\(\int \frac {(c+d x) (a x^2+b x^3)^{3/2}}{x^5} \, dx\) [253]

Optimal result

Integrand size = 24, antiderivative size = 114 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^5} \, dx=\frac {(3 b c+2 a d) \sqrt {a x^2+b x^3}}{x}-\frac {c \left (a x^2+b x^3\right )^{3/2}}{x^4}+\frac {2 d \left (a x^2+b x^3\right )^{3/2}}{3 x^3}-\sqrt {a} (3 b c+2 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.77 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^5} \, dx=\frac {(a+b x) (2 b x (3 c+d x)+a (-3 c+8 d x))-3 \sqrt {a} (3 b c+2 a d) x \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{3 \sqrt {x^2 (a+b x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1944, 1927, 1927, 1914, 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.

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

\(\Big \downarrow \) 1944

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

\(\Big \downarrow \) 1927

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

\(\Big \downarrow \) 1927

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

\(\Big \downarrow \) 1914

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

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 1914

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[2/(2 - n) 
Subst[Int[1/(1 - a*x^2), x], x, x/Sqrt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, 
n}, x] && NeQ[n, 2]
 

rule 1927

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a* 
(n - j)*(p/(c^j*(m + n*p + 1)))   Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1) 
, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (Int 
egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + n*p + 1, 0]
 

rule 1944

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j 
+ b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b 
*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1))   Int[(e*x)^(m + n)*(a*x^ 
j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j 
+ n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 
] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( 
GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 
, 0]
 

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88

method	result	size
pseudoelliptic	\(\frac {b^{3} x^{4} \left (a d -\frac {3 b c}{8}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-3 \left (\frac {b^{2} x^{2} \left (4 d x +c \right ) a^{\frac {3}{2}}}{12}+b x \left (\frac {14 d x}{9}+c \right ) a^{\frac {5}{2}}+\left (\frac {8 d x}{9}+\frac {2 c}{3}\right ) a^{\frac {7}{2}}-\frac {\sqrt {a}\, b^{3} c \,x^{3}}{8}\right ) \sqrt {b x +a}}{8 a^{\frac {5}{2}} x^{4}}\)	\(100\)
risch	\(-\frac {a c \sqrt {x^{2} \left (b x +a \right )}}{x^{2}}+\frac {\left (\frac {2 \left (b x +a \right )^{\frac {3}{2}} d}{3}+2 \sqrt {b x +a}\, a d +2 \sqrt {b x +a}\, b c -\left (2 a d +3 b c \right ) \sqrt {a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )\right ) \sqrt {x^{2} \left (b x +a \right )}}{x \sqrt {b x +a}}\)	\(101\)
default	\(\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (2 \left (b x +a \right )^{\frac {3}{2}} d x \sqrt {a}+6 a^{\frac {3}{2}} d x \sqrt {b x +a}+6 b c x \sqrt {a}\, \sqrt {b x +a}-6 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{2} d x -9 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a b c x -3 a^{\frac {3}{2}} c \sqrt {b x +a}\right )}{3 x^{4} \left (b x +a \right )^{\frac {3}{2}} \sqrt {a}}\)	\(123\)

Input:

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

Output:

1/8*(b^3*x^4*(a*d-3/8*b*c)*arctanh((b*x+a)^(1/2)/a^(1/2))-3*(1/12*b^2*x^2* 
(4*d*x+c)*a^(3/2)+b*x*(14/9*d*x+c)*a^(5/2)+(8/9*d*x+2/3*c)*a^(7/2)-1/8*a^( 
1/2)*b^3*c*x^3)*(b*x+a)^(1/2))/a^(5/2)/x^4
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.67 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^5} \, dx=\left [\frac {3 \, {\left (3 \, b c + 2 \, a d\right )} \sqrt {a} x^{2} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (2 \, b d x^{2} - 3 \, a c + 2 \, {\left (3 \, b c + 4 \, a d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{6 \, x^{2}}, \frac {3 \, {\left (3 \, b c + 2 \, a d\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (2 \, b d x^{2} - 3 \, a c + 2 \, {\left (3 \, b c + 4 \, a d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{3 \, x^{2}}\right ] \] Input:

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

Output:

[1/6*(3*(3*b*c + 2*a*d)*sqrt(a)*x^2*log((b*x^2 + 2*a*x - 2*sqrt(b*x^3 + a* 
x^2)*sqrt(a))/x^2) + 2*(2*b*d*x^2 - 3*a*c + 2*(3*b*c + 4*a*d)*x)*sqrt(b*x^ 
3 + a*x^2))/x^2, 1/3*(3*(3*b*c + 2*a*d)*sqrt(-a)*x^2*arctan(sqrt(b*x^3 + a 
*x^2)*sqrt(-a)/(b*x^2 + a*x)) + (2*b*d*x^2 - 3*a*c + 2*(3*b*c + 4*a*d)*x)* 
sqrt(b*x^3 + a*x^2))/x^2]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^5} \, dx=-\frac {1}{3} \, {\left (\frac {3 \, \sqrt {b x + a} a c \mathrm {sgn}\left (x\right )}{b x} - \frac {3 \, {\left (3 \, a b c \mathrm {sgn}\left (x\right ) + 2 \, a^{2} d \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} b} - \frac {2 \, {\left (3 \, \sqrt {b x + a} b^{3} c \mathrm {sgn}\left (x\right ) + {\left (b x + a\right )}^{\frac {3}{2}} b^{2} d \mathrm {sgn}\left (x\right ) + 3 \, \sqrt {b x + a} a b^{2} d \mathrm {sgn}\left (x\right )\right )}}{b^{3}}\right )} b \] Input:

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

Output:

-1/3*(3*sqrt(b*x + a)*a*c*sgn(x)/(b*x) - 3*(3*a*b*c*sgn(x) + 2*a^2*d*sgn(x 
))*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*b) - 2*(3*sqrt(b*x + a)*b^3*c* 
sgn(x) + (b*x + a)^(3/2)*b^2*d*sgn(x) + 3*sqrt(b*x + a)*a*b^2*d*sgn(x))/b^ 
3)*b
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.08 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^5} \, dx=\frac {-6 \sqrt {b x +a}\, a c +16 \sqrt {b x +a}\, a d x +12 \sqrt {b x +a}\, b c x +4 \sqrt {b x +a}\, b d \,x^{2}+6 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a d x +9 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b c x -6 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a d x -9 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b c x}{6 x} \] Input:

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

Output:

( - 6*sqrt(a + b*x)*a*c + 16*sqrt(a + b*x)*a*d*x + 12*sqrt(a + b*x)*b*c*x 
+ 4*sqrt(a + b*x)*b*d*x**2 + 6*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*a*d*x 
+ 9*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*b*c*x - 6*sqrt(a)*log(sqrt(a + b* 
x) + sqrt(a))*a*d*x - 9*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b*c*x)/(6*x)