\(\int \frac {a+b x^2}{x^4 (c+d x)^{3/2}} \, dx\) [505]

Optimal result

Integrand size = 20, antiderivative size = 139 \[ \int \frac {a+b x^2}{x^4 (c+d x)^{3/2}} \, dx=-\frac {2 d \left (b c^2+a d^2\right )}{c^4 \sqrt {c+d x}}-\frac {a \sqrt {c+d x}}{3 c^2 x^3}+\frac {11 a d \sqrt {c+d x}}{12 c^3 x^2}-\frac {\left (8 b c^2+19 a d^2\right ) \sqrt {c+d x}}{8 c^4 x}+\frac {d \left (24 b c^2+35 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{8 c^{9/2}} \] Output:

-2*d*(a*d^2+b*c^2)/c^4/(d*x+c)^(1/2)-1/3*a*(d*x+c)^(1/2)/c^2/x^3+11/12*a*d 
*(d*x+c)^(1/2)/c^3/x^2-1/8*(19*a*d^2+8*b*c^2)*(d*x+c)^(1/2)/c^4/x+1/8*d*(3 
5*a*d^2+24*b*c^2)*arctanh((d*x+c)^(1/2)/c^(1/2))/c^(9/2)
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.77 \[ \int \frac {a+b x^2}{x^4 (c+d x)^{3/2}} \, dx=-\frac {24 b c^2 x^2 (c+3 d x)+a \left (8 c^3-14 c^2 d x+35 c d^2 x^2+105 d^3 x^3\right )}{24 c^4 x^3 \sqrt {c+d x}}+\frac {d \left (24 b c^2+35 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{8 c^{9/2}} \] Input:

Integrate[(a + b*x^2)/(x^4*(c + d*x)^(3/2)),x]
 

Output:

-1/24*(24*b*c^2*x^2*(c + 3*d*x) + a*(8*c^3 - 14*c^2*d*x + 35*c*d^2*x^2 + 1 
05*d^3*x^3))/(c^4*x^3*Sqrt[c + d*x]) + (d*(24*b*c^2 + 35*a*d^2)*ArcTanh[Sq 
rt[c + d*x]/Sqrt[c]])/(8*c^(9/2))
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.17, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {517, 1582, 25, 361, 27, 361, 25, 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.

\(\displaystyle \int \frac {a+b x^2}{x^4 (c+d x)^{3/2}} \, dx\)

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 1582

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 361

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 361

\(\displaystyle 2 d \left (\frac {\frac {3 \left (-\frac {1}{2} \int \frac {16 \left (b c^2+a d^2\right )+\left (\frac {19 a d^2}{c}+8 b c\right ) (c+d x)}{d x (c+d x)}d\sqrt {c+d x}-\frac {\sqrt {c+d x} \left (19 a d^2+8 b c^2\right )}{2 c d x}\right )}{4 c}+\frac {11 a \sqrt {c+d x}}{4 c x^2}}{6 c^2}-\frac {a \sqrt {c+d x}}{6 c^2 d x^3}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 2 d \left (\frac {\frac {3 \left (\frac {1}{2} \int -\frac {16 \left (b c^2+a d^2\right )+\left (\frac {19 a d^2}{c}+8 b c\right ) (c+d x)}{d x (c+d x)}d\sqrt {c+d x}-\frac {\sqrt {c+d x} \left (19 a d^2+8 b c^2\right )}{2 c d x}\right )}{4 c}+\frac {11 a \sqrt {c+d x}}{4 c x^2}}{6 c^2}-\frac {a \sqrt {c+d x}}{6 c^2 d x^3}\right )\)

\(\Big \downarrow \) 359

\(\displaystyle 2 d \left (\frac {\frac {3 \left (\frac {1}{2} \left (\frac {\left (35 a d^2+24 b c^2\right ) \int -\frac {1}{d x}d\sqrt {c+d x}}{c}-\frac {16 \left (a d^2+b c^2\right )}{c \sqrt {c+d x}}\right )-\frac {\sqrt {c+d x} \left (19 a d^2+8 b c^2\right )}{2 c d x}\right )}{4 c}+\frac {11 a \sqrt {c+d x}}{4 c x^2}}{6 c^2}-\frac {a \sqrt {c+d x}}{6 c^2 d x^3}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle 2 d \left (\frac {\frac {3 \left (\frac {1}{2} \left (\frac {\left (35 a d^2+24 b c^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{3/2}}-\frac {16 \left (a d^2+b c^2\right )}{c \sqrt {c+d x}}\right )-\frac {\sqrt {c+d x} \left (19 a d^2+8 b c^2\right )}{2 c d x}\right )}{4 c}+\frac {11 a \sqrt {c+d x}}{4 c x^2}}{6 c^2}-\frac {a \sqrt {c+d x}}{6 c^2 d x^3}\right )\)

Input:

Int[(a + b*x^2)/(x^4*(c + d*x)^(3/2)),x]
 

Output:

2*d*(-1/6*(a*Sqrt[c + d*x])/(c^2*d*x^3) + ((11*a*Sqrt[c + d*x])/(4*c*x^2) 
+ (3*(-1/2*((8*b*c^2 + 19*a*d^2)*Sqrt[c + d*x])/(c*d*x) + ((-16*(b*c^2 + a 
*d^2))/(c*Sqrt[c + d*x]) + ((24*b*c^2 + 35*a*d^2)*ArcTanh[Sqrt[c + d*x]/Sq 
rt[c]])/c^(3/2))/2))/(4*c))/(6*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 = 0.28 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.75

method	result	size
pseudoelliptic	\(\frac {\frac {35 d \,x^{3} \sqrt {d x +c}\, \left (a \,d^{2}+\frac {24 b \,c^{2}}{35}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{8}+\frac {7 d x \left (-\frac {36 b \,x^{2}}{7}+a \right ) c^{\frac {5}{2}}}{12}+\frac {\left (-3 b \,x^{2}-a \right ) c^{\frac {7}{2}}}{3}-\frac {35 d^{2} x^{2} \left (d x \sqrt {c}+\frac {c^{\frac {3}{2}}}{3}\right ) a}{8}}{c^{\frac {9}{2}} \sqrt {d x +c}\, x^{3}}\)	\(104\)
risch	\(-\frac {\sqrt {d x +c}\, \left (57 a \,d^{2} x^{2}+24 b \,c^{2} x^{2}-22 a d x c +8 a \,c^{2}\right )}{24 c^{4} x^{3}}-\frac {d \left (-\frac {2 \left (-16 a \,d^{2}-16 b \,c^{2}\right )}{\sqrt {d x +c}}-\frac {2 \left (35 a \,d^{2}+24 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\sqrt {c}}\right )}{16 c^{4}}\)	\(107\)
derivativedivides	\(2 d \left (\frac {-\frac {\left (\frac {19 a \,d^{2}}{16}+\frac {b \,c^{2}}{2}\right ) \left (d x +c \right )^{\frac {5}{2}}+\left (-\frac {17}{6} a \,d^{2} c -b \,c^{3}\right ) \left (d x +c \right )^{\frac {3}{2}}+\left (\frac {29}{16} a \,c^{2} d^{2}+\frac {1}{2} b \,c^{4}\right ) \sqrt {d x +c}}{d^{3} x^{3}}+\frac {\left (35 a \,d^{2}+24 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{16 \sqrt {c}}}{c^{4}}-\frac {a \,d^{2}+b \,c^{2}}{c^{4} \sqrt {d x +c}}\right )\)	\(139\)
default	\(2 d \left (\frac {-\frac {\left (\frac {19 a \,d^{2}}{16}+\frac {b \,c^{2}}{2}\right ) \left (d x +c \right )^{\frac {5}{2}}+\left (-\frac {17}{6} a \,d^{2} c -b \,c^{3}\right ) \left (d x +c \right )^{\frac {3}{2}}+\left (\frac {29}{16} a \,c^{2} d^{2}+\frac {1}{2} b \,c^{4}\right ) \sqrt {d x +c}}{d^{3} x^{3}}+\frac {\left (35 a \,d^{2}+24 b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{16 \sqrt {c}}}{c^{4}}-\frac {a \,d^{2}+b \,c^{2}}{c^{4} \sqrt {d x +c}}\right )\)	\(139\)

Input:

int((b*x^2+a)/x^4/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

7/12*(15/2*d*x^3*(d*x+c)^(1/2)*(a*d^2+24/35*b*c^2)*arctanh((d*x+c)^(1/2)/c 
^(1/2))+d*x*(-36/7*b*x^2+a)*c^(5/2)+4/7*(-3*b*x^2-a)*c^(7/2)-15/2*d^2*x^2* 
(d*x*c^(1/2)+1/3*c^(3/2))*a)/(d*x+c)^(1/2)/c^(9/2)/x^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.18 \[ \int \frac {a+b x^2}{x^4 (c+d x)^{3/2}} \, dx=\left [\frac {3 \, {\left ({\left (24 \, b c^{2} d^{2} + 35 \, a d^{4}\right )} x^{4} + {\left (24 \, b c^{3} d + 35 \, a c d^{3}\right )} x^{3}\right )} \sqrt {c} \log \left (\frac {d x + 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (14 \, a c^{3} d x - 8 \, a c^{4} - 3 \, {\left (24 \, b c^{3} d + 35 \, a c d^{3}\right )} x^{3} - {\left (24 \, b c^{4} + 35 \, a c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{48 \, {\left (c^{5} d x^{4} + c^{6} x^{3}\right )}}, -\frac {3 \, {\left ({\left (24 \, b c^{2} d^{2} + 35 \, a d^{4}\right )} x^{4} + {\left (24 \, b c^{3} d + 35 \, a c d^{3}\right )} x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) - {\left (14 \, a c^{3} d x - 8 \, a c^{4} - 3 \, {\left (24 \, b c^{3} d + 35 \, a c d^{3}\right )} x^{3} - {\left (24 \, b c^{4} + 35 \, a c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x + c}}{24 \, {\left (c^{5} d x^{4} + c^{6} x^{3}\right )}}\right ] \] Input:

integrate((b*x^2+a)/x^4/(d*x+c)^(3/2),x, algorithm="fricas")
 

Output:

[1/48*(3*((24*b*c^2*d^2 + 35*a*d^4)*x^4 + (24*b*c^3*d + 35*a*c*d^3)*x^3)*s 
qrt(c)*log((d*x + 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*(14*a*c^3*d*x - 8* 
a*c^4 - 3*(24*b*c^3*d + 35*a*c*d^3)*x^3 - (24*b*c^4 + 35*a*c^2*d^2)*x^2)*s 
qrt(d*x + c))/(c^5*d*x^4 + c^6*x^3), -1/24*(3*((24*b*c^2*d^2 + 35*a*d^4)*x 
^4 + (24*b*c^3*d + 35*a*c*d^3)*x^3)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x + c) 
) - (14*a*c^3*d*x - 8*a*c^4 - 3*(24*b*c^3*d + 35*a*c*d^3)*x^3 - (24*b*c^4 
+ 35*a*c^2*d^2)*x^2)*sqrt(d*x + c))/(c^5*d*x^4 + c^6*x^3)]
 

Sympy [A] (verification not implemented)

Time = 103.97 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.53 \[ \int \frac {a+b x^2}{x^4 (c+d x)^{3/2}} \, dx=a \left (- \frac {1}{3 c \sqrt {d} x^{\frac {7}{2}} \sqrt {\frac {c}{d x} + 1}} + \frac {7 \sqrt {d}}{12 c^{2} x^{\frac {5}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {35 d^{\frac {3}{2}}}{24 c^{3} x^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {35 d^{\frac {5}{2}}}{8 c^{4} \sqrt {x} \sqrt {\frac {c}{d x} + 1}} + \frac {35 d^{3} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{8 c^{\frac {9}{2}}}\right ) + b \left (- \frac {1}{c \sqrt {d} x^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}} - \frac {3 \sqrt {d}}{c^{2} \sqrt {x} \sqrt {\frac {c}{d x} + 1}} + \frac {3 d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{c^{\frac {5}{2}}}\right ) \] Input:

integrate((b*x**2+a)/x**4/(d*x+c)**(3/2),x)
 

Output:

a*(-1/(3*c*sqrt(d)*x**(7/2)*sqrt(c/(d*x) + 1)) + 7*sqrt(d)/(12*c**2*x**(5/ 
2)*sqrt(c/(d*x) + 1)) - 35*d**(3/2)/(24*c**3*x**(3/2)*sqrt(c/(d*x) + 1)) - 
 35*d**(5/2)/(8*c**4*sqrt(x)*sqrt(c/(d*x) + 1)) + 35*d**3*asinh(sqrt(c)/(s 
qrt(d)*sqrt(x)))/(8*c**(9/2))) + b*(-1/(c*sqrt(d)*x**(3/2)*sqrt(c/(d*x) + 
1)) - 3*sqrt(d)/(c**2*sqrt(x)*sqrt(c/(d*x) + 1)) + 3*d*asinh(sqrt(c)/(sqrt 
(d)*sqrt(x)))/c**(5/2))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.46 \[ \int \frac {a+b x^2}{x^4 (c+d x)^{3/2}} \, dx=\frac {1}{48} \, d^{3} {\left (\frac {2 \, {\left (48 \, b c^{5} + 48 \, a c^{3} d^{2} - 3 \, {\left (24 \, b c^{2} + 35 \, a d^{2}\right )} {\left (d x + c\right )}^{3} + 8 \, {\left (24 \, b c^{3} + 35 \, a c d^{2}\right )} {\left (d x + c\right )}^{2} - 21 \, {\left (8 \, b c^{4} + 11 \, a c^{2} d^{2}\right )} {\left (d x + c\right )}\right )}}{{\left (d x + c\right )}^{\frac {7}{2}} c^{4} d^{2} - 3 \, {\left (d x + c\right )}^{\frac {5}{2}} c^{5} d^{2} + 3 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{6} d^{2} - \sqrt {d x + c} c^{7} d^{2}} - \frac {3 \, {\left (24 \, b c^{2} + 35 \, a d^{2}\right )} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{c^{\frac {9}{2}} d^{2}}\right )} \] Input:

integrate((b*x^2+a)/x^4/(d*x+c)^(3/2),x, algorithm="maxima")
 

Output:

1/48*d^3*(2*(48*b*c^5 + 48*a*c^3*d^2 - 3*(24*b*c^2 + 35*a*d^2)*(d*x + c)^3 
 + 8*(24*b*c^3 + 35*a*c*d^2)*(d*x + c)^2 - 21*(8*b*c^4 + 11*a*c^2*d^2)*(d* 
x + c))/((d*x + c)^(7/2)*c^4*d^2 - 3*(d*x + c)^(5/2)*c^5*d^2 + 3*(d*x + c) 
^(3/2)*c^6*d^2 - sqrt(d*x + c)*c^7*d^2) - 3*(24*b*c^2 + 35*a*d^2)*log((sqr 
t(d*x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c)))/(c^(9/2)*d^2))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.15 \[ \int \frac {a+b x^2}{x^4 (c+d x)^{3/2}} \, dx=-\frac {{\left (24 \, b c^{2} d + 35 \, a d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{8 \, \sqrt {-c} c^{4}} - \frac {2 \, {\left (b c^{2} d + a d^{3}\right )}}{\sqrt {d x + c} c^{4}} - \frac {24 \, {\left (d x + c\right )}^{\frac {5}{2}} b c^{2} d - 48 \, {\left (d x + c\right )}^{\frac {3}{2}} b c^{3} d + 24 \, \sqrt {d x + c} b c^{4} d + 57 \, {\left (d x + c\right )}^{\frac {5}{2}} a d^{3} - 136 \, {\left (d x + c\right )}^{\frac {3}{2}} a c d^{3} + 87 \, \sqrt {d x + c} a c^{2} d^{3}}{24 \, c^{4} d^{3} x^{3}} \] Input:

integrate((b*x^2+a)/x^4/(d*x+c)^(3/2),x, algorithm="giac")
 

Output:

-1/8*(24*b*c^2*d + 35*a*d^3)*arctan(sqrt(d*x + c)/sqrt(-c))/(sqrt(-c)*c^4) 
 - 2*(b*c^2*d + a*d^3)/(sqrt(d*x + c)*c^4) - 1/24*(24*(d*x + c)^(5/2)*b*c^ 
2*d - 48*(d*x + c)^(3/2)*b*c^3*d + 24*sqrt(d*x + c)*b*c^4*d + 57*(d*x + c) 
^(5/2)*a*d^3 - 136*(d*x + c)^(3/2)*a*c*d^3 + 87*sqrt(d*x + c)*a*c^2*d^3)/( 
c^4*d^3*x^3)
 

Mupad [B] (verification not implemented)

Time = 8.89 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.24 \[ \int \frac {a+b x^2}{x^4 (c+d x)^{3/2}} \, dx=\frac {d\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}}{\sqrt {c}}\right )\,\left (24\,b\,c^2+35\,a\,d^2\right )}{8\,c^{9/2}}-\frac {\frac {2\,\left (b\,c^2\,d+a\,d^3\right )}{c}+\frac {\left (24\,b\,c^2\,d+35\,a\,d^3\right )\,{\left (c+d\,x\right )}^2}{3\,c^3}-\frac {\left (24\,b\,c^2\,d+35\,a\,d^3\right )\,{\left (c+d\,x\right )}^3}{8\,c^4}-\frac {7\,\left (8\,b\,c^2\,d+11\,a\,d^3\right )\,\left (c+d\,x\right )}{8\,c^2}}{3\,c\,{\left (c+d\,x\right )}^{5/2}-{\left (c+d\,x\right )}^{7/2}+c^3\,\sqrt {c+d\,x}-3\,c^2\,{\left (c+d\,x\right )}^{3/2}} \] Input:

int((a + b*x^2)/(x^4*(c + d*x)^(3/2)),x)
 

Output:

(d*atanh((c + d*x)^(1/2)/c^(1/2))*(35*a*d^2 + 24*b*c^2))/(8*c^(9/2)) - ((2 
*(a*d^3 + b*c^2*d))/c + ((35*a*d^3 + 24*b*c^2*d)*(c + d*x)^2)/(3*c^3) - (( 
35*a*d^3 + 24*b*c^2*d)*(c + d*x)^3)/(8*c^4) - (7*(11*a*d^3 + 8*b*c^2*d)*(c 
 + d*x))/(8*c^2))/(3*c*(c + d*x)^(5/2) - (c + d*x)^(7/2) + c^3*(c + d*x)^( 
1/2) - 3*c^2*(c + d*x)^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.34 \[ \int \frac {a+b x^2}{x^4 (c+d x)^{3/2}} \, dx=\frac {-105 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a \,d^{3} x^{3}-72 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) b \,c^{2} d \,x^{3}+105 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a \,d^{3} x^{3}+72 \sqrt {c}\, \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) b \,c^{2} d \,x^{3}-16 a \,c^{4}+28 a \,c^{3} d x -70 a \,c^{2} d^{2} x^{2}-210 a c \,d^{3} x^{3}-48 b \,c^{4} x^{2}-144 b \,c^{3} d \,x^{3}}{48 \sqrt {d x +c}\, c^{5} x^{3}} \] Input:

int((b*x^2+a)/x^4/(d*x+c)^(3/2),x)
 

Output:

( - 105*sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*a*d**3*x**3 - 7 
2*sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*b*c**2*d*x**3 + 105*s 
qrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) + sqrt(c))*a*d**3*x**3 + 72*sqrt(c) 
*sqrt(c + d*x)*log(sqrt(c + d*x) + sqrt(c))*b*c**2*d*x**3 - 16*a*c**4 + 28 
*a*c**3*d*x - 70*a*c**2*d**2*x**2 - 210*a*c*d**3*x**3 - 48*b*c**4*x**2 - 1 
44*b*c**3*d*x**3)/(48*sqrt(c + d*x)*c**5*x**3)