\(\int \frac {1}{(c x)^{3/2} (a x+b x^2)^{5/2}} \, dx\) [120]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 21, antiderivative size = 207 \[ \int \frac {1}{(c x)^{3/2} \left (a x+b x^2\right )^{5/2}} \, dx=-\frac {1}{3 a (c x)^{3/2} \left (a x+b x^2\right )^{3/2}}+\frac {3 b}{4 a^2 c \sqrt {c x} \left (a x+b x^2\right )^{3/2}}-\frac {21 b^2 \sqrt {c x}}{8 a^3 c^2 \left (a x+b x^2\right )^{3/2}}-\frac {35 b^3 (c x)^{3/2}}{8 a^4 c^3 \left (a x+b x^2\right )^{3/2}}-\frac {105 b^3 \sqrt {c x}}{8 a^5 c^2 \sqrt {a x+b x^2}}+\frac {105 b^3 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a x+b x^2}}{\sqrt {a} \sqrt {c x}}\right )}{8 a^{11/2} c^{3/2}} \] Output:

-1/3/a/(c*x)^(3/2)/(b*x^2+a*x)^(3/2)+3/4*b/a^2/c/(c*x)^(1/2)/(b*x^2+a*x)^( 
3/2)-21/8*b^2*(c*x)^(1/2)/a^3/c^2/(b*x^2+a*x)^(3/2)-35/8*b^3*(c*x)^(3/2)/a 
^4/c^3/(b*x^2+a*x)^(3/2)-105/8*b^3*(c*x)^(1/2)/a^5/c^2/(b*x^2+a*x)^(1/2)+1 
05/8*b^3*arctanh(c^(1/2)*(b*x^2+a*x)^(1/2)/a^(1/2)/(c*x)^(1/2))/a^(11/2)/c 
^(3/2)
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(c x)^{3/2} \left (a x+b x^2\right )^{5/2}} \, dx=\frac {-\sqrt {a} \left (8 a^4-18 a^3 b x+63 a^2 b^2 x^2+420 a b^3 x^3+315 b^4 x^4\right )+315 b^3 x^3 (a+b x)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{24 a^{11/2} (c x)^{3/2} (x (a+b x))^{3/2}} \] Input:

Integrate[1/((c*x)^(3/2)*(a*x + b*x^2)^(5/2)),x]
 

Output:

(-(Sqrt[a]*(8*a^4 - 18*a^3*b*x + 63*a^2*b^2*x^2 + 420*a*b^3*x^3 + 315*b^4* 
x^4)) + 315*b^3*x^3*(a + b*x)^(3/2)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(24*a^ 
(11/2)*(c*x)^(3/2)*(x*(a + b*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1135, 1135, 1132, 1135, 1132, 1136, 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.

\(\displaystyle \int \frac {1}{(c x)^{3/2} \left (a x+b x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1135

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

\(\Big \downarrow \) 1135

\(\displaystyle -\frac {3 b \left (-\frac {7 b \int \frac {\sqrt {c x}}{\left (b x^2+a x\right )^{5/2}}dx}{4 a c}-\frac {1}{2 a \sqrt {c x} \left (a x+b x^2\right )^{3/2}}\right )}{2 a c}-\frac {1}{3 a (c x)^{3/2} \left (a x+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1132

\(\displaystyle -\frac {3 b \left (-\frac {7 b \left (\frac {5 c \int \frac {1}{\sqrt {c x} \left (b x^2+a x\right )^{3/2}}dx}{3 a}+\frac {2 \sqrt {c x}}{3 a \left (a x+b x^2\right )^{3/2}}\right )}{4 a c}-\frac {1}{2 a \sqrt {c x} \left (a x+b x^2\right )^{3/2}}\right )}{2 a c}-\frac {1}{3 a (c x)^{3/2} \left (a x+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1135

\(\displaystyle -\frac {3 b \left (-\frac {7 b \left (\frac {5 c \left (-\frac {3 b \int \frac {\sqrt {c x}}{\left (b x^2+a x\right )^{3/2}}dx}{2 a c}-\frac {1}{a \sqrt {c x} \sqrt {a x+b x^2}}\right )}{3 a}+\frac {2 \sqrt {c x}}{3 a \left (a x+b x^2\right )^{3/2}}\right )}{4 a c}-\frac {1}{2 a \sqrt {c x} \left (a x+b x^2\right )^{3/2}}\right )}{2 a c}-\frac {1}{3 a (c x)^{3/2} \left (a x+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1132

\(\displaystyle -\frac {3 b \left (-\frac {7 b \left (\frac {5 c \left (-\frac {3 b \left (\frac {c \int \frac {1}{\sqrt {c x} \sqrt {b x^2+a x}}dx}{a}+\frac {2 \sqrt {c x}}{a \sqrt {a x+b x^2}}\right )}{2 a c}-\frac {1}{a \sqrt {c x} \sqrt {a x+b x^2}}\right )}{3 a}+\frac {2 \sqrt {c x}}{3 a \left (a x+b x^2\right )^{3/2}}\right )}{4 a c}-\frac {1}{2 a \sqrt {c x} \left (a x+b x^2\right )^{3/2}}\right )}{2 a c}-\frac {1}{3 a (c x)^{3/2} \left (a x+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1136

\(\displaystyle -\frac {3 b \left (-\frac {7 b \left (\frac {5 c \left (-\frac {3 b \left (\frac {2 c^2 \int \frac {1}{\frac {c \left (b x^2+a x\right )}{x}-a c}d\frac {\sqrt {b x^2+a x}}{\sqrt {c x}}}{a}+\frac {2 \sqrt {c x}}{a \sqrt {a x+b x^2}}\right )}{2 a c}-\frac {1}{a \sqrt {c x} \sqrt {a x+b x^2}}\right )}{3 a}+\frac {2 \sqrt {c x}}{3 a \left (a x+b x^2\right )^{3/2}}\right )}{4 a c}-\frac {1}{2 a \sqrt {c x} \left (a x+b x^2\right )^{3/2}}\right )}{2 a c}-\frac {1}{3 a (c x)^{3/2} \left (a x+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {3 b \left (-\frac {7 b \left (\frac {5 c \left (-\frac {3 b \left (\frac {2 \sqrt {c x}}{a \sqrt {a x+b x^2}}-\frac {2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a x+b x^2}}{\sqrt {a} \sqrt {c x}}\right )}{a^{3/2}}\right )}{2 a c}-\frac {1}{a \sqrt {c x} \sqrt {a x+b x^2}}\right )}{3 a}+\frac {2 \sqrt {c x}}{3 a \left (a x+b x^2\right )^{3/2}}\right )}{4 a c}-\frac {1}{2 a \sqrt {c x} \left (a x+b x^2\right )^{3/2}}\right )}{2 a c}-\frac {1}{3 a (c x)^{3/2} \left (a x+b x^2\right )^{3/2}}\)

Input:

Int[1/((c*x)^(3/2)*(a*x + b*x^2)^(5/2)),x]
 

Output:

-1/3*1/(a*(c*x)^(3/2)*(a*x + b*x^2)^(3/2)) - (3*b*(-1/2*1/(a*Sqrt[c*x]*(a* 
x + b*x^2)^(3/2)) - (7*b*((2*Sqrt[c*x])/(3*a*(a*x + b*x^2)^(3/2)) + (5*c*( 
-(1/(a*Sqrt[c*x]*Sqrt[a*x + b*x^2])) - (3*b*((2*Sqrt[c*x])/(a*Sqrt[a*x + b 
*x^2]) - (2*Sqrt[c]*ArcTanh[(Sqrt[c]*Sqrt[a*x + b*x^2])/(Sqrt[a]*Sqrt[c*x] 
)])/a^(3/2)))/(2*a*c)))/(3*a)))/(4*a*c)))/(2*a*c)
 

Defintions of rubi rules used

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 1132
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(2*c*d - b*e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(e*(p + 
 1)*(b^2 - 4*a*c))), x] - Simp[(2*c*d - b*e)*((m + 2*p + 2)/((p + 1)*(b^2 - 
 4*a*c)))   Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; Free 
Q[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && LtQ 
[0, m, 1] && IntegerQ[2*p]
 

rule 1135
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(-e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((m + p + 1)*(2* 
c*d - b*e))), x] + Simp[c*((m + 2*p + 2)/((m + p + 1)*(2*c*d - b*e)))   Int 
[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, 
 x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && I 
ntegerQ[2*p]
 

rule 1136
Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x 
_Symbol] :> Simp[2*e   Subst[Int[1/(2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + 
 b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 
- b*d*e + a*e^2, 0]
 
Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.66

method result size
risch \(-\frac {\left (b x +a \right ) \left (123 b^{2} x^{2}-34 a b x +8 a^{2}\right )}{24 a^{5} x^{2} c \sqrt {c x}\, \sqrt {x \left (b x +a \right )}}-\frac {b^{3} \left (\frac {128}{\sqrt {c b x +a c}}+\frac {32 a c}{3 \left (c b x +a c \right )^{\frac {3}{2}}}-\frac {210 \,\operatorname {arctanh}\left (\frac {\sqrt {c b x +a c}}{\sqrt {a c}}\right )}{\sqrt {a c}}\right ) \sqrt {c \left (b x +a \right )}\, x}{16 a^{5} c \sqrt {c x}\, \sqrt {x \left (b x +a \right )}}\) \(137\)
default \(\frac {\sqrt {x \left (b x +a \right )}\, \left (315 \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (b x +a \right )}}{\sqrt {a c}}\right ) b^{4} x^{4} \sqrt {c \left (b x +a \right )}+315 \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (b x +a \right )}}{\sqrt {a c}}\right ) a \,b^{3} x^{3} \sqrt {c \left (b x +a \right )}-315 \sqrt {a c}\, b^{4} x^{4}-420 \sqrt {a c}\, a \,b^{3} x^{3}-63 \sqrt {a c}\, a^{2} b^{2} x^{2}+18 \sqrt {a c}\, a^{3} b x -8 \sqrt {a c}\, a^{4}\right )}{24 c \,x^{3} \sqrt {c x}\, \left (b x +a \right )^{2} a^{5} \sqrt {a c}}\) \(171\)

Input:

int(1/(c*x)^(3/2)/(b*x^2+a*x)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/24*(b*x+a)*(123*b^2*x^2-34*a*b*x+8*a^2)/a^5/x^2/c/(c*x)^(1/2)/(x*(b*x+a 
))^(1/2)-1/16*b^3/a^5*(128/(b*c*x+a*c)^(1/2)+32/3*a*c/(b*c*x+a*c)^(3/2)-21 
0/(a*c)^(1/2)*arctanh((b*c*x+a*c)^(1/2)/(a*c)^(1/2)))/c*(c*(b*x+a))^(1/2)/ 
(c*x)^(1/2)/(x*(b*x+a))^(1/2)*x
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.69 \[ \int \frac {1}{(c x)^{3/2} \left (a x+b x^2\right )^{5/2}} \, dx=\left [\frac {315 \, {\left (b^{5} x^{6} + 2 \, a b^{4} x^{5} + a^{2} b^{3} x^{4}\right )} \sqrt {a c} \log \left (-\frac {b c x^{2} + 2 \, a c x + 2 \, \sqrt {b x^{2} + a x} \sqrt {a c} \sqrt {c x}}{x^{2}}\right ) - 2 \, {\left (315 \, a b^{4} x^{4} + 420 \, a^{2} b^{3} x^{3} + 63 \, a^{3} b^{2} x^{2} - 18 \, a^{4} b x + 8 \, a^{5}\right )} \sqrt {b x^{2} + a x} \sqrt {c x}}{48 \, {\left (a^{6} b^{2} c^{2} x^{6} + 2 \, a^{7} b c^{2} x^{5} + a^{8} c^{2} x^{4}\right )}}, -\frac {315 \, {\left (b^{5} x^{6} + 2 \, a b^{4} x^{5} + a^{2} b^{3} x^{4}\right )} \sqrt {-a c} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-a c} \sqrt {c x}}{a c x}\right ) + {\left (315 \, a b^{4} x^{4} + 420 \, a^{2} b^{3} x^{3} + 63 \, a^{3} b^{2} x^{2} - 18 \, a^{4} b x + 8 \, a^{5}\right )} \sqrt {b x^{2} + a x} \sqrt {c x}}{24 \, {\left (a^{6} b^{2} c^{2} x^{6} + 2 \, a^{7} b c^{2} x^{5} + a^{8} c^{2} x^{4}\right )}}\right ] \] Input:

integrate(1/(c*x)^(3/2)/(b*x^2+a*x)^(5/2),x, algorithm="fricas")
 

Output:

[1/48*(315*(b^5*x^6 + 2*a*b^4*x^5 + a^2*b^3*x^4)*sqrt(a*c)*log(-(b*c*x^2 + 
 2*a*c*x + 2*sqrt(b*x^2 + a*x)*sqrt(a*c)*sqrt(c*x))/x^2) - 2*(315*a*b^4*x^ 
4 + 420*a^2*b^3*x^3 + 63*a^3*b^2*x^2 - 18*a^4*b*x + 8*a^5)*sqrt(b*x^2 + a* 
x)*sqrt(c*x))/(a^6*b^2*c^2*x^6 + 2*a^7*b*c^2*x^5 + a^8*c^2*x^4), -1/24*(31 
5*(b^5*x^6 + 2*a*b^4*x^5 + a^2*b^3*x^4)*sqrt(-a*c)*arctan(sqrt(b*x^2 + a*x 
)*sqrt(-a*c)*sqrt(c*x)/(a*c*x)) + (315*a*b^4*x^4 + 420*a^2*b^3*x^3 + 63*a^ 
3*b^2*x^2 - 18*a^4*b*x + 8*a^5)*sqrt(b*x^2 + a*x)*sqrt(c*x))/(a^6*b^2*c^2* 
x^6 + 2*a^7*b*c^2*x^5 + a^8*c^2*x^4)]
 

Sympy [F]

\[ \int \frac {1}{(c x)^{3/2} \left (a x+b x^2\right )^{5/2}} \, dx=\int \frac {1}{\left (c x\right )^{\frac {3}{2}} \left (x \left (a + b x\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(c*x)**(3/2)/(b*x**2+a*x)**(5/2),x)
 

Output:

Integral(1/((c*x)**(3/2)*(x*(a + b*x))**(5/2)), x)
 

Maxima [F]

\[ \int \frac {1}{(c x)^{3/2} \left (a x+b x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a x\right )}^{\frac {5}{2}} \left (c x\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(c*x)^(3/2)/(b*x^2+a*x)^(5/2),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a*x)^(5/2)*(c*x)^(3/2)), x)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(c x)^{3/2} \left (a x+b x^2\right )^{5/2}} \, dx=-\frac {1}{24} \, c^{5} {\left (\frac {315 \, b^{3} \arctan \left (\frac {\sqrt {b c x + a c}}{\sqrt {-a c}}\right )}{\sqrt {-a c} a^{5} c^{5} {\left | c \right |}} + \frac {16 \, a^{4} b^{3} c^{4} + 144 \, {\left (b c x + a c\right )} a^{3} b^{3} c^{3} - 693 \, {\left (b c x + a c\right )}^{2} a^{2} b^{3} c^{2} + 840 \, {\left (b c x + a c\right )}^{3} a b^{3} c - 315 \, {\left (b c x + a c\right )}^{4} b^{3}}{{\left (\sqrt {b c x + a c} a c - {\left (b c x + a c\right )}^{\frac {3}{2}}\right )}^{3} a^{5} c^{5} {\left | c \right |}}\right )} \] Input:

integrate(1/(c*x)^(3/2)/(b*x^2+a*x)^(5/2),x, algorithm="giac")
 

Output:

-1/24*c^5*(315*b^3*arctan(sqrt(b*c*x + a*c)/sqrt(-a*c))/(sqrt(-a*c)*a^5*c^ 
5*abs(c)) + (16*a^4*b^3*c^4 + 144*(b*c*x + a*c)*a^3*b^3*c^3 - 693*(b*c*x + 
 a*c)^2*a^2*b^3*c^2 + 840*(b*c*x + a*c)^3*a*b^3*c - 315*(b*c*x + a*c)^4*b^ 
3)/((sqrt(b*c*x + a*c)*a*c - (b*c*x + a*c)^(3/2))^3*a^5*c^5*abs(c)))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(c x)^{3/2} \left (a x+b x^2\right )^{5/2}} \, dx=\int \frac {1}{{\left (b\,x^2+a\,x\right )}^{5/2}\,{\left (c\,x\right )}^{3/2}} \,d x \] Input:

int(1/((a*x + b*x^2)^(5/2)*(c*x)^(3/2)),x)
 

Output:

int(1/((a*x + b*x^2)^(5/2)*(c*x)^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(c x)^{3/2} \left (a x+b x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (-315 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{3} x^{3}-315 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{4} x^{4}+315 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{3} x^{3}+315 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{4} x^{4}-16 a^{5}+36 a^{4} b x -126 a^{3} b^{2} x^{2}-840 a^{2} b^{3} x^{3}-630 a \,b^{4} x^{4}\right )}{48 \sqrt {b x +a}\, a^{6} c^{2} x^{3} \left (b x +a \right )} \] Input:

int(1/(c*x)^(3/2)/(b*x^2+a*x)^(5/2),x)
 

Output:

(sqrt(c)*( - 315*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*a*b**3 
*x**3 - 315*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*b**4*x**4 + 
 315*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*a*b**3*x**3 + 315* 
sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*b**4*x**4 - 16*a**5 + 3 
6*a**4*b*x - 126*a**3*b**2*x**2 - 840*a**2*b**3*x**3 - 630*a*b**4*x**4))/( 
48*sqrt(a + b*x)*a**6*c**2*x**3*(a + b*x))