\(\int \frac {(a+b x^2)^2}{x^5 \sqrt {c+d x}} \, dx\) [546]

Optimal result

Integrand size = 22, antiderivative size = 165 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \sqrt {c+d x}} \, dx=-\frac {a^2 \sqrt {c+d x}}{4 c x^4}+\frac {7 a^2 d \sqrt {c+d x}}{24 c^2 x^3}-\frac {a \left (96 b c^2+35 a d^2\right ) \sqrt {c+d x}}{96 c^3 x^2}+\frac {a d \left (96 b c^2+35 a d^2\right ) \sqrt {c+d x}}{64 c^4 x}-\frac {\left (128 b^2 c^4+96 a b c^2 d^2+35 a^2 d^4\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{64 c^{9/2}} \] Output:

-1/4*a^2*(d*x+c)^(1/2)/c/x^4+7/24*a^2*d*(d*x+c)^(1/2)/c^2/x^3-1/96*a*(35*a 
*d^2+96*b*c^2)*(d*x+c)^(1/2)/c^3/x^2+1/64*a*d*(35*a*d^2+96*b*c^2)*(d*x+c)^ 
(1/2)/c^4/x-1/64*(35*a^2*d^4+96*a*b*c^2*d^2+128*b^2*c^4)*arctanh((d*x+c)^( 
1/2)/c^(1/2))/c^(9/2)
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.75 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \sqrt {c+d x}} \, dx=\frac {\frac {a \sqrt {c} \sqrt {c+d x} \left (96 b c^2 x^2 (-2 c+3 d x)+a \left (-48 c^3+56 c^2 d x-70 c d^2 x^2+105 d^3 x^3\right )\right )}{x^4}-3 \left (128 b^2 c^4+96 a b c^2 d^2+35 a^2 d^4\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{192 c^{9/2}} \] Input:

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

Output:

((a*Sqrt[c]*Sqrt[c + d*x]*(96*b*c^2*x^2*(-2*c + 3*d*x) + a*(-48*c^3 + 56*c 
^2*d*x - 70*c*d^2*x^2 + 105*d^3*x^3)))/x^4 - 3*(128*b^2*c^4 + 96*a*b*c^2*d 
^2 + 35*a^2*d^4)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(192*c^(9/2))
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.16, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {517, 25, 1471, 25, 2345, 25, 1471, 27, 298, 219}

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.

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

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1471

\(\displaystyle 2 \left (\frac {\int -\frac {8 b^2 c^4+16 a b d^2 c^2+24 b^2 (c+d x)^2 c^2-8 b^2 (c+d x)^3 c-8 b \left (3 b c^2+2 a d^2\right ) (c+d x) c+7 a^2 d^4}{d^4 x^4}d\sqrt {c+d x}}{8 c}-\frac {a^2 \sqrt {c+d x}}{8 c x^4}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 2 \left (-\frac {\int \frac {8 b^2 c^4+16 a b d^2 c^2+24 b^2 (c+d x)^2 c^2-8 b^2 (c+d x)^3 c-8 b \left (3 b c^2+2 a d^2\right ) (c+d x) c+7 a^2 d^4}{d^4 x^4}d\sqrt {c+d x}}{8 c}-\frac {a^2 \sqrt {c+d x}}{8 c x^4}\right )\)

\(\Big \downarrow \) 2345

\(\displaystyle 2 \left (-\frac {-\frac {\int \frac {48 b^2 c^4-96 b^2 (c+d x) c^3+96 a b d^2 c^2+48 b^2 (c+d x)^2 c^2+35 a^2 d^4}{d^3 x^3}d\sqrt {c+d x}}{6 c}-\frac {7 a^2 d \sqrt {c+d x}}{6 c x^3}}{8 c}-\frac {a^2 \sqrt {c+d x}}{8 c x^4}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 2 \left (-\frac {\frac {\int -\frac {48 b^2 c^4-96 b^2 (c+d x) c^3+96 a b d^2 c^2+48 b^2 (c+d x)^2 c^2+35 a^2 d^4}{d^3 x^3}d\sqrt {c+d x}}{6 c}-\frac {7 a^2 d \sqrt {c+d x}}{6 c x^3}}{8 c}-\frac {a^2 \sqrt {c+d x}}{8 c x^4}\right )\)

\(\Big \downarrow \) 1471

\(\displaystyle 2 \left (-\frac {\frac {\frac {a \sqrt {c+d x} \left (35 a d^2+96 b c^2\right )}{4 c x^2}-\frac {\int -\frac {3 \left (\left (8 b c^2+5 a d^2\right ) \left (8 b c^2+7 a d^2\right )-64 b^2 c^3 (c+d x)\right )}{d^2 x^2}d\sqrt {c+d x}}{4 c}}{6 c}-\frac {7 a^2 d \sqrt {c+d x}}{6 c x^3}}{8 c}-\frac {a^2 \sqrt {c+d x}}{8 c x^4}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \left (-\frac {\frac {\frac {3 \int \frac {\left (8 b c^2+5 a d^2\right ) \left (8 b c^2+7 a d^2\right )-64 b^2 c^3 (c+d x)}{d^2 x^2}d\sqrt {c+d x}}{4 c}+\frac {a \sqrt {c+d x} \left (35 a d^2+96 b c^2\right )}{4 c x^2}}{6 c}-\frac {7 a^2 d \sqrt {c+d x}}{6 c x^3}}{8 c}-\frac {a^2 \sqrt {c+d x}}{8 c x^4}\right )\)

\(\Big \downarrow \) 298

\(\displaystyle 2 \left (-\frac {\frac {\frac {3 \left (\frac {\left (35 a^2 d^4+96 a b c^2 d^2+128 b^2 c^4\right ) \int -\frac {1}{d x}d\sqrt {c+d x}}{2 c}-\frac {a d \sqrt {c+d x} \left (35 a d^2+96 b c^2\right )}{2 c x}\right )}{4 c}+\frac {a \sqrt {c+d x} \left (35 a d^2+96 b c^2\right )}{4 c x^2}}{6 c}-\frac {7 a^2 d \sqrt {c+d x}}{6 c x^3}}{8 c}-\frac {a^2 \sqrt {c+d x}}{8 c x^4}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle 2 \left (-\frac {\frac {\frac {3 \left (\frac {\left (35 a^2 d^4+96 a b c^2 d^2+128 b^2 c^4\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{2 c^{3/2}}-\frac {a d \sqrt {c+d x} \left (35 a d^2+96 b c^2\right )}{2 c x}\right )}{4 c}+\frac {a \sqrt {c+d x} \left (35 a d^2+96 b c^2\right )}{4 c x^2}}{6 c}-\frac {7 a^2 d \sqrt {c+d x}}{6 c x^3}}{8 c}-\frac {a^2 \sqrt {c+d x}}{8 c x^4}\right )\)

Input:

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

Output:

2*(-1/8*(a^2*Sqrt[c + d*x])/(c*x^4) - ((-7*a^2*d*Sqrt[c + d*x])/(6*c*x^3) 
+ ((a*(96*b*c^2 + 35*a*d^2)*Sqrt[c + d*x])/(4*c*x^2) + (3*(-1/2*(a*d*(96*b 
*c^2 + 35*a*d^2)*Sqrt[c + d*x])/(c*x) + ((128*b^2*c^4 + 96*a*b*c^2*d^2 + 3 
5*a^2*d^4)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(2*c^(3/2))))/(4*c))/(6*c))/(8* 
c))
 

Defintions of rubi rules used

rule 25

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

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

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 1471

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 
, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x 
, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q 
 + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), 
x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 
2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
 

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 = 0.36 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.68

method	result	size
pseudoelliptic	\(\frac {-\frac {35 x^{4} \left (a^{2} d^{4}+\frac {96}{35} b \,c^{2} d^{2} a +\frac {128}{35} b^{2} c^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{64}+\frac {7 \left (d x \left (\frac {36 b \,x^{2}}{7}+a \right ) c^{\frac {5}{2}}+\frac {6 \left (-4 b \,x^{2}-a \right ) c^{\frac {7}{2}}}{7}+\frac {15 d^{2} x^{2} \left (d x \sqrt {c}-\frac {2 c^{\frac {3}{2}}}{3}\right ) a}{8}\right ) \sqrt {d x +c}\, a}{24}}{c^{\frac {9}{2}} x^{4}}\)	\(113\)
risch	\(-\frac {\sqrt {d x +c}\, a \left (-105 a \,x^{3} d^{3}-288 b \,c^{2} d \,x^{3}+70 a \,d^{2} x^{2} c +192 b \,c^{3} x^{2}-56 a d x \,c^{2}+48 c^{3} a \right )}{192 c^{4} x^{4}}-\frac {\left (35 a^{2} d^{4}+96 b \,c^{2} d^{2} a +128 b^{2} c^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{64 c^{\frac {9}{2}}}\)	\(115\)
derivativedivides	\(-\frac {2 \left (-\frac {a \,d^{2} \left (35 a \,d^{2}+96 b \,c^{2}\right ) \left (d x +c \right )^{\frac {7}{2}}}{128 c^{4}}+\frac {11 a \,d^{2} \left (35 a \,d^{2}+96 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{384 c^{3}}-\frac {a \,d^{2} \left (511 a \,d^{2}+1248 b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{384 c^{2}}+\frac {a \,d^{2} \left (93 a \,d^{2}+160 b \,c^{2}\right ) \sqrt {d x +c}}{128 c}\right )}{d^{4} x^{4}}-\frac {\left (35 a^{2} d^{4}+96 b \,c^{2} d^{2} a +128 b^{2} c^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{64 c^{\frac {9}{2}}}\)	\(171\)
default	\(-\frac {2 \left (-\frac {a \,d^{2} \left (35 a \,d^{2}+96 b \,c^{2}\right ) \left (d x +c \right )^{\frac {7}{2}}}{128 c^{4}}+\frac {11 a \,d^{2} \left (35 a \,d^{2}+96 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{384 c^{3}}-\frac {a \,d^{2} \left (511 a \,d^{2}+1248 b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{384 c^{2}}+\frac {a \,d^{2} \left (93 a \,d^{2}+160 b \,c^{2}\right ) \sqrt {d x +c}}{128 c}\right )}{d^{4} x^{4}}-\frac {\left (35 a^{2} d^{4}+96 b \,c^{2} d^{2} a +128 b^{2} c^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{64 c^{\frac {9}{2}}}\)	\(171\)

Input:

int((b*x^2+a)^2/x^5/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

7/24/c^(9/2)*(-15/8*x^4*(a^2*d^4+96/35*b*c^2*d^2*a+128/35*b^2*c^4)*arctanh 
((d*x+c)^(1/2)/c^(1/2))+(d*x*(36/7*b*x^2+a)*c^(5/2)+6/7*(-4*b*x^2-a)*c^(7/ 
2)+15/8*d^2*x^2*(d*x*c^(1/2)-2/3*c^(3/2))*a)*(d*x+c)^(1/2)*a)/x^4
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.68 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \sqrt {c+d x}} \, dx=\left [\frac {3 \, {\left (128 \, b^{2} c^{4} + 96 \, a b c^{2} d^{2} + 35 \, a^{2} d^{4}\right )} \sqrt {c} x^{4} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (56 \, a^{2} c^{3} d x - 48 \, a^{2} c^{4} + 3 \, {\left (96 \, a b c^{3} d + 35 \, a^{2} c d^{3}\right )} x^{3} - 2 \, {\left (96 \, a b c^{4} + 35 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{384 \, c^{5} x^{4}}, \frac {3 \, {\left (128 \, b^{2} c^{4} + 96 \, a b c^{2} d^{2} + 35 \, a^{2} d^{4}\right )} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) + {\left (56 \, a^{2} c^{3} d x - 48 \, a^{2} c^{4} + 3 \, {\left (96 \, a b c^{3} d + 35 \, a^{2} c d^{3}\right )} x^{3} - 2 \, {\left (96 \, a b c^{4} + 35 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{192 \, c^{5} x^{4}}\right ] \] Input:

integrate((b*x^2+a)^2/x^5/(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

[1/384*(3*(128*b^2*c^4 + 96*a*b*c^2*d^2 + 35*a^2*d^4)*sqrt(c)*x^4*log((d*x 
 - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*(56*a^2*c^3*d*x - 48*a^2*c^4 + 3* 
(96*a*b*c^3*d + 35*a^2*c*d^3)*x^3 - 2*(96*a*b*c^4 + 35*a^2*c^2*d^2)*x^2)*s 
qrt(d*x + c))/(c^5*x^4), 1/192*(3*(128*b^2*c^4 + 96*a*b*c^2*d^2 + 35*a^2*d 
^4)*sqrt(-c)*x^4*arctan(sqrt(-c)/sqrt(d*x + c)) + (56*a^2*c^3*d*x - 48*a^2 
*c^4 + 3*(96*a*b*c^3*d + 35*a^2*c*d^3)*x^3 - 2*(96*a*b*c^4 + 35*a^2*c^2*d^ 
2)*x^2)*sqrt(d*x + c))/(c^5*x^4)]
 

Sympy [A] (verification not implemented)

Time = 160.38 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.98 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \sqrt {c+d x}} \, dx=- \frac {a^{2}}{4 \sqrt {d} x^{\frac {9}{2}} \sqrt {\frac {c}{d x} + 1}} + \frac {a^{2} \sqrt {d}}{24 c x^{\frac {7}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {7 a^{2} d^{\frac {3}{2}}}{96 c^{2} x^{\frac {5}{2}} \sqrt {\frac {c}{d x} + 1}} + \frac {35 a^{2} d^{\frac {5}{2}}}{192 c^{3} x^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}} + \frac {35 a^{2} d^{\frac {7}{2}}}{64 c^{4} \sqrt {x} \sqrt {\frac {c}{d x} + 1}} - \frac {35 a^{2} d^{4} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{64 c^{\frac {9}{2}}} - \frac {a b}{\sqrt {d} x^{\frac {5}{2}} \sqrt {\frac {c}{d x} + 1}} + \frac {a b \sqrt {d}}{2 c x^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}} + \frac {3 a b d^{\frac {3}{2}}}{2 c^{2} \sqrt {x} \sqrt {\frac {c}{d x} + 1}} - \frac {3 a b d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{2 c^{\frac {5}{2}}} + b^{2} \left (\begin {cases} \frac {2 \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{\sqrt {- c}} & \text {for}\: d \neq 0 \\\frac {\log {\left (x \right )}}{\sqrt {c}} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((b*x**2+a)**2/x**5/(d*x+c)**(1/2),x)
 

Output:

-a**2/(4*sqrt(d)*x**(9/2)*sqrt(c/(d*x) + 1)) + a**2*sqrt(d)/(24*c*x**(7/2) 
*sqrt(c/(d*x) + 1)) - 7*a**2*d**(3/2)/(96*c**2*x**(5/2)*sqrt(c/(d*x) + 1)) 
 + 35*a**2*d**(5/2)/(192*c**3*x**(3/2)*sqrt(c/(d*x) + 1)) + 35*a**2*d**(7/ 
2)/(64*c**4*sqrt(x)*sqrt(c/(d*x) + 1)) - 35*a**2*d**4*asinh(sqrt(c)/(sqrt( 
d)*sqrt(x)))/(64*c**(9/2)) - a*b/(sqrt(d)*x**(5/2)*sqrt(c/(d*x) + 1)) + a* 
b*sqrt(d)/(2*c*x**(3/2)*sqrt(c/(d*x) + 1)) + 3*a*b*d**(3/2)/(2*c**2*sqrt(x 
)*sqrt(c/(d*x) + 1)) - 3*a*b*d**2*asinh(sqrt(c)/(sqrt(d)*sqrt(x)))/(2*c**( 
5/2)) + b**2*Piecewise((2*atan(sqrt(c + d*x)/sqrt(-c))/sqrt(-c), Ne(d, 0)) 
, (log(x)/sqrt(c), True))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.48 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \sqrt {c+d x}} \, dx=\frac {1}{384} \, d^{4} {\left (\frac {2 \, {\left (3 \, {\left (96 \, a b c^{2} + 35 \, a^{2} d^{2}\right )} {\left (d x + c\right )}^{\frac {7}{2}} - 11 \, {\left (96 \, a b c^{3} + 35 \, a^{2} c d^{2}\right )} {\left (d x + c\right )}^{\frac {5}{2}} + {\left (1248 \, a b c^{4} + 511 \, a^{2} c^{2} d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} - 3 \, {\left (160 \, a b c^{5} + 93 \, a^{2} c^{3} d^{2}\right )} \sqrt {d x + c}\right )}}{{\left (d x + c\right )}^{4} c^{4} d^{2} - 4 \, {\left (d x + c\right )}^{3} c^{5} d^{2} + 6 \, {\left (d x + c\right )}^{2} c^{6} d^{2} - 4 \, {\left (d x + c\right )} c^{7} d^{2} + c^{8} d^{2}} + \frac {3 \, {\left (128 \, b^{2} c^{4} + 96 \, a b c^{2} d^{2} + 35 \, a^{2} d^{4}\right )} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{c^{\frac {9}{2}} d^{4}}\right )} \] Input:

integrate((b*x^2+a)^2/x^5/(d*x+c)^(1/2),x, algorithm="maxima")
 

Output:

1/384*d^4*(2*(3*(96*a*b*c^2 + 35*a^2*d^2)*(d*x + c)^(7/2) - 11*(96*a*b*c^3 
 + 35*a^2*c*d^2)*(d*x + c)^(5/2) + (1248*a*b*c^4 + 511*a^2*c^2*d^2)*(d*x + 
 c)^(3/2) - 3*(160*a*b*c^5 + 93*a^2*c^3*d^2)*sqrt(d*x + c))/((d*x + c)^4*c 
^4*d^2 - 4*(d*x + c)^3*c^5*d^2 + 6*(d*x + c)^2*c^6*d^2 - 4*(d*x + c)*c^7*d 
^2 + c^8*d^2) + 3*(128*b^2*c^4 + 96*a*b*c^2*d^2 + 35*a^2*d^4)*log((sqrt(d* 
x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c)))/(c^(9/2)*d^4))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \sqrt {c+d x}} \, dx=\frac {\frac {3 \, {\left (128 \, b^{2} c^{4} d + 96 \, a b c^{2} d^{3} + 35 \, a^{2} d^{5}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{4}} + \frac {288 \, {\left (d x + c\right )}^{\frac {7}{2}} a b c^{2} d^{3} - 1056 \, {\left (d x + c\right )}^{\frac {5}{2}} a b c^{3} d^{3} + 1248 \, {\left (d x + c\right )}^{\frac {3}{2}} a b c^{4} d^{3} - 480 \, \sqrt {d x + c} a b c^{5} d^{3} + 105 \, {\left (d x + c\right )}^{\frac {7}{2}} a^{2} d^{5} - 385 \, {\left (d x + c\right )}^{\frac {5}{2}} a^{2} c d^{5} + 511 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} c^{2} d^{5} - 279 \, \sqrt {d x + c} a^{2} c^{3} d^{5}}{c^{4} d^{4} x^{4}}}{192 \, d} \] Input:

integrate((b*x^2+a)^2/x^5/(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

1/192*(3*(128*b^2*c^4*d + 96*a*b*c^2*d^3 + 35*a^2*d^5)*arctan(sqrt(d*x + c 
)/sqrt(-c))/(sqrt(-c)*c^4) + (288*(d*x + c)^(7/2)*a*b*c^2*d^3 - 1056*(d*x 
+ c)^(5/2)*a*b*c^3*d^3 + 1248*(d*x + c)^(3/2)*a*b*c^4*d^3 - 480*sqrt(d*x + 
 c)*a*b*c^5*d^3 + 105*(d*x + c)^(7/2)*a^2*d^5 - 385*(d*x + c)^(5/2)*a^2*c* 
d^5 + 511*(d*x + c)^(3/2)*a^2*c^2*d^5 - 279*sqrt(d*x + c)*a^2*c^3*d^5)/(c^ 
4*d^4*x^4))/d
 

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.32 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \sqrt {c+d x}} \, dx=-\frac {\frac {11\,\left (35\,a^2\,d^4+96\,b\,a\,c^2\,d^2\right )\,{\left (c+d\,x\right )}^{5/2}}{192\,c^3}-\frac {\left (35\,a^2\,d^4+96\,b\,a\,c^2\,d^2\right )\,{\left (c+d\,x\right )}^{7/2}}{64\,c^4}+\frac {\left (93\,a^2\,d^4+160\,b\,a\,c^2\,d^2\right )\,\sqrt {c+d\,x}}{64\,c}-\frac {\left (511\,a^2\,d^4+1248\,b\,a\,c^2\,d^2\right )\,{\left (c+d\,x\right )}^{3/2}}{192\,c^2}}{{\left (c+d\,x\right )}^4-4\,c^3\,\left (c+d\,x\right )-4\,c\,{\left (c+d\,x\right )}^3+6\,c^2\,{\left (c+d\,x\right )}^2+c^4}-\frac {\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}}{\sqrt {c}}\right )\,\left (35\,a^2\,d^4+96\,a\,b\,c^2\,d^2+128\,b^2\,c^4\right )}{64\,c^{9/2}} \] Input:

int((a + b*x^2)^2/(x^5*(c + d*x)^(1/2)),x)
 

Output:

- ((11*(35*a^2*d^4 + 96*a*b*c^2*d^2)*(c + d*x)^(5/2))/(192*c^3) - ((35*a^2 
*d^4 + 96*a*b*c^2*d^2)*(c + d*x)^(7/2))/(64*c^4) + ((93*a^2*d^4 + 160*a*b* 
c^2*d^2)*(c + d*x)^(1/2))/(64*c) - ((511*a^2*d^4 + 1248*a*b*c^2*d^2)*(c + 
d*x)^(3/2))/(192*c^2))/((c + d*x)^4 - 4*c^3*(c + d*x) - 4*c*(c + d*x)^3 + 
6*c^2*(c + d*x)^2 + c^4) - (atanh((c + d*x)^(1/2)/c^(1/2))*(35*a^2*d^4 + 1 
28*b^2*c^4 + 96*a*b*c^2*d^2))/(64*c^(9/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.56 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \sqrt {c+d x}} \, dx=\frac {-96 \sqrt {d x +c}\, a^{2} c^{4}+112 \sqrt {d x +c}\, a^{2} c^{3} d x -140 \sqrt {d x +c}\, a^{2} c^{2} d^{2} x^{2}+210 \sqrt {d x +c}\, a^{2} c \,d^{3} x^{3}-384 \sqrt {d x +c}\, a b \,c^{4} x^{2}+576 \sqrt {d x +c}\, a b \,c^{3} d \,x^{3}+105 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a^{2} d^{4} x^{4}+288 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a b \,c^{2} d^{2} x^{4}+384 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) b^{2} c^{4} x^{4}-105 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a^{2} d^{4} x^{4}-288 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a b \,c^{2} d^{2} x^{4}-384 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) b^{2} c^{4} x^{4}}{384 c^{5} x^{4}} \] Input:

int((b*x^2+a)^2/x^5/(d*x+c)^(1/2),x)
 

Output:

( - 96*sqrt(c + d*x)*a**2*c**4 + 112*sqrt(c + d*x)*a**2*c**3*d*x - 140*sqr 
t(c + d*x)*a**2*c**2*d**2*x**2 + 210*sqrt(c + d*x)*a**2*c*d**3*x**3 - 384* 
sqrt(c + d*x)*a*b*c**4*x**2 + 576*sqrt(c + d*x)*a*b*c**3*d*x**3 + 105*sqrt 
(c)*log(sqrt(c + d*x) - sqrt(c))*a**2*d**4*x**4 + 288*sqrt(c)*log(sqrt(c + 
 d*x) - sqrt(c))*a*b*c**2*d**2*x**4 + 384*sqrt(c)*log(sqrt(c + d*x) - sqrt 
(c))*b**2*c**4*x**4 - 105*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a**2*d**4*x 
**4 - 288*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a*b*c**2*d**2*x**4 - 384*sq 
rt(c)*log(sqrt(c + d*x) + sqrt(c))*b**2*c**4*x**4)/(384*c**5*x**4)