\(\int \frac {a+b x^2}{x^5 (c+d x)^{3/2}} \, dx\) [506]

Optimal result

Integrand size = 20, antiderivative size = 176 \[ \int \frac {a+b x^2}{x^5 (c+d x)^{3/2}} \, dx=\frac {2 d^2 \left (b c^2+a d^2\right )}{c^5 \sqrt {c+d x}}-\frac {a \sqrt {c+d x}}{4 c^2 x^4}+\frac {5 a d \sqrt {c+d x}}{8 c^3 x^3}-\frac {\left (16 b c^2+41 a d^2\right ) \sqrt {c+d x}}{32 c^4 x^2}+\frac {d \left (112 b c^2+187 a d^2\right ) \sqrt {c+d x}}{64 c^5 x}-\frac {15 d^2 \left (16 b c^2+21 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{64 c^{11/2}} \] Output:

2*d^2*(a*d^2+b*c^2)/c^5/(d*x+c)^(1/2)-1/4*a*(d*x+c)^(1/2)/c^2/x^4+5/8*a*d* 
(d*x+c)^(1/2)/c^3/x^3-1/32*(41*a*d^2+16*b*c^2)*(d*x+c)^(1/2)/c^4/x^2+1/64* 
d*(187*a*d^2+112*b*c^2)*(d*x+c)^(1/2)/c^5/x-15/64*d^2*(21*a*d^2+16*b*c^2)* 
arctanh((d*x+c)^(1/2)/c^(1/2))/c^(11/2)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.76 \[ \int \frac {a+b x^2}{x^5 (c+d x)^{3/2}} \, dx=\frac {\frac {\sqrt {c} \left (16 b c^2 x^2 \left (-2 c^2+5 c d x+15 d^2 x^2\right )+a \left (-16 c^4+24 c^3 d x-42 c^2 d^2 x^2+105 c d^3 x^3+315 d^4 x^4\right )\right )}{x^4 \sqrt {c+d x}}-15 d^2 \left (16 b c^2+21 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{64 c^{11/2}} \] Input:

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

Output:

((Sqrt[c]*(16*b*c^2*x^2*(-2*c^2 + 5*c*d*x + 15*d^2*x^2) + a*(-16*c^4 + 24* 
c^3*d*x - 42*c^2*d^2*x^2 + 105*c*d^3*x^3 + 315*d^4*x^4)))/(x^4*Sqrt[c + d* 
x]) - 15*d^2*(16*b*c^2 + 21*a*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(64*c^( 
11/2))
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.19, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.650, Rules used = {517, 25, 1582, 25, 361, 27, 361, 25, 361, 25, 27, 359, 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.

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

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1582

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 361

\(\displaystyle 2 d^2 \left (-\frac {-\frac {1}{6} \int \frac {3 \left (25 a (c+d x) d^2+16 c \left (b c^2+a d^2\right )\right )}{c d^3 x^3 (c+d x)}d\sqrt {c+d x}-\frac {5 a \sqrt {c+d x}}{2 c d x^3}}{8 c^2}-\frac {a \sqrt {c+d x}}{8 c^2 d^2 x^4}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 d^2 \left (-\frac {\frac {\int -\frac {25 a (c+d x) d^2+16 c \left (b c^2+a d^2\right )}{d^3 x^3 (c+d x)}d\sqrt {c+d x}}{2 c}-\frac {5 a \sqrt {c+d x}}{2 c d x^3}}{8 c^2}-\frac {a \sqrt {c+d x}}{8 c^2 d^2 x^4}\right )\)

\(\Big \downarrow \) 361

\(\displaystyle 2 d^2 \left (-\frac {\frac {\frac {\sqrt {c+d x} \left (41 a d^2+16 b c^2\right )}{4 c d^2 x^2}-\frac {1}{4} \int -\frac {64 \left (b c^2+a d^2\right )+3 \left (\frac {41 a d^2}{c}+16 b c\right ) (c+d x)}{d^2 x^2 (c+d x)}d\sqrt {c+d x}}{2 c}-\frac {5 a \sqrt {c+d x}}{2 c d x^3}}{8 c^2}-\frac {a \sqrt {c+d x}}{8 c^2 d^2 x^4}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 2 d^2 \left (-\frac {\frac {\frac {1}{4} \int \frac {64 \left (b c^2+a d^2\right )+3 \left (\frac {41 a d^2}{c}+16 b c\right ) (c+d x)}{d^2 x^2 (c+d x)}d\sqrt {c+d x}+\frac {\sqrt {c+d x} \left (41 a d^2+16 b c^2\right )}{4 c d^2 x^2}}{2 c}-\frac {5 a \sqrt {c+d x}}{2 c d x^3}}{8 c^2}-\frac {a \sqrt {c+d x}}{8 c^2 d^2 x^4}\right )\)

\(\Big \downarrow \) 361

\(\displaystyle 2 d^2 \left (-\frac {\frac {\frac {1}{4} \left (-\frac {1}{2} \int \frac {128 \left (b c^2+a d^2\right )+c \left (\frac {187 a d^2}{c^2}+112 b\right ) (c+d x)}{c d x (c+d x)}d\sqrt {c+d x}-\frac {\sqrt {c+d x} \left (\frac {187 a d^2}{c^2}+112 b\right )}{2 d x}\right )+\frac {\sqrt {c+d x} \left (41 a d^2+16 b c^2\right )}{4 c d^2 x^2}}{2 c}-\frac {5 a \sqrt {c+d x}}{2 c d x^3}}{8 c^2}-\frac {a \sqrt {c+d x}}{8 c^2 d^2 x^4}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 2 d^2 \left (-\frac {\frac {\frac {1}{4} \left (\frac {1}{2} \int -\frac {128 \left (b c^2+a d^2\right )+c \left (\frac {187 a d^2}{c^2}+112 b\right ) (c+d x)}{c d x (c+d x)}d\sqrt {c+d x}-\frac {\sqrt {c+d x} \left (\frac {187 a d^2}{c^2}+112 b\right )}{2 d x}\right )+\frac {\sqrt {c+d x} \left (41 a d^2+16 b c^2\right )}{4 c d^2 x^2}}{2 c}-\frac {5 a \sqrt {c+d x}}{2 c d x^3}}{8 c^2}-\frac {a \sqrt {c+d x}}{8 c^2 d^2 x^4}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 d^2 \left (-\frac {\frac {\frac {1}{4} \left (\frac {\int -\frac {128 \left (b c^2+a d^2\right )+c \left (\frac {187 a d^2}{c^2}+112 b\right ) (c+d x)}{d x (c+d x)}d\sqrt {c+d x}}{2 c}-\frac {\sqrt {c+d x} \left (\frac {187 a d^2}{c^2}+112 b\right )}{2 d x}\right )+\frac {\sqrt {c+d x} \left (41 a d^2+16 b c^2\right )}{4 c d^2 x^2}}{2 c}-\frac {5 a \sqrt {c+d x}}{2 c d x^3}}{8 c^2}-\frac {a \sqrt {c+d x}}{8 c^2 d^2 x^4}\right )\)

\(\Big \downarrow \) 359

\(\displaystyle 2 d^2 \left (-\frac {\frac {\frac {1}{4} \left (\frac {\frac {15 \left (21 a d^2+16 b c^2\right ) \int -\frac {1}{d x}d\sqrt {c+d x}}{c}-\frac {128 \left (a d^2+b c^2\right )}{c \sqrt {c+d x}}}{2 c}-\frac {\sqrt {c+d x} \left (\frac {187 a d^2}{c^2}+112 b\right )}{2 d x}\right )+\frac {\sqrt {c+d x} \left (41 a d^2+16 b c^2\right )}{4 c d^2 x^2}}{2 c}-\frac {5 a \sqrt {c+d x}}{2 c d x^3}}{8 c^2}-\frac {a \sqrt {c+d x}}{8 c^2 d^2 x^4}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle 2 d^2 \left (-\frac {\frac {\frac {1}{4} \left (\frac {\frac {15 \left (21 a d^2+16 b c^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{3/2}}-\frac {128 \left (a d^2+b c^2\right )}{c \sqrt {c+d x}}}{2 c}-\frac {\sqrt {c+d x} \left (\frac {187 a d^2}{c^2}+112 b\right )}{2 d x}\right )+\frac {\sqrt {c+d x} \left (41 a d^2+16 b c^2\right )}{4 c d^2 x^2}}{2 c}-\frac {5 a \sqrt {c+d x}}{2 c d x^3}}{8 c^2}-\frac {a \sqrt {c+d x}}{8 c^2 d^2 x^4}\right )\)

Input:

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

Output:

2*d^2*(-1/8*(a*Sqrt[c + d*x])/(c^2*d^2*x^4) - ((-5*a*Sqrt[c + d*x])/(2*c*d 
*x^3) + (((16*b*c^2 + 41*a*d^2)*Sqrt[c + d*x])/(4*c*d^2*x^2) + (-1/2*((112 
*b + (187*a*d^2)/c^2)*Sqrt[c + d*x])/(d*x) + ((-128*(b*c^2 + a*d^2))/(c*Sq 
rt[c + d*x]) + (15*(16*b*c^2 + 21*a*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/c 
^(3/2))/(2*c))/4)/(2*c))/(8*c^2))
 

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 359

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

rule 361

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : 
> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p 
+ 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1))   Int[x^m*(a + b*x^2)^(p + 1)*E 
xpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c 
- a*d)*x^(-m + 2))/(a + b*x^2)] - ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], 
 x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ILtQ[m/ 
2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 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 1582

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

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.71

method	result	size
pseudoelliptic	\(\frac {-\frac {315 d^{2} x^{4} \left (a \,d^{2}+\frac {16 b \,c^{2}}{21}\right ) \sqrt {d x +c}\, \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{64}-\frac {21 d^{2} x^{2} \left (-\frac {40 b \,x^{2}}{7}+a \right ) c^{\frac {5}{2}}}{32}+\frac {3 x d \left (\frac {10 b \,x^{2}}{3}+a \right ) c^{\frac {7}{2}}}{8}+\frac {\left (-2 b \,x^{2}-a \right ) c^{\frac {9}{2}}}{4}+\frac {315 d^{3} x^{3} \left (d x \sqrt {c}+\frac {c^{\frac {3}{2}}}{3}\right ) a}{64}}{c^{\frac {11}{2}} \sqrt {d x +c}\, x^{4}}\)	\(125\)
risch	\(-\frac {\sqrt {d x +c}\, \left (-187 a \,x^{3} d^{3}-112 b \,c^{2} d \,x^{3}+82 a \,d^{2} x^{2} c +32 b \,c^{3} x^{2}-40 a d x \,c^{2}+16 c^{3} a \right )}{64 c^{5} x^{4}}+\frac {d^{2} \left (-\frac {2 \left (-128 a \,d^{2}-128 b \,c^{2}\right )}{\sqrt {d x +c}}-\frac {2 \left (315 a \,d^{2}+240 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\sqrt {c}}\right )}{128 c^{5}}\)	\(131\)
derivativedivides	\(2 d^{2} \left (-\frac {\frac {\left (-\frac {187 a \,d^{2}}{128}-\frac {7 b \,c^{2}}{8}\right ) \left (d x +c \right )^{\frac {7}{2}}+\left (\frac {643}{128} a \,d^{2} c +\frac {23}{8} b \,c^{3}\right ) \left (d x +c \right )^{\frac {5}{2}}+\left (-\frac {765}{128} a \,c^{2} d^{2}-\frac {25}{8} b \,c^{4}\right ) \left (d x +c \right )^{\frac {3}{2}}+\left (\frac {325}{128} a \,c^{3} d^{2}+\frac {9}{8} b \,c^{5}\right ) \sqrt {d x +c}}{d^{4} x^{4}}+\frac {15 \left (21 a \,d^{2}+16 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{128 \sqrt {c}}}{c^{5}}-\frac {-a \,d^{2}-b \,c^{2}}{c^{5} \sqrt {d x +c}}\right )\)	\(167\)
default	\(2 d^{2} \left (-\frac {\frac {\left (-\frac {187 a \,d^{2}}{128}-\frac {7 b \,c^{2}}{8}\right ) \left (d x +c \right )^{\frac {7}{2}}+\left (\frac {643}{128} a \,d^{2} c +\frac {23}{8} b \,c^{3}\right ) \left (d x +c \right )^{\frac {5}{2}}+\left (-\frac {765}{128} a \,c^{2} d^{2}-\frac {25}{8} b \,c^{4}\right ) \left (d x +c \right )^{\frac {3}{2}}+\left (\frac {325}{128} a \,c^{3} d^{2}+\frac {9}{8} b \,c^{5}\right ) \sqrt {d x +c}}{d^{4} x^{4}}+\frac {15 \left (21 a \,d^{2}+16 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{128 \sqrt {c}}}{c^{5}}-\frac {-a \,d^{2}-b \,c^{2}}{c^{5} \sqrt {d x +c}}\right )\)	\(167\)

Input:

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

Output:

3/8/c^(11/2)*(-105/8*d^2*x^4*(a*d^2+16/21*b*c^2)*(d*x+c)^(1/2)*arctanh((d* 
x+c)^(1/2)/c^(1/2))-7/4*d^2*x^2*(-40/7*b*x^2+a)*c^(5/2)+x*d*(10/3*b*x^2+a) 
*c^(7/2)+2/3*(-2*b*x^2-a)*c^(9/2)+105/8*d^3*x^3*(d*x*c^(1/2)+1/3*c^(3/2))* 
a)/(d*x+c)^(1/2)/x^4
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.01 \[ \int \frac {a+b x^2}{x^5 (c+d x)^{3/2}} \, dx=\left [\frac {15 \, {\left ({\left (16 \, b c^{2} d^{3} + 21 \, a d^{5}\right )} x^{5} + {\left (16 \, b c^{3} d^{2} + 21 \, a c d^{4}\right )} x^{4}\right )} \sqrt {c} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (24 \, a c^{4} d x - 16 \, a c^{5} + 15 \, {\left (16 \, b c^{3} d^{2} + 21 \, a c d^{4}\right )} x^{4} + 5 \, {\left (16 \, b c^{4} d + 21 \, a c^{2} d^{3}\right )} x^{3} - 2 \, {\left (16 \, b c^{5} + 21 \, a c^{3} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{128 \, {\left (c^{6} d x^{5} + c^{7} x^{4}\right )}}, \frac {15 \, {\left ({\left (16 \, b c^{2} d^{3} + 21 \, a d^{5}\right )} x^{5} + {\left (16 \, b c^{3} d^{2} + 21 \, a c d^{4}\right )} x^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) + {\left (24 \, a c^{4} d x - 16 \, a c^{5} + 15 \, {\left (16 \, b c^{3} d^{2} + 21 \, a c d^{4}\right )} x^{4} + 5 \, {\left (16 \, b c^{4} d + 21 \, a c^{2} d^{3}\right )} x^{3} - 2 \, {\left (16 \, b c^{5} + 21 \, a c^{3} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{64 \, {\left (c^{6} d x^{5} + c^{7} x^{4}\right )}}\right ] \] Input:

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

Output:

[1/128*(15*((16*b*c^2*d^3 + 21*a*d^5)*x^5 + (16*b*c^3*d^2 + 21*a*c*d^4)*x^ 
4)*sqrt(c)*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*(24*a*c^4*d*x 
- 16*a*c^5 + 15*(16*b*c^3*d^2 + 21*a*c*d^4)*x^4 + 5*(16*b*c^4*d + 21*a*c^2 
*d^3)*x^3 - 2*(16*b*c^5 + 21*a*c^3*d^2)*x^2)*sqrt(d*x + c))/(c^6*d*x^5 + c 
^7*x^4), 1/64*(15*((16*b*c^2*d^3 + 21*a*d^5)*x^5 + (16*b*c^3*d^2 + 21*a*c* 
d^4)*x^4)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x + c)) + (24*a*c^4*d*x - 16*a*c 
^5 + 15*(16*b*c^3*d^2 + 21*a*c*d^4)*x^4 + 5*(16*b*c^4*d + 21*a*c^2*d^3)*x^ 
3 - 2*(16*b*c^5 + 21*a*c^3*d^2)*x^2)*sqrt(d*x + c))/(c^6*d*x^5 + c^7*x^4)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.38 \[ \int \frac {a+b x^2}{x^5 (c+d x)^{3/2}} \, dx=\frac {1}{128} \, d^{4} {\left (\frac {2 \, {\left (128 \, b c^{6} + 128 \, a c^{4} d^{2} + 15 \, {\left (16 \, b c^{2} + 21 \, a d^{2}\right )} {\left (d x + c\right )}^{4} - 55 \, {\left (16 \, b c^{3} + 21 \, a c d^{2}\right )} {\left (d x + c\right )}^{3} + 73 \, {\left (16 \, b c^{4} + 21 \, a c^{2} d^{2}\right )} {\left (d x + c\right )}^{2} - {\left (656 \, b c^{5} + 837 \, a c^{3} d^{2}\right )} {\left (d x + c\right )}\right )}}{{\left (d x + c\right )}^{\frac {9}{2}} c^{5} d^{2} - 4 \, {\left (d x + c\right )}^{\frac {7}{2}} c^{6} d^{2} + 6 \, {\left (d x + c\right )}^{\frac {5}{2}} c^{7} d^{2} - 4 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{8} d^{2} + \sqrt {d x + c} c^{9} d^{2}} + \frac {15 \, {\left (16 \, b c^{2} + 21 \, a d^{2}\right )} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{c^{\frac {11}{2}} d^{2}}\right )} \] Input:

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

Output:

1/128*d^4*(2*(128*b*c^6 + 128*a*c^4*d^2 + 15*(16*b*c^2 + 21*a*d^2)*(d*x + 
c)^4 - 55*(16*b*c^3 + 21*a*c*d^2)*(d*x + c)^3 + 73*(16*b*c^4 + 21*a*c^2*d^ 
2)*(d*x + c)^2 - (656*b*c^5 + 837*a*c^3*d^2)*(d*x + c))/((d*x + c)^(9/2)*c 
^5*d^2 - 4*(d*x + c)^(7/2)*c^6*d^2 + 6*(d*x + c)^(5/2)*c^7*d^2 - 4*(d*x + 
c)^(3/2)*c^8*d^2 + sqrt(d*x + c)*c^9*d^2) + 15*(16*b*c^2 + 21*a*d^2)*log(( 
sqrt(d*x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c)))/(c^(11/2)*d^2))
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.15 \[ \int \frac {a+b x^2}{x^5 (c+d x)^{3/2}} \, dx=\frac {15 \, {\left (16 \, b c^{2} d^{2} + 21 \, a d^{4}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{64 \, \sqrt {-c} c^{5}} + \frac {2 \, {\left (b c^{2} d^{2} + a d^{4}\right )}}{\sqrt {d x + c} c^{5}} + \frac {112 \, {\left (d x + c\right )}^{\frac {7}{2}} b c^{2} d^{2} - 368 \, {\left (d x + c\right )}^{\frac {5}{2}} b c^{3} d^{2} + 400 \, {\left (d x + c\right )}^{\frac {3}{2}} b c^{4} d^{2} - 144 \, \sqrt {d x + c} b c^{5} d^{2} + 187 \, {\left (d x + c\right )}^{\frac {7}{2}} a d^{4} - 643 \, {\left (d x + c\right )}^{\frac {5}{2}} a c d^{4} + 765 \, {\left (d x + c\right )}^{\frac {3}{2}} a c^{2} d^{4} - 325 \, \sqrt {d x + c} a c^{3} d^{4}}{64 \, c^{5} d^{4} x^{4}} \] Input:

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

Output:

15/64*(16*b*c^2*d^2 + 21*a*d^4)*arctan(sqrt(d*x + c)/sqrt(-c))/(sqrt(-c)*c 
^5) + 2*(b*c^2*d^2 + a*d^4)/(sqrt(d*x + c)*c^5) + 1/64*(112*(d*x + c)^(7/2 
)*b*c^2*d^2 - 368*(d*x + c)^(5/2)*b*c^3*d^2 + 400*(d*x + c)^(3/2)*b*c^4*d^ 
2 - 144*sqrt(d*x + c)*b*c^5*d^2 + 187*(d*x + c)^(7/2)*a*d^4 - 643*(d*x + c 
)^(5/2)*a*c*d^4 + 765*(d*x + c)^(3/2)*a*c^2*d^4 - 325*sqrt(d*x + c)*a*c^3* 
d^4)/(c^5*d^4*x^4)
 

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.25 \[ \int \frac {a+b x^2}{x^5 (c+d x)^{3/2}} \, dx=\frac {\frac {2\,\left (b\,c^2\,d^2+a\,d^4\right )}{c}-\frac {\left (656\,b\,c^2\,d^2+837\,a\,d^4\right )\,\left (c+d\,x\right )}{64\,c^2}+\frac {73\,\left (16\,b\,c^2\,d^2+21\,a\,d^4\right )\,{\left (c+d\,x\right )}^2}{64\,c^3}-\frac {55\,\left (16\,b\,c^2\,d^2+21\,a\,d^4\right )\,{\left (c+d\,x\right )}^3}{64\,c^4}+\frac {15\,\left (16\,b\,c^2\,d^2+21\,a\,d^4\right )\,{\left (c+d\,x\right )}^4}{64\,c^5}}{{\left (c+d\,x\right )}^{9/2}-4\,c\,{\left (c+d\,x\right )}^{7/2}+c^4\,\sqrt {c+d\,x}-4\,c^3\,{\left (c+d\,x\right )}^{3/2}+6\,c^2\,{\left (c+d\,x\right )}^{5/2}}-\frac {15\,d^2\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}}{\sqrt {c}}\right )\,\left (16\,b\,c^2+21\,a\,d^2\right )}{64\,c^{11/2}} \] Input:

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

Output:

((2*(a*d^4 + b*c^2*d^2))/c - ((837*a*d^4 + 656*b*c^2*d^2)*(c + d*x))/(64*c 
^2) + (73*(21*a*d^4 + 16*b*c^2*d^2)*(c + d*x)^2)/(64*c^3) - (55*(21*a*d^4 
+ 16*b*c^2*d^2)*(c + d*x)^3)/(64*c^4) + (15*(21*a*d^4 + 16*b*c^2*d^2)*(c + 
 d*x)^4)/(64*c^5))/((c + d*x)^(9/2) - 4*c*(c + d*x)^(7/2) + c^4*(c + d*x)^ 
(1/2) - 4*c^3*(c + d*x)^(3/2) + 6*c^2*(c + d*x)^(5/2)) - (15*d^2*atanh((c 
+ d*x)^(1/2)/c^(1/2))*(21*a*d^2 + 16*b*c^2))/(64*c^(11/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.22 \[ \int \frac {a+b x^2}{x^5 (c+d x)^{3/2}} \, dx=\frac {315 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a \,d^{4} x^{4}+240 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) b \,c^{2} d^{2} x^{4}-315 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a \,d^{4} x^{4}-240 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) b \,c^{2} d^{2} x^{4}-32 a \,c^{5}+48 a \,c^{4} d x -84 a \,c^{3} d^{2} x^{2}+210 a \,c^{2} d^{3} x^{3}+630 a c \,d^{4} x^{4}-64 b \,c^{5} x^{2}+160 b \,c^{4} d \,x^{3}+480 b \,c^{3} d^{2} x^{4}}{128 \sqrt {d x +c}\, c^{6} x^{4}} \] Input:

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

Output:

(315*sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*a*d**4*x**4 + 240* 
sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*b*c**2*d**2*x**4 - 315* 
sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) + sqrt(c))*a*d**4*x**4 - 240*sqrt( 
c)*sqrt(c + d*x)*log(sqrt(c + d*x) + sqrt(c))*b*c**2*d**2*x**4 - 32*a*c**5 
 + 48*a*c**4*d*x - 84*a*c**3*d**2*x**2 + 210*a*c**2*d**3*x**3 + 630*a*c*d* 
*4*x**4 - 64*b*c**5*x**2 + 160*b*c**4*d*x**3 + 480*b*c**3*d**2*x**4)/(128* 
sqrt(c + d*x)*c**6*x**4)