\(\int \frac {(c+d x)^{5/2} (a+b x^2)}{x^5} \, dx\) [488]

Optimal result

Integrand size = 20, antiderivative size = 158 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^5} \, dx=2 b d^2 \sqrt {c+d x}-\frac {a c^2 \sqrt {c+d x}}{4 x^4}-\frac {17 a c d \sqrt {c+d x}}{24 x^3}-\frac {\left (48 b c^2+59 a d^2\right ) \sqrt {c+d x}}{96 x^2}-\frac {d \left (144 b c^2+5 a d^2\right ) \sqrt {c+d x}}{64 c x}-\frac {5 d^2 \left (48 b c^2-a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{64 c^{3/2}} \] Output:

2*b*d^2*(d*x+c)^(1/2)-1/4*a*c^2*(d*x+c)^(1/2)/x^4-17/24*a*c*d*(d*x+c)^(1/2 
)/x^3-1/96*(59*a*d^2+48*b*c^2)*(d*x+c)^(1/2)/x^2-1/64*d*(5*a*d^2+144*b*c^2 
)*(d*x+c)^(1/2)/c/x-5/64*d^2*(-a*d^2+48*b*c^2)*arctanh((d*x+c)^(1/2)/c^(1/ 
2))/c^(3/2)
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.75 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^5} \, dx=-\frac {\sqrt {c+d x} \left (48 b c x^2 \left (2 c^2+9 c d x-8 d^2 x^2\right )+a \left (48 c^3+136 c^2 d x+118 c d^2 x^2+15 d^3 x^3\right )\right )}{192 c x^4}+\frac {5 d^2 \left (-48 b c^2+a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{64 c^{3/2}} \] Input:

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

Output:

-1/192*(Sqrt[c + d*x]*(48*b*c*x^2*(2*c^2 + 9*c*d*x - 8*d^2*x^2) + a*(48*c^ 
3 + 136*c^2*d*x + 118*c*d^2*x^2 + 15*d^3*x^3)))/(c*x^4) + (5*d^2*(-48*b*c^ 
2 + a*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(64*c^(3/2))
 

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.22, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {517, 25, 1580, 25, 2345, 2345, 27, 1471, 27, 299, 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.

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

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1580

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2345

\(\displaystyle 2 d^2 \left (\frac {1}{8} \left (-\frac {\int -\frac {48 b c (c+d x)^3+48 a c d^2 (c+d x)+11 a c^2 d^2}{d^3 x^3}d\sqrt {c+d x}}{6 c}-\frac {17 a c \sqrt {c+d x}}{6 d x^3}\right )-\frac {a c^2 \sqrt {c+d x}}{8 d^2 x^4}\right )\)

\(\Big \downarrow \) 2345

\(\displaystyle 2 d^2 \left (\frac {1}{8} \left (-\frac {\frac {c \sqrt {c+d x} \left (59 a d^2+48 b c^2\right )}{4 d^2 x^2}-\frac {\int \frac {3 \left (64 b (c+d x) c^3+64 b (c+d x)^2 c^2+\left (16 b c^2+5 a d^2\right ) c^2\right )}{d^2 x^2}d\sqrt {c+d x}}{4 c}}{6 c}-\frac {17 a c \sqrt {c+d x}}{6 d x^3}\right )-\frac {a c^2 \sqrt {c+d x}}{8 d^2 x^4}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 d^2 \left (\frac {1}{8} \left (-\frac {\frac {c \sqrt {c+d x} \left (59 a d^2+48 b c^2\right )}{4 d^2 x^2}-\frac {3 \int \frac {64 b (c+d x) c^3+64 b (c+d x)^2 c^2+\left (16 b c^2+5 a d^2\right ) c^2}{d^2 x^2}d\sqrt {c+d x}}{4 c}}{6 c}-\frac {17 a c \sqrt {c+d x}}{6 d x^3}\right )-\frac {a c^2 \sqrt {c+d x}}{8 d^2 x^4}\right )\)

\(\Big \downarrow \) 1471

\(\displaystyle 2 d^2 \left (\frac {1}{8} \left (-\frac {\frac {c \sqrt {c+d x} \left (59 a d^2+48 b c^2\right )}{4 d^2 x^2}-\frac {3 \left (-\frac {\int -\frac {c^2 \left (112 b c^2+128 b (c+d x) c-5 a d^2\right )}{d x}d\sqrt {c+d x}}{2 c}-\frac {c \sqrt {c+d x} \left (5 a d^2+144 b c^2\right )}{2 d x}\right )}{4 c}}{6 c}-\frac {17 a c \sqrt {c+d x}}{6 d x^3}\right )-\frac {a c^2 \sqrt {c+d x}}{8 d^2 x^4}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 d^2 \left (\frac {1}{8} \left (-\frac {\frac {c \sqrt {c+d x} \left (59 a d^2+48 b c^2\right )}{4 d^2 x^2}-\frac {3 \left (-\frac {1}{2} c \int -\frac {112 b c^2+128 b (c+d x) c-5 a d^2}{d x}d\sqrt {c+d x}-\frac {c \sqrt {c+d x} \left (5 a d^2+144 b c^2\right )}{2 d x}\right )}{4 c}}{6 c}-\frac {17 a c \sqrt {c+d x}}{6 d x^3}\right )-\frac {a c^2 \sqrt {c+d x}}{8 d^2 x^4}\right )\)

\(\Big \downarrow \) 299

\(\displaystyle 2 d^2 \left (\frac {1}{8} \left (-\frac {\frac {c \sqrt {c+d x} \left (59 a d^2+48 b c^2\right )}{4 d^2 x^2}-\frac {3 \left (-\frac {1}{2} c \left (5 \left (48 b c^2-a d^2\right ) \int -\frac {1}{d x}d\sqrt {c+d x}-128 b c \sqrt {c+d x}\right )-\frac {c \sqrt {c+d x} \left (5 a d^2+144 b c^2\right )}{2 d x}\right )}{4 c}}{6 c}-\frac {17 a c \sqrt {c+d x}}{6 d x^3}\right )-\frac {a c^2 \sqrt {c+d x}}{8 d^2 x^4}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle 2 d^2 \left (\frac {1}{8} \left (-\frac {\frac {c \sqrt {c+d x} \left (59 a d^2+48 b c^2\right )}{4 d^2 x^2}-\frac {3 \left (-\frac {1}{2} c \left (\frac {5 \left (48 b c^2-a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\sqrt {c}}-128 b c \sqrt {c+d x}\right )-\frac {c \sqrt {c+d x} \left (5 a d^2+144 b c^2\right )}{2 d x}\right )}{4 c}}{6 c}-\frac {17 a c \sqrt {c+d x}}{6 d x^3}\right )-\frac {a c^2 \sqrt {c+d x}}{8 d^2 x^4}\right )\)

Input:

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

Output:

2*d^2*(-1/8*(a*c^2*Sqrt[c + d*x])/(d^2*x^4) + ((-17*a*c*Sqrt[c + d*x])/(6* 
d*x^3) - ((c*(48*b*c^2 + 59*a*d^2)*Sqrt[c + d*x])/(4*d^2*x^2) - (3*(-1/2*( 
c*(144*b*c^2 + 5*a*d^2)*Sqrt[c + d*x])/(d*x) - (c*(-128*b*c*Sqrt[c + d*x] 
+ (5*(48*b*c^2 - a*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/Sqrt[c]))/2))/(4*c 
))/(6*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 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 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 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 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 = 109, normalized size of antiderivative = 0.69

method	result	size
pseudoelliptic	\(-\frac {17 \left (-\frac {15 d^{2} x^{4} \left (a \,d^{2}-48 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{136}+\sqrt {d x +c}\, \left (\frac {59 d^{2} x^{2} \left (-\frac {192 b \,x^{2}}{59}+a \right ) c^{\frac {3}{2}}}{68}+d x \left (\frac {54 b \,x^{2}}{17}+a \right ) c^{\frac {5}{2}}+\frac {6 \left (2 b \,x^{2}+a \right ) c^{\frac {7}{2}}}{17}+\frac {15 \sqrt {c}\, a \,d^{3} x^{3}}{136}\right )\right )}{24 c^{\frac {3}{2}} x^{4}}\)	\(109\)
risch	\(-\frac {\sqrt {d x +c}\, \left (15 a \,x^{3} d^{3}+432 b \,c^{2} d \,x^{3}+118 a \,d^{2} x^{2} c +96 b \,c^{3} x^{2}+136 a d x \,c^{2}+48 c^{3} a \right )}{192 x^{4} c}-\frac {d^{2} \left (-256 b c \sqrt {d x +c}+\frac {2 \left (-5 a \,d^{2}+240 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\sqrt {c}}\right )}{128 c}\)	\(120\)
derivativedivides	\(2 d^{2} \left (b \sqrt {d x +c}-\frac {\frac {\left (5 a \,d^{2}+144 b \,c^{2}\right ) \left (d x +c \right )^{\frac {7}{2}}}{128 c}+\left (-\frac {25 b \,c^{2}}{8}+\frac {73 a \,d^{2}}{384}\right ) \left (d x +c \right )^{\frac {5}{2}}-\frac {c \left (55 a \,d^{2}-1104 b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{384}+\left (-\frac {7}{8} b \,c^{4}+\frac {5}{128} a \,c^{2} d^{2}\right ) \sqrt {d x +c}}{d^{4} x^{4}}+\frac {5 \left (a \,d^{2}-48 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{128 c^{\frac {3}{2}}}\right )\)	\(147\)
default	\(2 d^{2} \left (b \sqrt {d x +c}-\frac {\frac {\left (5 a \,d^{2}+144 b \,c^{2}\right ) \left (d x +c \right )^{\frac {7}{2}}}{128 c}+\left (-\frac {25 b \,c^{2}}{8}+\frac {73 a \,d^{2}}{384}\right ) \left (d x +c \right )^{\frac {5}{2}}-\frac {c \left (55 a \,d^{2}-1104 b \,c^{2}\right ) \left (d x +c \right )^{\frac {3}{2}}}{384}+\left (-\frac {7}{8} b \,c^{4}+\frac {5}{128} a \,c^{2} d^{2}\right ) \sqrt {d x +c}}{d^{4} x^{4}}+\frac {5 \left (a \,d^{2}-48 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{128 c^{\frac {3}{2}}}\right )\)	\(147\)

Input:

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

Output:

-17/24*(-15/136*d^2*x^4*(a*d^2-48*b*c^2)*arctanh((d*x+c)^(1/2)/c^(1/2))+(d 
*x+c)^(1/2)*(59/68*d^2*x^2*(-192/59*b*x^2+a)*c^(3/2)+d*x*(54/17*b*x^2+a)*c 
^(5/2)+6/17*(2*b*x^2+a)*c^(7/2)+15/136*c^(1/2)*a*d^3*x^3))/c^(3/2)/x^4
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.65 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^5} \, dx=\left [-\frac {15 \, {\left (48 \, b c^{2} d^{2} - a d^{4}\right )} \sqrt {c} x^{4} \log \left (\frac {d x + 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (384 \, b c^{2} d^{2} x^{4} - 136 \, a c^{3} d x - 48 \, a c^{4} - 3 \, {\left (144 \, b c^{3} d + 5 \, a c d^{3}\right )} x^{3} - 2 \, {\left (48 \, b c^{4} + 59 \, a c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{384 \, c^{2} x^{4}}, \frac {15 \, {\left (48 \, b c^{2} d^{2} - a d^{4}\right )} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) + {\left (384 \, b c^{2} d^{2} x^{4} - 136 \, a c^{3} d x - 48 \, a c^{4} - 3 \, {\left (144 \, b c^{3} d + 5 \, a c d^{3}\right )} x^{3} - 2 \, {\left (48 \, b c^{4} + 59 \, a c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{192 \, c^{2} x^{4}}\right ] \] Input:

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

Output:

[-1/384*(15*(48*b*c^2*d^2 - a*d^4)*sqrt(c)*x^4*log((d*x + 2*sqrt(d*x + c)* 
sqrt(c) + 2*c)/x) - 2*(384*b*c^2*d^2*x^4 - 136*a*c^3*d*x - 48*a*c^4 - 3*(1 
44*b*c^3*d + 5*a*c*d^3)*x^3 - 2*(48*b*c^4 + 59*a*c^2*d^2)*x^2)*sqrt(d*x + 
c))/(c^2*x^4), 1/192*(15*(48*b*c^2*d^2 - a*d^4)*sqrt(-c)*x^4*arctan(sqrt(- 
c)/sqrt(d*x + c)) + (384*b*c^2*d^2*x^4 - 136*a*c^3*d*x - 48*a*c^4 - 3*(144 
*b*c^3*d + 5*a*c*d^3)*x^3 - 2*(48*b*c^4 + 59*a*c^2*d^2)*x^2)*sqrt(d*x + c) 
)/(c^2*x^4)]
 

Sympy [A] (verification not implemented)

Time = 147.36 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.24 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^5} \, dx=- \frac {a c^{3}}{4 \sqrt {d} x^{\frac {9}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {23 a c^{2} \sqrt {d}}{24 x^{\frac {7}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {127 a c d^{\frac {3}{2}}}{96 x^{\frac {5}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {133 a d^{\frac {5}{2}}}{192 x^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {5 a d^{\frac {7}{2}}}{64 c \sqrt {x} \sqrt {\frac {c}{d x} + 1}} + \frac {5 a d^{4} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{64 c^{\frac {3}{2}}} - \frac {7 b \sqrt {c} d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{4} - \frac {b c^{3}}{2 \sqrt {d} x^{\frac {5}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {3 b c^{2} \sqrt {d}}{4 x^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {2 b c d^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}}{\sqrt {x}} - \frac {b c d^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {\frac {c}{d x} + 1}} + b d^{2} \left (\begin {cases} \frac {2 c \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 \sqrt {c + d x} & \text {for}\: d \neq 0 \\\sqrt {c} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

-a*c**3/(4*sqrt(d)*x**(9/2)*sqrt(c/(d*x) + 1)) - 23*a*c**2*sqrt(d)/(24*x** 
(7/2)*sqrt(c/(d*x) + 1)) - 127*a*c*d**(3/2)/(96*x**(5/2)*sqrt(c/(d*x) + 1) 
) - 133*a*d**(5/2)/(192*x**(3/2)*sqrt(c/(d*x) + 1)) - 5*a*d**(7/2)/(64*c*s 
qrt(x)*sqrt(c/(d*x) + 1)) + 5*a*d**4*asinh(sqrt(c)/(sqrt(d)*sqrt(x)))/(64* 
c**(3/2)) - 7*b*sqrt(c)*d**2*asinh(sqrt(c)/(sqrt(d)*sqrt(x)))/4 - b*c**3/( 
2*sqrt(d)*x**(5/2)*sqrt(c/(d*x) + 1)) - 3*b*c**2*sqrt(d)/(4*x**(3/2)*sqrt( 
c/(d*x) + 1)) - 2*b*c*d**(3/2)*sqrt(c/(d*x) + 1)/sqrt(x) - b*c*d**(3/2)/(4 
*sqrt(x)*sqrt(c/(d*x) + 1)) + b*d**2*Piecewise((2*c*atan(sqrt(c + d*x)/sqr 
t(-c))/sqrt(-c) + 2*sqrt(c + d*x), Ne(d, 0)), (sqrt(c)*log(x), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.46 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^5} \, dx=-\frac {1}{384} \, d^{4} {\left (\frac {2 \, {\left (3 \, {\left (144 \, b c^{2} + 5 \, a d^{2}\right )} {\left (d x + c\right )}^{\frac {7}{2}} - {\left (1200 \, b c^{3} - 73 \, a c d^{2}\right )} {\left (d x + c\right )}^{\frac {5}{2}} + {\left (1104 \, b c^{4} - 55 \, a c^{2} d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} - 3 \, {\left (112 \, b c^{5} - 5 \, a c^{3} d^{2}\right )} \sqrt {d x + c}\right )}}{{\left (d x + c\right )}^{4} c d^{2} - 4 \, {\left (d x + c\right )}^{3} c^{2} d^{2} + 6 \, {\left (d x + c\right )}^{2} c^{3} d^{2} - 4 \, {\left (d x + c\right )} c^{4} d^{2} + c^{5} d^{2}} - \frac {768 \, \sqrt {d x + c} b}{d^{2}} - \frac {15 \, {\left (48 \, b c^{2} - a d^{2}\right )} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{c^{\frac {3}{2}} d^{2}}\right )} \] Input:

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

Output:

-1/384*d^4*(2*(3*(144*b*c^2 + 5*a*d^2)*(d*x + c)^(7/2) - (1200*b*c^3 - 73* 
a*c*d^2)*(d*x + c)^(5/2) + (1104*b*c^4 - 55*a*c^2*d^2)*(d*x + c)^(3/2) - 3 
*(112*b*c^5 - 5*a*c^3*d^2)*sqrt(d*x + c))/((d*x + c)^4*c*d^2 - 4*(d*x + c) 
^3*c^2*d^2 + 6*(d*x + c)^2*c^3*d^2 - 4*(d*x + c)*c^4*d^2 + c^5*d^2) - 768* 
sqrt(d*x + c)*b/d^2 - 15*(48*b*c^2 - a*d^2)*log((sqrt(d*x + c) - sqrt(c))/ 
(sqrt(d*x + c) + sqrt(c)))/(c^(3/2)*d^2))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.23 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^5} \, dx=\frac {384 \, \sqrt {d x + c} b d^{3} + \frac {15 \, {\left (48 \, b c^{2} d^{3} - a d^{5}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c} - \frac {432 \, {\left (d x + c\right )}^{\frac {7}{2}} b c^{2} d^{3} - 1200 \, {\left (d x + c\right )}^{\frac {5}{2}} b c^{3} d^{3} + 1104 \, {\left (d x + c\right )}^{\frac {3}{2}} b c^{4} d^{3} - 336 \, \sqrt {d x + c} b c^{5} d^{3} + 15 \, {\left (d x + c\right )}^{\frac {7}{2}} a d^{5} + 73 \, {\left (d x + c\right )}^{\frac {5}{2}} a c d^{5} - 55 \, {\left (d x + c\right )}^{\frac {3}{2}} a c^{2} d^{5} + 15 \, \sqrt {d x + c} a c^{3} d^{5}}{c d^{4} x^{4}}}{192 \, d} \] Input:

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

Output:

1/192*(384*sqrt(d*x + c)*b*d^3 + 15*(48*b*c^2*d^3 - a*d^5)*arctan(sqrt(d*x 
 + c)/sqrt(-c))/(sqrt(-c)*c) - (432*(d*x + c)^(7/2)*b*c^2*d^3 - 1200*(d*x 
+ c)^(5/2)*b*c^3*d^3 + 1104*(d*x + c)^(3/2)*b*c^4*d^3 - 336*sqrt(d*x + c)* 
b*c^5*d^3 + 15*(d*x + c)^(7/2)*a*d^5 + 73*(d*x + c)^(5/2)*a*c*d^5 - 55*(d* 
x + c)^(3/2)*a*c^2*d^5 + 15*sqrt(d*x + c)*a*c^3*d^5)/(c*d^4*x^4))/d
 

Mupad [B] (verification not implemented)

Time = 8.09 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.25 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^5} \, dx=2\,b\,d^2\,\sqrt {c+d\,x}-\frac {\left (\frac {23\,b\,c^3\,d^2}{4}-\frac {55\,a\,c\,d^4}{192}\right )\,{\left (c+d\,x\right )}^{3/2}+\left (\frac {5\,a\,c^2\,d^4}{64}-\frac {7\,b\,c^4\,d^2}{4}\right )\,\sqrt {c+d\,x}+\left (\frac {73\,a\,d^4}{192}-\frac {25\,b\,c^2\,d^2}{4}\right )\,{\left (c+d\,x\right )}^{5/2}+\frac {\left (144\,b\,c^2\,d^2+5\,a\,d^4\right )\,{\left (c+d\,x\right )}^{7/2}}{64\,c}}{{\left (c+d\,x\right )}^4-4\,c^3\,\left (c+d\,x\right )-4\,c\,{\left (c+d\,x\right )}^3+6\,c^2\,{\left (c+d\,x\right )}^2+c^4}+\frac {5\,d^2\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}}{\sqrt {c}}\right )\,\left (a\,d^2-48\,b\,c^2\right )}{64\,c^{3/2}} \] Input:

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

Output:

2*b*d^2*(c + d*x)^(1/2) - (((23*b*c^3*d^2)/4 - (55*a*c*d^4)/192)*(c + d*x) 
^(3/2) + ((5*a*c^2*d^4)/64 - (7*b*c^4*d^2)/4)*(c + d*x)^(1/2) + ((73*a*d^4 
)/192 - (25*b*c^2*d^2)/4)*(c + d*x)^(5/2) + ((5*a*d^4 + 144*b*c^2*d^2)*(c 
+ d*x)^(7/2))/(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) + (5*d^2*atanh((c + d*x)^(1/2)/c^(1/2))*(a*d^2 - 4 
8*b*c^2))/(64*c^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.34 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x^5} \, dx=\frac {-96 \sqrt {d x +c}\, a \,c^{4}-272 \sqrt {d x +c}\, a \,c^{3} d x -236 \sqrt {d x +c}\, a \,c^{2} d^{2} x^{2}-30 \sqrt {d x +c}\, a c \,d^{3} x^{3}-192 \sqrt {d x +c}\, b \,c^{4} x^{2}-864 \sqrt {d x +c}\, b \,c^{3} d \,x^{3}+768 \sqrt {d x +c}\, b \,c^{2} d^{2} x^{4}-15 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a \,d^{4} x^{4}+720 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) b \,c^{2} d^{2} x^{4}+15 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a \,d^{4} x^{4}-720 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) b \,c^{2} d^{2} x^{4}}{384 c^{2} x^{4}} \] Input:

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

Output:

( - 96*sqrt(c + d*x)*a*c**4 - 272*sqrt(c + d*x)*a*c**3*d*x - 236*sqrt(c + 
d*x)*a*c**2*d**2*x**2 - 30*sqrt(c + d*x)*a*c*d**3*x**3 - 192*sqrt(c + d*x) 
*b*c**4*x**2 - 864*sqrt(c + d*x)*b*c**3*d*x**3 + 768*sqrt(c + d*x)*b*c**2* 
d**2*x**4 - 15*sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*a*d**4*x**4 + 720*sqrt 
(c)*log(sqrt(c + d*x) - sqrt(c))*b*c**2*d**2*x**4 + 15*sqrt(c)*log(sqrt(c 
+ d*x) + sqrt(c))*a*d**4*x**4 - 720*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*b 
*c**2*d**2*x**4)/(384*c**2*x**4)