\(\int \frac {(c+d x)^{3/2} (a+b x^2)}{(e x)^{5/2}} \, dx\) [756]

Optimal result

Integrand size = 24, antiderivative size = 170 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=-\frac {\left (b c^2+4 a d^2\right ) \sqrt {c+d x}}{2 d e^2 \sqrt {e x}}+\frac {b c \sqrt {e x} \sqrt {c+d x}}{4 e^3}-\frac {2 a (c+d x)^{3/2}}{3 e (e x)^{3/2}}+\frac {b (c+d x)^{5/2}}{2 d e^2 \sqrt {e x}}+\frac {\left (3 b c^2+8 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{4 \sqrt {d} e^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.64 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\frac {x \left (\sqrt {d} \sqrt {c+d x} \left (3 b x^2 (5 c+2 d x)-8 a (c+4 d x)\right )+6 \left (3 b c^2+8 a d^2\right ) x^{3/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {x}}{-\sqrt {c}+\sqrt {c+d x}}\right )\right )}{12 \sqrt {d} (e x)^{5/2}} \] Input:

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

Output:

(x*(Sqrt[d]*Sqrt[c + d*x]*(3*b*x^2*(5*c + 2*d*x) - 8*a*(c + 4*d*x)) + 6*(3 
*b*c^2 + 8*a*d^2)*x^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[x])/(-Sqrt[c] + Sqrt[c + d 
*x])]))/(12*Sqrt[d]*(e*x)^(5/2))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {520, 27, 87, 60, 60, 65, 221}

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.

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

\(\Big \downarrow \) 520

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\left (8 a d^2+3 b c^2\right ) \left (\frac {3}{4} c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{\sqrt {d} \sqrt {e}}+\frac {\sqrt {e x} \sqrt {c+d x}}{e}\right )+\frac {\sqrt {e x} (c+d x)^{3/2}}{2 e}\right )}{c e}-\frac {4 a d (c+d x)^{5/2}}{c e \sqrt {e x}}}{3 c e}-\frac {2 a (c+d x)^{5/2}}{3 c e (e x)^{3/2}}\)

Input:

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

Output:

(-2*a*(c + d*x)^(5/2))/(3*c*e*(e*x)^(3/2)) + ((-4*a*d*(c + d*x)^(5/2))/(c* 
e*Sqrt[e*x]) + ((3*b*c^2 + 8*a*d^2)*((Sqrt[e*x]*(c + d*x)^(3/2))/(2*e) + ( 
3*c*((Sqrt[e*x]*Sqrt[c + d*x])/e + (c*ArcTanh[(Sqrt[d]*Sqrt[e*x])/(Sqrt[e] 
*Sqrt[c + d*x])])/(Sqrt[d]*Sqrt[e])))/4))/(c*e))/(3*c*e)
 

Defintions of rubi rules used

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 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 65

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 0]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 520

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2)^p, e*x, x], R = Pol 
ynomialRemainder[(a + b*x^2)^p, e*x, x]}, Simp[R*(e*x)^(m + 1)*((c + d*x)^( 
n + 1)/((m + 1)*(e*c))), x] + Simp[1/((m + 1)*(e*c))   Int[(e*x)^(m + 1)*(c 
 + d*x)^n*ExpandToSum[(m + 1)*(e*c)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr 
eeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[m, -1] &&  !IntegerQ[n]
 

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.70

method	result	size
risch	\(-\frac {\sqrt {d x +c}\, \left (-6 b d \,x^{3}-15 b c \,x^{2}+32 a d x +8 a c \right )}{12 x \,e^{2} \sqrt {e x}}+\frac {\left (a \,d^{2}+\frac {3 b \,c^{2}}{8}\right ) \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right ) \sqrt {\left (d x +c \right ) e x}}{\sqrt {d e}\, e^{2} \sqrt {e x}\, \sqrt {d x +c}}\)	\(119\)
default	\(\frac {\sqrt {d x +c}\, \left (24 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) a \,d^{2} e \,x^{2}+9 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b \,c^{2} e \,x^{2}+12 b d \,x^{3} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+30 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, b c \,x^{2}-64 a d x \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}-16 a c \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\right )}{24 e^{2} x \sqrt {e x}\, \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}}\)	\(208\)

Input:

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

Output:

-1/12*(d*x+c)^(1/2)*(-6*b*d*x^3-15*b*c*x^2+32*a*d*x+8*a*c)/x/e^2/(e*x)^(1/ 
2)+(a*d^2+3/8*b*c^2)*ln((1/2*c*e+d*e*x)/(d*e)^(1/2)+(d*e*x^2+c*e*x)^(1/2)) 
/(d*e)^(1/2)/e^2*((d*x+c)*e*x)^(1/2)/(e*x)^(1/2)/(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.29 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\left [\frac {3 \, {\left (3 \, b c^{2} + 8 \, a d^{2}\right )} \sqrt {d e} x^{2} \log \left (2 \, d e x + c e + 2 \, \sqrt {d e} \sqrt {d x + c} \sqrt {e x}\right ) + 2 \, {\left (6 \, b d^{2} x^{3} + 15 \, b c d x^{2} - 32 \, a d^{2} x - 8 \, a c d\right )} \sqrt {d x + c} \sqrt {e x}}{24 \, d e^{3} x^{2}}, -\frac {3 \, {\left (3 \, b c^{2} + 8 \, a d^{2}\right )} \sqrt {-d e} x^{2} \arctan \left (\frac {\sqrt {-d e} \sqrt {d x + c} \sqrt {e x}}{d e x + c e}\right ) - {\left (6 \, b d^{2} x^{3} + 15 \, b c d x^{2} - 32 \, a d^{2} x - 8 \, a c d\right )} \sqrt {d x + c} \sqrt {e x}}{12 \, d e^{3} x^{2}}\right ] \] Input:

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

Output:

[1/24*(3*(3*b*c^2 + 8*a*d^2)*sqrt(d*e)*x^2*log(2*d*e*x + c*e + 2*sqrt(d*e) 
*sqrt(d*x + c)*sqrt(e*x)) + 2*(6*b*d^2*x^3 + 15*b*c*d*x^2 - 32*a*d^2*x - 8 
*a*c*d)*sqrt(d*x + c)*sqrt(e*x))/(d*e^3*x^2), -1/12*(3*(3*b*c^2 + 8*a*d^2) 
*sqrt(-d*e)*x^2*arctan(sqrt(-d*e)*sqrt(d*x + c)*sqrt(e*x)/(d*e*x + c*e)) - 
 (6*b*d^2*x^3 + 15*b*c*d*x^2 - 32*a*d^2*x - 8*a*c*d)*sqrt(d*x + c)*sqrt(e* 
x))/(d*e^3*x^2)]
 

Sympy [A] (verification not implemented)

Time = 8.60 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.78 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=- \frac {2 a \sqrt {c} d}{e^{\frac {5}{2}} \sqrt {x} \sqrt {1 + \frac {d x}{c}}} - \frac {2 a c \sqrt {d} \sqrt {\frac {c}{d x} + 1}}{3 e^{\frac {5}{2}} x} - \frac {2 a d^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}}{3 e^{\frac {5}{2}}} + \frac {2 a d^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{e^{\frac {5}{2}}} - \frac {2 a d^{2} \sqrt {x}}{\sqrt {c} e^{\frac {5}{2}} \sqrt {1 + \frac {d x}{c}}} + \frac {b c^{\frac {3}{2}} \sqrt {x} \sqrt {1 + \frac {d x}{c}}}{e^{\frac {5}{2}}} + \frac {b c^{\frac {3}{2}} \sqrt {x}}{4 e^{\frac {5}{2}} \sqrt {1 + \frac {d x}{c}}} + \frac {3 b \sqrt {c} d x^{\frac {3}{2}}}{4 e^{\frac {5}{2}} \sqrt {1 + \frac {d x}{c}}} + \frac {3 b c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{4 \sqrt {d} e^{\frac {5}{2}}} + \frac {b d^{2} x^{\frac {5}{2}}}{2 \sqrt {c} e^{\frac {5}{2}} \sqrt {1 + \frac {d x}{c}}} \] Input:

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

Output:

-2*a*sqrt(c)*d/(e**(5/2)*sqrt(x)*sqrt(1 + d*x/c)) - 2*a*c*sqrt(d)*sqrt(c/( 
d*x) + 1)/(3*e**(5/2)*x) - 2*a*d**(3/2)*sqrt(c/(d*x) + 1)/(3*e**(5/2)) + 2 
*a*d**(3/2)*asinh(sqrt(d)*sqrt(x)/sqrt(c))/e**(5/2) - 2*a*d**2*sqrt(x)/(sq 
rt(c)*e**(5/2)*sqrt(1 + d*x/c)) + b*c**(3/2)*sqrt(x)*sqrt(1 + d*x/c)/e**(5 
/2) + b*c**(3/2)*sqrt(x)/(4*e**(5/2)*sqrt(1 + d*x/c)) + 3*b*sqrt(c)*d*x**( 
3/2)/(4*e**(5/2)*sqrt(1 + d*x/c)) + 3*b*c**2*asinh(sqrt(d)*sqrt(x)/sqrt(c) 
)/(4*sqrt(d)*e**(5/2)) + b*d**2*x**(5/2)/(2*sqrt(c)*e**(5/2)*sqrt(1 + d*x/ 
c))
 

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.17 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\frac {d^{3} {\left (\frac {{\left ({\left (3 \, {\left (\frac {2 \, {\left (d x + c\right )} b e}{d} - \frac {b c e}{d}\right )} {\left (d x + c\right )} - \frac {4 \, {\left (3 \, b c^{3} d^{3} e^{2} + 8 \, a c d^{5} e^{2}\right )}}{c d^{4} e}\right )} {\left (d x + c\right )} + \frac {3 \, {\left (3 \, b c^{4} d^{3} e^{2} + 8 \, a c^{2} d^{5} e^{2}\right )}}{c d^{4} e}\right )} \sqrt {d x + c}}{{\left ({\left (d x + c\right )} d e - c d e\right )}^{\frac {3}{2}}} - \frac {3 \, {\left (3 \, b c^{2} + 8 \, a d^{2}\right )} \log \left ({\left | -\sqrt {d e} \sqrt {d x + c} + \sqrt {{\left (d x + c\right )} d e - c d e} \right |}\right )}{\sqrt {d e} d^{2}}\right )}}{12 \, e^{2} {\left | d \right |}} \] Input:

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

Output:

1/12*d^3*(((3*(2*(d*x + c)*b*e/d - b*c*e/d)*(d*x + c) - 4*(3*b*c^3*d^3*e^2 
 + 8*a*c*d^5*e^2)/(c*d^4*e))*(d*x + c) + 3*(3*b*c^4*d^3*e^2 + 8*a*c^2*d^5* 
e^2)/(c*d^4*e))*sqrt(d*x + c)/((d*x + c)*d*e - c*d*e)^(3/2) - 3*(3*b*c^2 + 
 8*a*d^2)*log(abs(-sqrt(d*e)*sqrt(d*x + c) + sqrt((d*x + c)*d*e - c*d*e))) 
/(sqrt(d*e)*d^2))/(e^2*abs(d))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.85 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\frac {\sqrt {e}\, \left (-64 \sqrt {x}\, \sqrt {d x +c}\, a c d -256 \sqrt {x}\, \sqrt {d x +c}\, a \,d^{2} x +120 \sqrt {x}\, \sqrt {d x +c}\, b c d \,x^{2}+48 \sqrt {x}\, \sqrt {d x +c}\, b \,d^{2} x^{3}+192 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) a \,d^{2} x^{2}+72 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) b \,c^{2} x^{2}+15 \sqrt {d}\, b \,c^{2} x^{2}\right )}{96 d \,e^{3} x^{2}} \] Input:

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

Output:

(sqrt(e)*( - 64*sqrt(x)*sqrt(c + d*x)*a*c*d - 256*sqrt(x)*sqrt(c + d*x)*a* 
d**2*x + 120*sqrt(x)*sqrt(c + d*x)*b*c*d*x**2 + 48*sqrt(x)*sqrt(c + d*x)*b 
*d**2*x**3 + 192*sqrt(d)*log((sqrt(c + d*x) + sqrt(x)*sqrt(d))/sqrt(c))*a* 
d**2*x**2 + 72*sqrt(d)*log((sqrt(c + d*x) + sqrt(x)*sqrt(d))/sqrt(c))*b*c* 
*2*x**2 + 15*sqrt(d)*b*c**2*x**2))/(96*d*e**3*x**2)