\(\int \frac {(c+d x)^{3/2} (a+b x^2)^2}{x^5} \, dx\) [530]

Optimal result

Integrand size = 22, antiderivative size = 190 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )^2}{x^5} \, dx=2 b^2 c \sqrt {c+d x}-\frac {a^2 c \sqrt {c+d x}}{4 x^4}-\frac {3 a^2 d \sqrt {c+d x}}{8 x^3}-\frac {a \left (32 b c^2+a d^2\right ) \sqrt {c+d x}}{32 c x^2}-\frac {a d \left (160 b c^2-3 a d^2\right ) \sqrt {c+d x}}{64 c^2 x}+\frac {2}{3} b^2 (c+d x)^{3/2}-\frac {\left (128 b^2 c^4+96 a b c^2 d^2+3 a^2 d^4\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{64 c^{5/2}} \] Output:

2*b^2*c*(d*x+c)^(1/2)-1/4*a^2*c*(d*x+c)^(1/2)/x^4-3/8*a^2*d*(d*x+c)^(1/2)/ 
x^3-1/32*a*(a*d^2+32*b*c^2)*(d*x+c)^(1/2)/c/x^2-1/64*a*d*(-3*a*d^2+160*b*c 
^2)*(d*x+c)^(1/2)/c^2/x+2/3*b^2*(d*x+c)^(3/2)-1/64*(3*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^(5/2)
 

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.76 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )^2}{x^5} \, dx=\frac {\frac {\sqrt {c} \sqrt {c+d x} \left (128 b^2 c^2 x^4 (4 c+d x)-96 a b c^2 x^2 (2 c+5 d x)-3 a^2 \left (16 c^3+24 c^2 d x+2 c d^2 x^2-3 d^3 x^3\right )\right )}{x^4}-3 \left (128 b^2 c^4+96 a b c^2 d^2+3 a^2 d^4\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{192 c^{5/2}} \] Input:

Integrate[((c + d*x)^(3/2)*(a + b*x^2)^2)/x^5,x]
 

Output:

((Sqrt[c]*Sqrt[c + d*x]*(128*b^2*c^2*x^4*(4*c + d*x) - 96*a*b*c^2*x^2*(2*c 
 + 5*d*x) - 3*a^2*(16*c^3 + 24*c^2*d*x + 2*c*d^2*x^2 - 3*d^3*x^3)))/x^4 - 
3*(128*b^2*c^4 + 96*a*b*c^2*d^2 + 3*a^2*d^4)*ArcTanh[Sqrt[c + d*x]/Sqrt[c] 
])/(192*c^(5/2))
 

Rubi [A] (verified)

Time = 1.24 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.13, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {517, 25, 1580, 2345, 27, 2345, 2345, 27, 1468, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^2 (c+d x)^{3/2}}{x^5} \, dx\)

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1580

\(\displaystyle 2 \left (\frac {1}{8} \int \frac {8 b^2 (c+d x)^5-24 b^2 c (c+d x)^4+8 b \left (3 b c^2+2 a d^2\right ) (c+d x)^3-8 b c \left (b c^2+2 a d^2\right ) (c+d x)^2+8 a^2 d^4 (c+d x)+a^2 c d^4}{d^4 x^4}d\sqrt {c+d x}-\frac {a^2 c \sqrt {c+d x}}{8 x^4}\right )\)

\(\Big \downarrow \) 2345

\(\displaystyle 2 \left (\frac {1}{8} \left (-\frac {\int -\frac {3 \left (a^2 c d^4+16 b^2 c (c+d x)^4-32 b^2 c^2 (c+d x)^3+16 b c \left (b c^2+2 a d^2\right ) (c+d x)^2\right )}{d^3 x^3}d\sqrt {c+d x}}{6 c}-\frac {3 a^2 d \sqrt {c+d x}}{2 x^3}\right )-\frac {a^2 c \sqrt {c+d x}}{8 x^4}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \left (\frac {1}{8} \left (-\frac {\int -\frac {a^2 c d^4+16 b^2 c (c+d x)^4-32 b^2 c^2 (c+d x)^3+16 b c \left (b c^2+2 a d^2\right ) (c+d x)^2}{d^3 x^3}d\sqrt {c+d x}}{2 c}-\frac {3 a^2 d \sqrt {c+d x}}{2 x^3}\right )-\frac {a^2 c \sqrt {c+d x}}{8 x^4}\right )\)

\(\Big \downarrow \) 2345

\(\displaystyle 2 \left (\frac {1}{8} \left (-\frac {\frac {a \sqrt {c+d x} \left (a d^2+32 b c^2\right )}{4 x^2}-\frac {\int \frac {-64 b^2 (c+d x)^2 c^3+64 b^2 (c+d x)^3 c^2+128 a b d^2 (c+d x) c^2+a d^2 \left (32 b c^2-3 a d^2\right ) c}{d^2 x^2}d\sqrt {c+d x}}{4 c}}{2 c}-\frac {3 a^2 d \sqrt {c+d x}}{2 x^3}\right )-\frac {a^2 c \sqrt {c+d x}}{8 x^4}\right )\)

\(\Big \downarrow \) 2345

\(\displaystyle 2 \left (\frac {1}{8} \left (-\frac {\frac {a \sqrt {c+d x} \left (a d^2+32 b c^2\right )}{4 x^2}-\frac {-\frac {\int -\frac {c \left (3 a \left (32 b c^2+a d^2\right ) d^2+128 b^2 c^2 (c+d x)^2\right )}{d x}d\sqrt {c+d x}}{2 c}-\frac {a d \sqrt {c+d x} \left (160 b c^2-3 a d^2\right )}{2 x}}{4 c}}{2 c}-\frac {3 a^2 d \sqrt {c+d x}}{2 x^3}\right )-\frac {a^2 c \sqrt {c+d x}}{8 x^4}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \left (\frac {1}{8} \left (-\frac {\frac {a \sqrt {c+d x} \left (a d^2+32 b c^2\right )}{4 x^2}-\frac {-\frac {1}{2} \int -\frac {3 a \left (32 b c^2+a d^2\right ) d^2+128 b^2 c^2 (c+d x)^2}{d x}d\sqrt {c+d x}-\frac {a d \sqrt {c+d x} \left (160 b c^2-3 a d^2\right )}{2 x}}{4 c}}{2 c}-\frac {3 a^2 d \sqrt {c+d x}}{2 x^3}\right )-\frac {a^2 c \sqrt {c+d x}}{8 x^4}\right )\)

\(\Big \downarrow \) 1468

\(\displaystyle 2 \left (\frac {1}{8} \left (-\frac {\frac {a \sqrt {c+d x} \left (a d^2+32 b c^2\right )}{4 x^2}-\frac {-\frac {1}{2} \int \left (-128 b^2 c^3-128 b^2 (c+d x) c^2-\frac {128 b^2 c^4+96 a b d^2 c^2+3 a^2 d^4}{d x}\right )d\sqrt {c+d x}-\frac {a d \sqrt {c+d x} \left (160 b c^2-3 a d^2\right )}{2 x}}{4 c}}{2 c}-\frac {3 a^2 d \sqrt {c+d x}}{2 x^3}\right )-\frac {a^2 c \sqrt {c+d x}}{8 x^4}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (\frac {1}{8} \left (-\frac {\frac {a \sqrt {c+d x} \left (a d^2+32 b c^2\right )}{4 x^2}-\frac {\frac {1}{2} \left (-\frac {\left (3 a^2 d^4+96 a b c^2 d^2+128 b^2 c^4\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\sqrt {c}}+128 b^2 c^3 \sqrt {c+d x}+\frac {128}{3} b^2 c^2 (c+d x)^{3/2}\right )-\frac {a d \sqrt {c+d x} \left (160 b c^2-3 a d^2\right )}{2 x}}{4 c}}{2 c}-\frac {3 a^2 d \sqrt {c+d x}}{2 x^3}\right )-\frac {a^2 c \sqrt {c+d x}}{8 x^4}\right )\)

Input:

Int[((c + d*x)^(3/2)*(a + b*x^2)^2)/x^5,x]
 

Output:

2*(-1/8*(a^2*c*Sqrt[c + d*x])/x^4 + ((-3*a^2*d*Sqrt[c + d*x])/(2*x^3) - (( 
a*(32*b*c^2 + a*d^2)*Sqrt[c + d*x])/(4*x^2) - (-1/2*(a*d*(160*b*c^2 - 3*a* 
d^2)*Sqrt[c + d*x])/x + (128*b^2*c^3*Sqrt[c + d*x] + (128*b^2*c^2*(c + d*x 
)^(3/2))/3 - ((128*b^2*c^4 + 96*a*b*c^2*d^2 + 3*a^2*d^4)*ArcTanh[Sqrt[c + 
d*x]/Sqrt[c]])/Sqrt[c])/2)/(4*c))/(2*c))/8)
 

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 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 1468

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> I 
nt[ExpandIntegrand[(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e 
}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -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 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 = 0.31 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.71

method	result	size
pseudoelliptic	\(-\frac {3 \left (\frac {x^{4} \left (a^{2} d^{4}+32 b \,c^{2} d^{2} a +\frac {128}{3} b^{2} c^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{8}+\sqrt {d x +c}\, \left (d x \left (-\frac {16}{9} b^{2} x^{4}+\frac {20}{3} a b \,x^{2}+a^{2}\right ) c^{\frac {5}{2}}+\frac {2 \left (-\frac {32}{3} b^{2} x^{4}+4 a b \,x^{2}+a^{2}\right ) c^{\frac {7}{2}}}{3}-\frac {d^{2} x^{2} \left (d x \sqrt {c}-\frac {2 c^{\frac {3}{2}}}{3}\right ) a^{2}}{8}\right )\right )}{8 c^{\frac {5}{2}} x^{4}}\)	\(134\)
risch	\(-\frac {\sqrt {d x +c}\, a \left (-3 a \,x^{3} d^{3}+160 b \,c^{2} d \,x^{3}+2 a \,d^{2} x^{2} c +64 b \,c^{3} x^{2}+24 a d x \,c^{2}+16 c^{3} a \right )}{64 x^{4} c^{2}}+\frac {\frac {256 b^{2} c^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+256 c^{3} b^{2} \sqrt {d x +c}-\frac {2 \left (3 a^{2} d^{4}+96 b \,c^{2} d^{2} a +128 b^{2} c^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\sqrt {c}}}{128 c^{2}}\)	\(151\)
derivativedivides	\(\frac {2 b^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+2 b^{2} c \sqrt {d x +c}-\frac {2 \left (-\frac {a \,d^{2} \left (3 a \,d^{2}-160 b \,c^{2}\right ) \left (d x +c \right )^{\frac {7}{2}}}{128 c^{2}}+\frac {a \,d^{2} \left (11 a \,d^{2}-416 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{128 c}+\frac {11 a \,d^{2} \left (a \,d^{2}+32 b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{128}+\left (-\frac {3}{128} a^{2} c \,d^{4}-\frac {3}{4} a \,c^{3} d^{2} b \right ) \sqrt {d x +c}\right )}{d^{4} x^{4}}-\frac {\left (3 a^{2} d^{4}+96 b \,c^{2} d^{2} a +128 b^{2} c^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{64 c^{\frac {5}{2}}}\)	\(191\)
default	\(\frac {2 b^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+2 b^{2} c \sqrt {d x +c}-\frac {2 \left (-\frac {a \,d^{2} \left (3 a \,d^{2}-160 b \,c^{2}\right ) \left (d x +c \right )^{\frac {7}{2}}}{128 c^{2}}+\frac {a \,d^{2} \left (11 a \,d^{2}-416 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{128 c}+\frac {11 a \,d^{2} \left (a \,d^{2}+32 b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{128}+\left (-\frac {3}{128} a^{2} c \,d^{4}-\frac {3}{4} a \,c^{3} d^{2} b \right ) \sqrt {d x +c}\right )}{d^{4} x^{4}}-\frac {\left (3 a^{2} d^{4}+96 b \,c^{2} d^{2} a +128 b^{2} c^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{64 c^{\frac {5}{2}}}\)	\(191\)

Input:

int((d*x+c)^(3/2)*(b*x^2+a)^2/x^5,x,method=_RETURNVERBOSE)
 

Output:

-3/8*(1/8*x^4*(a^2*d^4+32*b*c^2*d^2*a+128/3*b^2*c^4)*arctanh((d*x+c)^(1/2) 
/c^(1/2))+(d*x+c)^(1/2)*(d*x*(-16/9*b^2*x^4+20/3*a*b*x^2+a^2)*c^(5/2)+2/3* 
(-32/3*b^2*x^4+4*a*b*x^2+a^2)*c^(7/2)-1/8*d^2*x^2*(d*x*c^(1/2)-2/3*c^(3/2) 
)*a^2))/c^(5/2)/x^4
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.69 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )^2}{x^5} \, dx=\left [\frac {3 \, {\left (128 \, b^{2} c^{4} + 96 \, a b c^{2} d^{2} + 3 \, a^{2} d^{4}\right )} \sqrt {c} x^{4} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (128 \, b^{2} c^{3} d x^{5} + 512 \, b^{2} c^{4} x^{4} - 72 \, a^{2} c^{3} d x - 48 \, a^{2} c^{4} - 3 \, {\left (160 \, a b c^{3} d - 3 \, a^{2} c d^{3}\right )} x^{3} - 6 \, {\left (32 \, a b c^{4} + a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{384 \, c^{3} x^{4}}, \frac {3 \, {\left (128 \, b^{2} c^{4} + 96 \, a b c^{2} d^{2} + 3 \, a^{2} d^{4}\right )} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) + {\left (128 \, b^{2} c^{3} d x^{5} + 512 \, b^{2} c^{4} x^{4} - 72 \, a^{2} c^{3} d x - 48 \, a^{2} c^{4} - 3 \, {\left (160 \, a b c^{3} d - 3 \, a^{2} c d^{3}\right )} x^{3} - 6 \, {\left (32 \, a b c^{4} + a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{192 \, c^{3} x^{4}}\right ] \] Input:

integrate((d*x+c)^(3/2)*(b*x^2+a)^2/x^5,x, algorithm="fricas")
 

Output:

[1/384*(3*(128*b^2*c^4 + 96*a*b*c^2*d^2 + 3*a^2*d^4)*sqrt(c)*x^4*log((d*x 
- 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*(128*b^2*c^3*d*x^5 + 512*b^2*c^4*x 
^4 - 72*a^2*c^3*d*x - 48*a^2*c^4 - 3*(160*a*b*c^3*d - 3*a^2*c*d^3)*x^3 - 6 
*(32*a*b*c^4 + a^2*c^2*d^2)*x^2)*sqrt(d*x + c))/(c^3*x^4), 1/192*(3*(128*b 
^2*c^4 + 96*a*b*c^2*d^2 + 3*a^2*d^4)*sqrt(-c)*x^4*arctan(sqrt(-c)/sqrt(d*x 
 + c)) + (128*b^2*c^3*d*x^5 + 512*b^2*c^4*x^4 - 72*a^2*c^3*d*x - 48*a^2*c^ 
4 - 3*(160*a*b*c^3*d - 3*a^2*c*d^3)*x^3 - 6*(32*a*b*c^4 + a^2*c^2*d^2)*x^2 
)*sqrt(d*x + c))/(c^3*x^4)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )^2}{x^5} \, dx=\text {Timed out} \] Input:

integrate((d*x+c)**(3/2)*(b*x**2+a)**2/x**5,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.44 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )^2}{x^5} \, dx=-\frac {1}{384} \, d^{4} {\left (\frac {6 \, {\left ({\left (160 \, a b c^{2} - 3 \, a^{2} d^{2}\right )} {\left (d x + c\right )}^{\frac {7}{2}} - {\left (416 \, a b c^{3} - 11 \, a^{2} c d^{2}\right )} {\left (d x + c\right )}^{\frac {5}{2}} + 11 \, {\left (32 \, a b c^{4} + a^{2} c^{2} d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} - 3 \, {\left (32 \, a b c^{5} + a^{2} c^{3} d^{2}\right )} \sqrt {d x + c}\right )}}{{\left (d x + c\right )}^{4} c^{2} d^{2} - 4 \, {\left (d x + c\right )}^{3} c^{3} d^{2} + 6 \, {\left (d x + c\right )}^{2} c^{4} d^{2} - 4 \, {\left (d x + c\right )} c^{5} d^{2} + c^{6} d^{2}} - \frac {256 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} b^{2} + 3 \, \sqrt {d x + c} b^{2} c\right )}}{d^{4}} - \frac {3 \, {\left (128 \, b^{2} c^{4} + 96 \, a b c^{2} d^{2} + 3 \, a^{2} d^{4}\right )} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{c^{\frac {5}{2}} d^{4}}\right )} \] Input:

integrate((d*x+c)^(3/2)*(b*x^2+a)^2/x^5,x, algorithm="maxima")
 

Output:

-1/384*d^4*(6*((160*a*b*c^2 - 3*a^2*d^2)*(d*x + c)^(7/2) - (416*a*b*c^3 - 
11*a^2*c*d^2)*(d*x + c)^(5/2) + 11*(32*a*b*c^4 + a^2*c^2*d^2)*(d*x + c)^(3 
/2) - 3*(32*a*b*c^5 + a^2*c^3*d^2)*sqrt(d*x + c))/((d*x + c)^4*c^2*d^2 - 4 
*(d*x + c)^3*c^3*d^2 + 6*(d*x + c)^2*c^4*d^2 - 4*(d*x + c)*c^5*d^2 + c^6*d 
^2) - 256*((d*x + c)^(3/2)*b^2 + 3*sqrt(d*x + c)*b^2*c)/d^4 - 3*(128*b^2*c 
^4 + 96*a*b*c^2*d^2 + 3*a^2*d^4)*log((sqrt(d*x + c) - sqrt(c))/(sqrt(d*x + 
 c) + sqrt(c)))/(c^(5/2)*d^4))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.22 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )^2}{x^5} \, dx=\frac {128 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} d + 384 \, \sqrt {d x + c} b^{2} c d + \frac {3 \, {\left (128 \, b^{2} c^{4} d + 96 \, a b c^{2} d^{3} + 3 \, a^{2} d^{5}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{2}} - \frac {3 \, {\left (160 \, {\left (d x + c\right )}^{\frac {7}{2}} a b c^{2} d^{3} - 416 \, {\left (d x + c\right )}^{\frac {5}{2}} a b c^{3} d^{3} + 352 \, {\left (d x + c\right )}^{\frac {3}{2}} a b c^{4} d^{3} - 96 \, \sqrt {d x + c} a b c^{5} d^{3} - 3 \, {\left (d x + c\right )}^{\frac {7}{2}} a^{2} d^{5} + 11 \, {\left (d x + c\right )}^{\frac {5}{2}} a^{2} c d^{5} + 11 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} c^{2} d^{5} - 3 \, \sqrt {d x + c} a^{2} c^{3} d^{5}\right )}}{c^{2} d^{4} x^{4}}}{192 \, d} \] Input:

integrate((d*x+c)^(3/2)*(b*x^2+a)^2/x^5,x, algorithm="giac")
 

Output:

1/192*(128*(d*x + c)^(3/2)*b^2*d + 384*sqrt(d*x + c)*b^2*c*d + 3*(128*b^2* 
c^4*d + 96*a*b*c^2*d^3 + 3*a^2*d^5)*arctan(sqrt(d*x + c)/sqrt(-c))/(sqrt(- 
c)*c^2) - 3*(160*(d*x + c)^(7/2)*a*b*c^2*d^3 - 416*(d*x + c)^(5/2)*a*b*c^3 
*d^3 + 352*(d*x + c)^(3/2)*a*b*c^4*d^3 - 96*sqrt(d*x + c)*a*b*c^5*d^3 - 3* 
(d*x + c)^(7/2)*a^2*d^5 + 11*(d*x + c)^(5/2)*a^2*c*d^5 + 11*(d*x + c)^(3/2 
)*a^2*c^2*d^5 - 3*sqrt(d*x + c)*a^2*c^3*d^5)/(c^2*d^4*x^4))/d
 

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.25 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )^2}{x^5} \, dx=\frac {2\,b^2\,{\left (c+d\,x\right )}^{3/2}}{3}+\frac {\left (\frac {3\,a^2\,c\,d^4}{64}+\frac {3\,b\,a\,c^3\,d^2}{2}\right )\,\sqrt {c+d\,x}-\left (\frac {11\,a^2\,d^4}{64}+\frac {11\,b\,a\,c^2\,d^2}{2}\right )\,{\left (c+d\,x\right )}^{3/2}+\frac {\left (3\,a^2\,d^4-160\,a\,b\,c^2\,d^2\right )\,{\left (c+d\,x\right )}^{7/2}}{64\,c^2}-\frac {\left (11\,a^2\,d^4-416\,a\,b\,c^2\,d^2\right )\,{\left (c+d\,x\right )}^{5/2}}{64\,c}}{{\left (c+d\,x\right )}^4-4\,c^3\,\left (c+d\,x\right )-4\,c\,{\left (c+d\,x\right )}^3+6\,c^2\,{\left (c+d\,x\right )}^2+c^4}+2\,b^2\,c\,\sqrt {c+d\,x}+\frac {\mathrm {atan}\left (\frac {\sqrt {c+d\,x}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,\left (3\,a^2\,d^4+96\,a\,b\,c^2\,d^2+128\,b^2\,c^4\right )\,1{}\mathrm {i}}{64\,c^{5/2}} \] Input:

int(((a + b*x^2)^2*(c + d*x)^(3/2))/x^5,x)
                                                                                    
                                                                                    
 

Output:

(2*b^2*(c + d*x)^(3/2))/3 + (((3*a^2*c*d^4)/64 + (3*a*b*c^3*d^2)/2)*(c + d 
*x)^(1/2) - ((11*a^2*d^4)/64 + (11*a*b*c^2*d^2)/2)*(c + d*x)^(3/2) + ((3*a 
^2*d^4 - 160*a*b*c^2*d^2)*(c + d*x)^(7/2))/(64*c^2) - ((11*a^2*d^4 - 416*a 
*b*c^2*d^2)*(c + d*x)^(5/2))/(64*c))/((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) + (atan(((c + d*x)^(1/2)*1i)/c^(1/2 
))*(3*a^2*d^4 + 128*b^2*c^4 + 96*a*b*c^2*d^2)*1i)/(64*c^(5/2)) + 2*b^2*c*( 
c + d*x)^(1/2)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.54 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )^2}{x^5} \, dx=\frac {-96 \sqrt {d x +c}\, a^{2} c^{4}-144 \sqrt {d x +c}\, a^{2} c^{3} d x -12 \sqrt {d x +c}\, a^{2} c^{2} d^{2} x^{2}+18 \sqrt {d x +c}\, a^{2} c \,d^{3} x^{3}-384 \sqrt {d x +c}\, a b \,c^{4} x^{2}-960 \sqrt {d x +c}\, a b \,c^{3} d \,x^{3}+1024 \sqrt {d x +c}\, b^{2} c^{4} x^{4}+256 \sqrt {d x +c}\, b^{2} c^{3} d \,x^{5}+9 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a^{2} d^{4} x^{4}+288 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a b \,c^{2} d^{2} x^{4}+384 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) b^{2} c^{4} x^{4}-9 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a^{2} d^{4} x^{4}-288 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a b \,c^{2} d^{2} x^{4}-384 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) b^{2} c^{4} x^{4}}{384 c^{3} x^{4}} \] Input:

int((d*x+c)^(3/2)*(b*x^2+a)^2/x^5,x)
 

Output:

( - 96*sqrt(c + d*x)*a**2*c**4 - 144*sqrt(c + d*x)*a**2*c**3*d*x - 12*sqrt 
(c + d*x)*a**2*c**2*d**2*x**2 + 18*sqrt(c + d*x)*a**2*c*d**3*x**3 - 384*sq 
rt(c + d*x)*a*b*c**4*x**2 - 960*sqrt(c + d*x)*a*b*c**3*d*x**3 + 1024*sqrt( 
c + d*x)*b**2*c**4*x**4 + 256*sqrt(c + d*x)*b**2*c**3*d*x**5 + 9*sqrt(c)*l 
og(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 - 9*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a**2*d**4*x**4 - 2 
88*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a*b*c**2*d**2*x**4 - 384*sqrt(c)*l 
og(sqrt(c + d*x) + sqrt(c))*b**2*c**4*x**4)/(384*c**3*x**4)