\(\int \frac {(c+d x)^{3/2} (a+b x^2)}{(e x)^{3/2}} \, dx\) [755]

Optimal result

Integrand size = 24, antiderivative size = 171 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{3/2}} \, dx=-\frac {\left (b c^2-24 a d^2\right ) \sqrt {e x} \sqrt {c+d x}}{8 d e^2}-\frac {2 a (c+d x)^{3/2}}{e \sqrt {e x}}-\frac {b c \sqrt {e x} (c+d x)^{3/2}}{12 d e^2}+\frac {b \sqrt {e x} (c+d x)^{5/2}}{3 d e^2}-\frac {c \left (b c^2-24 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{8 d^{3/2} e^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.69 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{3/2}} \, dx=\frac {x \left (\sqrt {d} \sqrt {c+d x} \left (24 a d (-2 c+d x)+b x \left (3 c^2+14 c d x+8 d^2 x^2\right )\right )+6 c \left (b c^2-24 a d^2\right ) \sqrt {x} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}-\sqrt {c+d x}}\right )\right )}{24 d^{3/2} (e x)^{3/2}} \] Input:

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

Output:

(x*(Sqrt[d]*Sqrt[c + d*x]*(24*a*d*(-2*c + d*x) + b*x*(3*c^2 + 14*c*d*x + 8 
*d^2*x^2)) + 6*c*(b*c^2 - 24*a*d^2)*Sqrt[x]*ArcTanh[(Sqrt[d]*Sqrt[x])/(Sqr 
t[c] - Sqrt[c + d*x])]))/(24*d^(3/2)*(e*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {520, 27, 90, 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.

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

\(\Big \downarrow \) 520

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

\(\displaystyle \frac {\frac {b c \sqrt {e x} (c+d x)^{5/2}}{3 d e}-\frac {\left (b c^2-24 a d^2\right ) \int \frac {(c+d x)^{3/2}}{\sqrt {e x}}dx}{6 d}}{c e}-\frac {2 a (c+d x)^{5/2}}{c e \sqrt {e x}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {b c \sqrt {e x} (c+d x)^{5/2}}{3 d e}-\frac {\left (b c^2-24 a d^2\right ) \left (\frac {3}{4} c \int \frac {\sqrt {c+d x}}{\sqrt {e x}}dx+\frac {\sqrt {e x} (c+d x)^{3/2}}{2 e}\right )}{6 d}}{c e}-\frac {2 a (c+d x)^{5/2}}{c e \sqrt {e x}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {b c \sqrt {e x} (c+d x)^{5/2}}{3 d e}-\frac {\left (b c^2-24 a d^2\right ) \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{\sqrt {e x} \sqrt {c+d x}}dx+\frac {\sqrt {e x} \sqrt {c+d x}}{e}\right )+\frac {\sqrt {e x} (c+d x)^{3/2}}{2 e}\right )}{6 d}}{c e}-\frac {2 a (c+d x)^{5/2}}{c e \sqrt {e x}}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {\frac {b c \sqrt {e x} (c+d x)^{5/2}}{3 d e}-\frac {\left (b c^2-24 a d^2\right ) \left (\frac {3}{4} c \left (c \int \frac {1}{e-\frac {d e x}{c+d x}}d\frac {\sqrt {e x}}{\sqrt {c+d x}}+\frac {\sqrt {e x} \sqrt {c+d x}}{e}\right )+\frac {\sqrt {e x} (c+d x)^{3/2}}{2 e}\right )}{6 d}}{c e}-\frac {2 a (c+d x)^{5/2}}{c e \sqrt {e x}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {b c \sqrt {e x} (c+d x)^{5/2}}{3 d e}-\frac {\left (b c^2-24 a d^2\right ) \left (\frac {3}{4} c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{\sqrt {d} \sqrt {e}}+\frac {\sqrt {e x} \sqrt {c+d x}}{e}\right )+\frac {\sqrt {e x} (c+d x)^{3/2}}{2 e}\right )}{6 d}}{c e}-\frac {2 a (c+d x)^{5/2}}{c e \sqrt {e x}}\)

Input:

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

Output:

(-2*a*(c + d*x)^(5/2))/(c*e*Sqrt[e*x]) + ((b*c*Sqrt[e*x]*(c + d*x)^(5/2))/ 
(3*d*e) - ((b*c^2 - 24*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))/(6*d))/(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 90

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

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 = 138, normalized size of antiderivative = 0.81

method	result	size
risch	\(-\frac {\sqrt {d x +c}\, \left (-8 b \,d^{2} x^{3}-14 b c d \,x^{2}-24 a \,d^{2} x -3 c^{2} b x +48 a c d \right )}{24 d e \sqrt {e x}}+\frac {c \left (24 a \,d^{2}-b \,c^{2}\right ) \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right ) \sqrt {\left (d x +c \right ) e x}}{16 d \sqrt {d e}\, e \sqrt {e x}\, \sqrt {d x +c}}\)	\(138\)
default	\(\frac {\sqrt {d x +c}\, \left (16 b \,d^{2} x^{3} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+72 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) a c \,d^{2} e x -3 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b \,c^{3} e x +28 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, b c d \,x^{2}+48 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, a \,d^{2} x +6 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, b \,c^{2} x -96 a c d \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\right )}{48 e d \sqrt {e x}\, \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}}\)	\(233\)

Input:

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

Output:

-1/24*(d*x+c)^(1/2)*(-8*b*d^2*x^3-14*b*c*d*x^2-24*a*d^2*x-3*b*c^2*x+48*a*c 
*d)/d/e/(e*x)^(1/2)+1/16*c*(24*a*d^2-b*c^2)/d*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*((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 = 240, normalized size of antiderivative = 1.40 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{3/2}} \, dx=\left [-\frac {3 \, {\left (b c^{3} - 24 \, a c d^{2}\right )} \sqrt {d e} x \log \left (2 \, d e x + c e + 2 \, \sqrt {d e} \sqrt {d x + c} \sqrt {e x}\right ) - 2 \, {\left (8 \, b d^{3} x^{3} + 14 \, b c d^{2} x^{2} - 48 \, a c d^{2} + 3 \, {\left (b c^{2} d + 8 \, a d^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {e x}}{48 \, d^{2} e^{2} x}, \frac {3 \, {\left (b c^{3} - 24 \, a c d^{2}\right )} \sqrt {-d e} x \arctan \left (\frac {\sqrt {-d e} \sqrt {d x + c} \sqrt {e x}}{d e x + c e}\right ) + {\left (8 \, b d^{3} x^{3} + 14 \, b c d^{2} x^{2} - 48 \, a c d^{2} + 3 \, {\left (b c^{2} d + 8 \, a d^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {e x}}{24 \, d^{2} e^{2} x}\right ] \] Input:

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

Output:

[-1/48*(3*(b*c^3 - 24*a*c*d^2)*sqrt(d*e)*x*log(2*d*e*x + c*e + 2*sqrt(d*e) 
*sqrt(d*x + c)*sqrt(e*x)) - 2*(8*b*d^3*x^3 + 14*b*c*d^2*x^2 - 48*a*c*d^2 + 
 3*(b*c^2*d + 8*a*d^3)*x)*sqrt(d*x + c)*sqrt(e*x))/(d^2*e^2*x), 1/24*(3*(b 
*c^3 - 24*a*c*d^2)*sqrt(-d*e)*x*arctan(sqrt(-d*e)*sqrt(d*x + c)*sqrt(e*x)/ 
(d*e*x + c*e)) + (8*b*d^3*x^3 + 14*b*c*d^2*x^2 - 48*a*c*d^2 + 3*(b*c^2*d + 
 8*a*d^3)*x)*sqrt(d*x + c)*sqrt(e*x))/(d^2*e^2*x)]
 

Sympy [A] (verification not implemented)

Time = 10.11 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.63 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{3/2}} \, dx=- \frac {2 a c^{\frac {3}{2}}}{e^{\frac {3}{2}} \sqrt {x} \sqrt {1 + \frac {d x}{c}}} + \frac {a \sqrt {c} d \sqrt {x} \sqrt {1 + \frac {d x}{c}}}{e^{\frac {3}{2}}} - \frac {2 a \sqrt {c} d \sqrt {x}}{e^{\frac {3}{2}} \sqrt {1 + \frac {d x}{c}}} + \frac {3 a c \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{e^{\frac {3}{2}}} + \frac {b c^{\frac {5}{2}} \sqrt {x}}{8 d e^{\frac {3}{2}} \sqrt {1 + \frac {d x}{c}}} + \frac {17 b c^{\frac {3}{2}} x^{\frac {3}{2}}}{24 e^{\frac {3}{2}} \sqrt {1 + \frac {d x}{c}}} + \frac {11 b \sqrt {c} d x^{\frac {5}{2}}}{12 e^{\frac {3}{2}} \sqrt {1 + \frac {d x}{c}}} - \frac {b c^{3} \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{8 d^{\frac {3}{2}} e^{\frac {3}{2}}} + \frac {b d^{2} x^{\frac {7}{2}}}{3 \sqrt {c} e^{\frac {3}{2}} \sqrt {1 + \frac {d x}{c}}} \] Input:

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

Output:

-2*a*c**(3/2)/(e**(3/2)*sqrt(x)*sqrt(1 + d*x/c)) + a*sqrt(c)*d*sqrt(x)*sqr 
t(1 + d*x/c)/e**(3/2) - 2*a*sqrt(c)*d*sqrt(x)/(e**(3/2)*sqrt(1 + d*x/c)) + 
 3*a*c*sqrt(d)*asinh(sqrt(d)*sqrt(x)/sqrt(c))/e**(3/2) + b*c**(5/2)*sqrt(x 
)/(8*d*e**(3/2)*sqrt(1 + d*x/c)) + 17*b*c**(3/2)*x**(3/2)/(24*e**(3/2)*sqr 
t(1 + d*x/c)) + 11*b*sqrt(c)*d*x**(5/2)/(12*e**(3/2)*sqrt(1 + d*x/c)) - b* 
c**3*asinh(sqrt(d)*sqrt(x)/sqrt(c))/(8*d**(3/2)*e**(3/2)) + b*d**2*x**(7/2 
)/(3*sqrt(c)*e**(3/2)*sqrt(1 + d*x/c))
 

Maxima [F(-2)]

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

integrate((d*x+c)^(3/2)*(b*x^2+a)/(e*x)^(3/2),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 = 0.16 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.98 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{3/2}} \, dx=\frac {{\left (\frac {{\left ({\left (d x + c\right )} {\left (2 \, {\left (d x + c\right )} {\left (\frac {4 \, {\left (d x + c\right )} b}{d^{2}} - \frac {5 \, b c}{d^{2}}\right )} - \frac {b c^{2} d^{4} - 24 \, a d^{6}}{d^{6}}\right )} + \frac {3 \, {\left (b c^{3} d^{4} - 24 \, a c d^{6}\right )}}{d^{6}}\right )} \sqrt {d x + c}}{\sqrt {{\left (d x + c\right )} d e - c d e}} + \frac {3 \, {\left (b c^{3} - 24 \, a c d^{2}\right )} \log \left ({\left | -\sqrt {d e} \sqrt {d x + c} + \sqrt {{\left (d x + c\right )} d e - c d e} \right |}\right )}{\sqrt {d e} d^{2}}\right )} d^{2}}{24 \, e {\left | d \right |}} \] Input:

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

Output:

1/24*(((d*x + c)*(2*(d*x + c)*(4*(d*x + c)*b/d^2 - 5*b*c/d^2) - (b*c^2*d^4 
 - 24*a*d^6)/d^6) + 3*(b*c^3*d^4 - 24*a*c*d^6)/d^6)*sqrt(d*x + c)/sqrt((d* 
x + c)*d*e - c*d*e) + 3*(b*c^3 - 24*a*c*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))*d^2/(e*abs(d))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.99 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{3/2}} \, dx=\frac {\sqrt {e}\, \left (-384 \sqrt {x}\, \sqrt {d x +c}\, a c \,d^{2}+192 \sqrt {x}\, \sqrt {d x +c}\, a \,d^{3} x +24 \sqrt {x}\, \sqrt {d x +c}\, b \,c^{2} d x +112 \sqrt {x}\, \sqrt {d x +c}\, b c \,d^{2} x^{2}+64 \sqrt {x}\, \sqrt {d x +c}\, b \,d^{3} x^{3}+576 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) a c \,d^{2} x -24 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) b \,c^{3} x -432 \sqrt {d}\, a c \,d^{2} x +3 \sqrt {d}\, b \,c^{3} x \right )}{192 d^{2} e^{2} x} \] Input:

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

Output:

(sqrt(e)*( - 384*sqrt(x)*sqrt(c + d*x)*a*c*d**2 + 192*sqrt(x)*sqrt(c + d*x 
)*a*d**3*x + 24*sqrt(x)*sqrt(c + d*x)*b*c**2*d*x + 112*sqrt(x)*sqrt(c + d* 
x)*b*c*d**2*x**2 + 64*sqrt(x)*sqrt(c + d*x)*b*d**3*x**3 + 576*sqrt(d)*log( 
(sqrt(c + d*x) + sqrt(x)*sqrt(d))/sqrt(c))*a*c*d**2*x - 24*sqrt(d)*log((sq 
rt(c + d*x) + sqrt(x)*sqrt(d))/sqrt(c))*b*c**3*x - 432*sqrt(d)*a*c*d**2*x 
+ 3*sqrt(d)*b*c**3*x))/(192*d**2*e**2*x)