\(\int \frac {(c+d x)^{3/2} (a+b x^2)}{x^5} \, dx\) [479]

Optimal result

Integrand size = 20, antiderivative size = 138 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{x^5} \, dx=-\frac {a c \sqrt {c+d x}}{4 x^4}-\frac {3 a d \sqrt {c+d x}}{8 x^3}-\frac {\left (16 b c^2+a d^2\right ) \sqrt {c+d x}}{32 c x^2}-\frac {d \left (80 b-\frac {3 a d^2}{c^2}\right ) \sqrt {c+d x}}{64 x}-\frac {3 d^2 \left (16 b c^2+a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{64 c^{5/2}} \] Output:

-1/4*a*c*(d*x+c)^(1/2)/x^4-3/8*a*d*(d*x+c)^(1/2)/x^3-1/32*(a*d^2+16*b*c^2) 
*(d*x+c)^(1/2)/c/x^2-1/64*d*(80*b-3*a*d^2/c^2)*(d*x+c)^(1/2)/x-3/64*d^2*(a 
*d^2+16*b*c^2)*arctanh((d*x+c)^(1/2)/c^(1/2))/c^(5/2)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.80 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{x^5} \, dx=-\frac {\sqrt {c+d x} \left (16 b c^2 x^2 (2 c+5 d x)+a \left (16 c^3+24 c^2 d x+2 c d^2 x^2-3 d^3 x^3\right )\right )}{64 c^2 x^4}-\frac {3 d^2 \left (16 b c^2+a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{64 c^{5/2}} \] Input:

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

Output:

-1/64*(Sqrt[c + d*x]*(16*b*c^2*x^2*(2*c + 5*d*x) + a*(16*c^3 + 24*c^2*d*x 
+ 2*c*d^2*x^2 - 3*d^3*x^3)))/(c^2*x^4) - (3*d^2*(16*b*c^2 + a*d^2)*ArcTanh 
[Sqrt[c + d*x]/Sqrt[c]])/(64*c^(5/2))
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.24, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {517, 25, 1580, 2345, 27, 1472, 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.

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

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1580

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

\(\Big \downarrow \) 2345

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1472

\(\displaystyle 2 d^2 \left (\frac {1}{8} \left (\frac {1}{2} \left (\frac {\int \frac {16 b c^2+64 b (c+d x) c-3 a d^2}{d^2 x^2}d\sqrt {c+d x}}{4 c}-\frac {\sqrt {c+d x} \left (a d^2+16 b c^2\right )}{4 c d^2 x^2}\right )-\frac {3 a \sqrt {c+d x}}{2 d x^3}\right )-\frac {a c \sqrt {c+d x}}{8 d^2 x^4}\right )\)

\(\Big \downarrow \) 298

\(\displaystyle 2 d^2 \left (\frac {1}{8} \left (\frac {1}{2} \left (\frac {-\frac {3 \left (a d^2+16 b c^2\right ) \int -\frac {1}{d x}d\sqrt {c+d x}}{2 c}-\frac {\sqrt {c+d x} \left (80 b c^2-3 a d^2\right )}{2 c d x}}{4 c}-\frac {\sqrt {c+d x} \left (a d^2+16 b c^2\right )}{4 c d^2 x^2}\right )-\frac {3 a \sqrt {c+d x}}{2 d x^3}\right )-\frac {a c \sqrt {c+d x}}{8 d^2 x^4}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle 2 d^2 \left (\frac {1}{8} \left (\frac {1}{2} \left (\frac {-\frac {3 \left (a d^2+16 b c^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{2 c^{3/2}}-\frac {\sqrt {c+d x} \left (80 b c^2-3 a d^2\right )}{2 c d x}}{4 c}-\frac {\sqrt {c+d x} \left (a d^2+16 b c^2\right )}{4 c d^2 x^2}\right )-\frac {3 a \sqrt {c+d x}}{2 d x^3}\right )-\frac {a c \sqrt {c+d x}}{8 d^2 x^4}\right )\)

Input:

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

Output:

2*d^2*(-1/8*(a*c*Sqrt[c + d*x])/(d^2*x^4) + ((-3*a*Sqrt[c + d*x])/(2*d*x^3 
) + (-1/4*((16*b*c^2 + a*d^2)*Sqrt[c + d*x])/(c*d^2*x^2) + (-1/2*((80*b*c^ 
2 - 3*a*d^2)*Sqrt[c + d*x])/(c*d*x) - (3*(16*b*c^2 + a*d^2)*ArcTanh[Sqrt[c 
 + d*x]/Sqrt[c]])/(2*c^(3/2)))/(4*c))/2)/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 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 1472

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Wi 
th[{Qx = PolynomialQuotient[(a + c*x^4)^p, d + e*x^2, x], R = Coeff[Polynom 
ialRemainder[(a + 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, c, d, 
e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
 

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 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 = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.72

method	result	size
pseudoelliptic	\(-\frac {3 \left (\frac {d^{2} x^{4} \left (a \,d^{2}+16 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{8}+\sqrt {d x +c}\, \left (d x \left (\frac {10 b \,x^{2}}{3}+a \right ) c^{\frac {5}{2}}+\frac {2 \left (2 b \,x^{2}+a \right ) c^{\frac {7}{2}}}{3}-\frac {d^{2} x^{2} \left (d x \sqrt {c}-\frac {2 c^{\frac {3}{2}}}{3}\right ) a}{8}\right )\right )}{8 c^{\frac {5}{2}} x^{4}}\)	\(99\)
risch	\(-\frac {\sqrt {d x +c}\, \left (-3 a \,x^{3} d^{3}+80 b \,c^{2} d \,x^{3}+2 a \,d^{2} x^{2} c +32 b \,c^{3} x^{2}+24 a d x \,c^{2}+16 c^{3} a \right )}{64 x^{4} c^{2}}-\frac {3 d^{2} \left (a \,d^{2}+16 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{64 c^{\frac {5}{2}}}\)	\(102\)
derivativedivides	\(2 d^{2} \left (-\frac {-\frac {\left (3 a \,d^{2}-80 b \,c^{2}\right ) \left (d x +c \right )^{\frac {7}{2}}}{128 c^{2}}+\frac {\left (11 a \,d^{2}-208 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{128 c}+\left (\frac {11 a \,d^{2}}{128}+\frac {11 b \,c^{2}}{8}\right ) \left (d x +c \right )^{\frac {3}{2}}+\left (-\frac {3}{8} b \,c^{3}-\frac {3}{128} a \,d^{2} c \right ) \sqrt {d x +c}}{d^{4} x^{4}}-\frac {3 \left (a \,d^{2}+16 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{128 c^{\frac {5}{2}}}\right )\)	\(138\)
default	\(2 d^{2} \left (-\frac {-\frac {\left (3 a \,d^{2}-80 b \,c^{2}\right ) \left (d x +c \right )^{\frac {7}{2}}}{128 c^{2}}+\frac {\left (11 a \,d^{2}-208 b \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{128 c}+\left (\frac {11 a \,d^{2}}{128}+\frac {11 b \,c^{2}}{8}\right ) \left (d x +c \right )^{\frac {3}{2}}+\left (-\frac {3}{8} b \,c^{3}-\frac {3}{128} a \,d^{2} c \right ) \sqrt {d x +c}}{d^{4} x^{4}}-\frac {3 \left (a \,d^{2}+16 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{128 c^{\frac {5}{2}}}\right )\)	\(138\)

Input:

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

Output:

-3/8/c^(5/2)*(1/8*d^2*x^4*(a*d^2+16*b*c^2)*arctanh((d*x+c)^(1/2)/c^(1/2))+ 
(d*x+c)^(1/2)*(d*x*(10/3*b*x^2+a)*c^(5/2)+2/3*(2*b*x^2+a)*c^(7/2)-1/8*d^2* 
x^2*(d*x*c^(1/2)-2/3*c^(3/2))*a))/x^4
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.67 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{x^5} \, dx=\left [\frac {3 \, {\left (16 \, b c^{2} d^{2} + a d^{4}\right )} \sqrt {c} x^{4} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (24 \, a c^{3} d x + 16 \, a c^{4} + {\left (80 \, b c^{3} d - 3 \, a c d^{3}\right )} x^{3} + 2 \, {\left (16 \, b c^{4} + a c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{128 \, c^{3} x^{4}}, \frac {3 \, {\left (16 \, b c^{2} d^{2} + a d^{4}\right )} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) - {\left (24 \, a c^{3} d x + 16 \, a c^{4} + {\left (80 \, b c^{3} d - 3 \, a c d^{3}\right )} x^{3} + 2 \, {\left (16 \, b c^{4} + a c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{64 \, c^{3} x^{4}}\right ] \] Input:

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

Output:

[1/128*(3*(16*b*c^2*d^2 + a*d^4)*sqrt(c)*x^4*log((d*x - 2*sqrt(d*x + c)*sq 
rt(c) + 2*c)/x) - 2*(24*a*c^3*d*x + 16*a*c^4 + (80*b*c^3*d - 3*a*c*d^3)*x^ 
3 + 2*(16*b*c^4 + a*c^2*d^2)*x^2)*sqrt(d*x + c))/(c^3*x^4), 1/64*(3*(16*b* 
c^2*d^2 + a*d^4)*sqrt(-c)*x^4*arctan(sqrt(-c)/sqrt(d*x + c)) - (24*a*c^3*d 
*x + 16*a*c^4 + (80*b*c^3*d - 3*a*c*d^3)*x^3 + 2*(16*b*c^4 + a*c^2*d^2)*x^ 
2)*sqrt(d*x + c))/(c^3*x^4)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (129) = 258\).

Time = 136.28 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.13 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{x^5} \, dx=- \frac {a c^{2}}{4 \sqrt {d} x^{\frac {9}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {5 a c \sqrt {d}}{8 x^{\frac {7}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {13 a d^{\frac {3}{2}}}{32 x^{\frac {5}{2}} \sqrt {\frac {c}{d x} + 1}} + \frac {a d^{\frac {5}{2}}}{64 c x^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}} + \frac {3 a d^{\frac {7}{2}}}{64 c^{2} \sqrt {x} \sqrt {\frac {c}{d x} + 1}} - \frac {3 a d^{4} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{64 c^{\frac {5}{2}}} - \frac {b c^{2}}{2 \sqrt {d} x^{\frac {5}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {3 b c \sqrt {d}}{4 x^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {b d^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}}{\sqrt {x}} - \frac {b d^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {\frac {c}{d x} + 1}} - \frac {3 b d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{4 \sqrt {c}} \] Input:

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

Output:

-a*c**2/(4*sqrt(d)*x**(9/2)*sqrt(c/(d*x) + 1)) - 5*a*c*sqrt(d)/(8*x**(7/2) 
*sqrt(c/(d*x) + 1)) - 13*a*d**(3/2)/(32*x**(5/2)*sqrt(c/(d*x) + 1)) + a*d* 
*(5/2)/(64*c*x**(3/2)*sqrt(c/(d*x) + 1)) + 3*a*d**(7/2)/(64*c**2*sqrt(x)*s 
qrt(c/(d*x) + 1)) - 3*a*d**4*asinh(sqrt(c)/(sqrt(d)*sqrt(x)))/(64*c**(5/2) 
) - b*c**2/(2*sqrt(d)*x**(5/2)*sqrt(c/(d*x) + 1)) - 3*b*c*sqrt(d)/(4*x**(3 
/2)*sqrt(c/(d*x) + 1)) - b*d**(3/2)*sqrt(c/(d*x) + 1)/sqrt(x) - b*d**(3/2) 
/(4*sqrt(x)*sqrt(c/(d*x) + 1)) - 3*b*d**2*asinh(sqrt(c)/(sqrt(d)*sqrt(x))) 
/(4*sqrt(c))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.57 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{x^5} \, dx=-\frac {1}{128} \, d^{4} {\left (\frac {2 \, {\left ({\left (80 \, b c^{2} - 3 \, a d^{2}\right )} {\left (d x + c\right )}^{\frac {7}{2}} - {\left (208 \, b c^{3} - 11 \, a c d^{2}\right )} {\left (d x + c\right )}^{\frac {5}{2}} + 11 \, {\left (16 \, b c^{4} + a c^{2} d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} - 3 \, {\left (16 \, b c^{5} + a c^{3} d^{2}\right )} \sqrt {d x + c}\right )}}{{\left (d x + c\right )}^{4} c^{2} d^{2} - 4 \, {\left (d x + c\right )}^{3} c^{3} d^{2} + 6 \, {\left (d x + c\right )}^{2} c^{4} d^{2} - 4 \, {\left (d x + c\right )} c^{5} d^{2} + c^{6} d^{2}} - \frac {3 \, {\left (16 \, b c^{2} + a d^{2}\right )} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{c^{\frac {5}{2}} d^{2}}\right )} \] Input:

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

Output:

-1/128*d^4*(2*((80*b*c^2 - 3*a*d^2)*(d*x + c)^(7/2) - (208*b*c^3 - 11*a*c* 
d^2)*(d*x + c)^(5/2) + 11*(16*b*c^4 + a*c^2*d^2)*(d*x + c)^(3/2) - 3*(16*b 
*c^5 + a*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) - 3*(16*b*c^2 
 + a*d^2)*log((sqrt(d*x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c)))/(c^(5/2 
)*d^2))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.30 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{x^5} \, dx=\frac {\frac {3 \, {\left (16 \, b c^{2} d^{3} + a d^{5}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{2}} - \frac {80 \, {\left (d x + c\right )}^{\frac {7}{2}} b c^{2} d^{3} - 208 \, {\left (d x + c\right )}^{\frac {5}{2}} b c^{3} d^{3} + 176 \, {\left (d x + c\right )}^{\frac {3}{2}} b c^{4} d^{3} - 48 \, \sqrt {d x + c} b c^{5} d^{3} - 3 \, {\left (d x + c\right )}^{\frac {7}{2}} a d^{5} + 11 \, {\left (d x + c\right )}^{\frac {5}{2}} a c d^{5} + 11 \, {\left (d x + c\right )}^{\frac {3}{2}} a c^{2} d^{5} - 3 \, \sqrt {d x + c} a c^{3} d^{5}}{c^{2} d^{4} x^{4}}}{64 \, d} \] Input:

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

Output:

1/64*(3*(16*b*c^2*d^3 + a*d^5)*arctan(sqrt(d*x + c)/sqrt(-c))/(sqrt(-c)*c^ 
2) - (80*(d*x + c)^(7/2)*b*c^2*d^3 - 208*(d*x + c)^(5/2)*b*c^3*d^3 + 176*( 
d*x + c)^(3/2)*b*c^4*d^3 - 48*sqrt(d*x + c)*b*c^5*d^3 - 3*(d*x + c)^(7/2)* 
a*d^5 + 11*(d*x + c)^(5/2)*a*c*d^5 + 11*(d*x + c)^(3/2)*a*c^2*d^5 - 3*sqrt 
(d*x + c)*a*c^3*d^5)/(c^2*d^4*x^4))/d
 

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.35 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{x^5} \, dx=\frac {\left (\frac {3\,b\,c^3\,d^2}{4}+\frac {3\,a\,c\,d^4}{64}\right )\,\sqrt {c+d\,x}-\left (\frac {11\,b\,c^2\,d^2}{4}+\frac {11\,a\,d^4}{64}\right )\,{\left (c+d\,x\right )}^{3/2}+\frac {\left (3\,a\,d^4-80\,b\,c^2\,d^2\right )\,{\left (c+d\,x\right )}^{7/2}}{64\,c^2}-\frac {\left (11\,a\,d^4-208\,b\,c^2\,d^2\right )\,{\left (c+d\,x\right )}^{5/2}}{64\,c}}{{\left (c+d\,x\right )}^4-4\,c^3\,\left (c+d\,x\right )-4\,c\,{\left (c+d\,x\right )}^3+6\,c^2\,{\left (c+d\,x\right )}^2+c^4}-\frac {3\,d^2\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}}{\sqrt {c}}\right )\,\left (16\,b\,c^2+a\,d^2\right )}{64\,c^{5/2}} \] Input:

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

Output:

(((3*b*c^3*d^2)/4 + (3*a*c*d^4)/64)*(c + d*x)^(1/2) - ((11*a*d^4)/64 + (11 
*b*c^2*d^2)/4)*(c + d*x)^(3/2) + ((3*a*d^4 - 80*b*c^2*d^2)*(c + d*x)^(7/2) 
)/(64*c^2) - ((11*a*d^4 - 208*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) - (3*d 
^2*atanh((c + d*x)^(1/2)/c^(1/2))*(a*d^2 + 16*b*c^2))/(64*c^(5/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.41 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{x^5} \, dx=\frac {-32 \sqrt {d x +c}\, a \,c^{4}-48 \sqrt {d x +c}\, a \,c^{3} d x -4 \sqrt {d x +c}\, a \,c^{2} d^{2} x^{2}+6 \sqrt {d x +c}\, a c \,d^{3} x^{3}-64 \sqrt {d x +c}\, b \,c^{4} x^{2}-160 \sqrt {d x +c}\, b \,c^{3} d \,x^{3}+3 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a \,d^{4} x^{4}+48 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) b \,c^{2} d^{2} x^{4}-3 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a \,d^{4} x^{4}-48 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) b \,c^{2} d^{2} x^{4}}{128 c^{3} x^{4}} \] Input:

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

Output:

( - 32*sqrt(c + d*x)*a*c**4 - 48*sqrt(c + d*x)*a*c**3*d*x - 4*sqrt(c + d*x 
)*a*c**2*d**2*x**2 + 6*sqrt(c + d*x)*a*c*d**3*x**3 - 64*sqrt(c + d*x)*b*c* 
*4*x**2 - 160*sqrt(c + d*x)*b*c**3*d*x**3 + 3*sqrt(c)*log(sqrt(c + d*x) - 
sqrt(c))*a*d**4*x**4 + 48*sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*b*c**2*d**2 
*x**4 - 3*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a*d**4*x**4 - 48*sqrt(c)*lo 
g(sqrt(c + d*x) + sqrt(c))*b*c**2*d**2*x**4)/(128*c**3*x**4)