x^4 + 8*b^2*c^3*d*x^3 - 15*a^2*c* 
d^3*x - 30*a^2*c^2*d^2 - 16*(b^2*c^4 - 15*a*b*c^2*d^2)*x^2)*sqrt(d*x + c)) 
/(c^2*d^2*x^2)]
 

Sympy [A] (verification not implemented)

Time = 76.55 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.54 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )^2}{x^3} \, dx=- \frac {a^{2} c}{2 \sqrt {d} x^{\frac {5}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {3 a^{2} \sqrt {d}}{4 x^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {a^{2} d^{\frac {3}{2}}}{4 c \sqrt {x} \sqrt {\frac {c}{d x} + 1}} + \frac {a^{2} d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{4 c^{\frac {3}{2}}} + 2 a b \left (\begin {cases} \frac {2 c \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 \sqrt {c + d x} & \text {for}\: d \neq 0 \\\sqrt {c} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + b^{2} \left (\begin {cases} - \frac {2 c \left (c + d x\right )^{\frac {3}{2}}}{3 d^{2}} + \frac {2 \left (c + d x\right )^{\frac {5}{2}}}{5 d^{2}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{2}}{2} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((d*x+c)**(1/2)*(b*x**2+a)**2/x**3,x)
 

Output:

-a**2*c/(2*sqrt(d)*x**(5/2)*sqrt(c/(d*x) + 1)) - 3*a**2*sqrt(d)/(4*x**(3/2 
)*sqrt(c/(d*x) + 1)) - a**2*d**(3/2)/(4*c*sqrt(x)*sqrt(c/(d*x) + 1)) + a** 
2*d**2*asinh(sqrt(c)/(sqrt(d)*sqrt(x)))/(4*c**(3/2)) + 2*a*b*Piecewise((2* 
c*atan(sqrt(c + d*x)/sqrt(-c))/sqrt(-c) + 2*sqrt(c + d*x), Ne(d, 0)), (sqr 
t(c)*log(x), True)) + b**2*Piecewise((-2*c*(c + d*x)**(3/2)/(3*d**2) + 2*( 
c + d*x)**(5/2)/(5*d**2), Ne(d, 0)), (sqrt(c)*x**2/2, True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )^2}{x^3} \, dx=-\frac {1}{120} \, d^{2} {\left (\frac {30 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} a^{2} + \sqrt {d x + c} a^{2} c\right )}}{{\left (d x + c\right )}^{2} c - 2 \, {\left (d x + c\right )} c^{2} + c^{3}} - \frac {15 \, {\left (16 \, b c^{2} - a d^{2}\right )} a \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{c^{\frac {3}{2}} d^{2}} - \frac {16 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} - 5 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} c + 30 \, \sqrt {d x + c} a b d^{2}\right )}}{d^{4}}\right )} \] Input:

integrate((d*x+c)^(1/2)*(b*x^2+a)^2/x^3,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

-1/120*d^2*(30*((d*x + c)^(3/2)*a^2 + sqrt(d*x + c)*a^2*c)/((d*x + c)^2*c 
- 2*(d*x + c)*c^2 + c^3) - 15*(16*b*c^2 - a*d^2)*a*log((sqrt(d*x + c) - sq 
rt(c))/(sqrt(d*x + c) + sqrt(c)))/(c^(3/2)*d^2) - 16*(3*(d*x + c)^(5/2)*b^ 
2 - 5*(d*x + c)^(3/2)*b^2*c + 30*sqrt(d*x + c)*a*b*d^2)/d^4)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.04 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )^2}{x^3} \, dx=\frac {\frac {15 \, {\left (16 \, a b c^{2} d - a^{2} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c} + \frac {8 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d^{4} - 5 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} c d^{4} + 30 \, \sqrt {d x + c} a b d^{6}\right )}}{d^{5}} - \frac {15 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} a^{2} d^{3} + \sqrt {d x + c} a^{2} c d^{3}\right )}}{c d^{2} x^{2}}}{60 \, d} \] Input:

integrate((d*x+c)^(1/2)*(b*x^2+a)^2/x^3,x, algorithm="giac")
 

Output:

1/60*(15*(16*a*b*c^2*d - a^2*d^3)*arctan(sqrt(d*x + c)/sqrt(-c))/(sqrt(-c) 
*c) + 8*(3*(d*x + c)^(5/2)*b^2*d^4 - 5*(d*x + c)^(3/2)*b^2*c*d^4 + 30*sqrt 
(d*x + c)*a*b*d^6)/d^5 - 15*((d*x + c)^(3/2)*a^2*d^3 + sqrt(d*x + c)*a^2*c 
*d^3)/(c*d^2*x^2))/d
 

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )^2}{x^3} \, dx=\left (\frac {12\,b^2\,c^2+4\,a\,b\,d^2}{d^2}-\frac {12\,b^2\,c^2}{d^2}\right )\,\sqrt {c+d\,x}-\frac {\frac {a^2\,d^2\,\sqrt {c+d\,x}}{4}+\frac {a^2\,d^2\,{\left (c+d\,x\right )}^{3/2}}{4\,c}}{{\left (c+d\,x\right )}^2-2\,c\,\left (c+d\,x\right )+c^2}+\frac {2\,b^2\,{\left (c+d\,x\right )}^{5/2}}{5\,d^2}-\frac {2\,b^2\,c\,{\left (c+d\,x\right )}^{3/2}}{3\,d^2}-\frac {a\,\mathrm {atan}\left (\frac {\sqrt {c+d\,x}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,\left (a\,d^2-16\,b\,c^2\right )\,1{}\mathrm {i}}{4\,c^{3/2}} \] Input:

int(((a + b*x^2)^2*(c + d*x)^(1/2))/x^3,x)
 

Output:

((12*b^2*c^2 + 4*a*b*d^2)/d^2 - (12*b^2*c^2)/d^2)*(c + d*x)^(1/2) - ((a^2* 
d^2*(c + d*x)^(1/2))/4 + (a^2*d^2*(c + d*x)^(3/2))/(4*c))/((c + d*x)^2 - 2 
*c*(c + d*x) + c^2) + (2*b^2*(c + d*x)^(5/2))/(5*d^2) - (2*b^2*c*(c + d*x) 
^(3/2))/(3*d^2) - (a*atan(((c + d*x)^(1/2)*1i)/c^(1/2))*(a*d^2 - 16*b*c^2) 
*1i)/(4*c^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.63 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )^2}{x^3} \, dx=\frac {-60 \sqrt {d x +c}\, a^{2} c^{2} d^{2}-30 \sqrt {d x +c}\, a^{2} c \,d^{3} x +480 \sqrt {d x +c}\, a b \,c^{2} d^{2} x^{2}-32 \sqrt {d x +c}\, b^{2} c^{4} x^{2}+16 \sqrt {d x +c}\, b^{2} c^{3} d \,x^{3}+48 \sqrt {d x +c}\, b^{2} c^{2} d^{2} x^{4}-15 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a^{2} d^{4} x^{2}+240 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a b \,c^{2} d^{2} x^{2}+15 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a^{2} d^{4} x^{2}-240 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a b \,c^{2} d^{2} x^{2}}{120 c^{2} d^{2} x^{2}} \] Input:

int((d*x+c)^(1/2)*(b*x^2+a)^2/x^3,x)
 

Output:

( - 60*sqrt(c + d*x)*a**2*c**2*d**2 - 30*sqrt(c + d*x)*a**2*c*d**3*x + 480 
*sqrt(c + d*x)*a*b*c**2*d**2*x**2 - 32*sqrt(c + d*x)*b**2*c**4*x**2 + 16*s 
qrt(c + d*x)*b**2*c**3*d*x**3 + 48*sqrt(c + d*x)*b**2*c**2*d**2*x**4 - 15* 
sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*a**2*d**4*x**2 + 240*sqrt(c)*log(sqrt 
(c + d*x) - sqrt(c))*a*b*c**2*d**2*x**2 + 15*sqrt(c)*log(sqrt(c + d*x) + s 
qrt(c))*a**2*d**4*x**2 - 240*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a*b*c**2 
*d**2*x**2)/(120*c**2*d**2*x**2)