\(\int \frac {(c+d x) (a x^2+b x^3)^{3/2}}{x} \, dx\) [249]

Optimal result

Integrand size = 24, antiderivative size = 131 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x} \, dx=\frac {2 a^2 (b c-a d) \left (a x^2+b x^3\right )^{5/2}}{5 b^4 x^5}-\frac {2 a (2 b c-3 a d) \left (a x^2+b x^3\right )^{7/2}}{7 b^4 x^7}+\frac {2 (b c-3 a d) \left (a x^2+b x^3\right )^{9/2}}{9 b^4 x^9}+\frac {2 d \left (a x^2+b x^3\right )^{11/2}}{11 b^4 x^{11}} \] Output:

2/5*a^2*(-a*d+b*c)*(b*x^3+a*x^2)^(5/2)/b^4/x^5-2/7*a*(-3*a*d+2*b*c)*(b*x^3 
+a*x^2)^(7/2)/b^4/x^7+2/9*(-3*a*d+b*c)*(b*x^3+a*x^2)^(9/2)/b^4/x^9+2/11*d* 
(b*x^3+a*x^2)^(11/2)/b^4/x^11
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.61 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x} \, dx=\frac {2 x (a+b x)^3 \left (-48 a^3 d+35 b^3 x^2 (11 c+9 d x)+8 a^2 b (11 c+15 d x)-10 a b^2 x (22 c+21 d x)\right )}{3465 b^4 \sqrt {x^2 (a+b x)}} \] Input:

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

Output:

(2*x*(a + b*x)^3*(-48*a^3*d + 35*b^3*x^2*(11*c + 9*d*x) + 8*a^2*b*(11*c + 
15*d*x) - 10*a*b^2*x*(22*c + 21*d*x)))/(3465*b^4*Sqrt[x^2*(a + b*x)])
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1945, 1922, 1922, 1920}

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.

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

\(\Big \downarrow \) 1945

\(\displaystyle \frac {(11 b c-6 a d) \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x}dx}{11 b}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}\)

\(\Big \downarrow \) 1922

\(\displaystyle \frac {(11 b c-6 a d) \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}-\frac {4 a \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^2}dx}{9 b}\right )}{11 b}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}\)

\(\Big \downarrow \) 1922

\(\displaystyle \frac {(11 b c-6 a d) \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{7 b x^4}-\frac {2 a \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^3}dx}{7 b}\right )}{9 b}\right )}{11 b}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}\)

\(\Big \downarrow \) 1920

\(\displaystyle \frac {\left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{7 b x^4}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{35 b^2 x^5}\right )}{9 b}\right ) (11 b c-6 a d)}{11 b}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}\)

Input:

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

Output:

(2*d*(a*x^2 + b*x^3)^(5/2))/(11*b*x^2) + ((11*b*c - 6*a*d)*((2*(a*x^2 + b* 
x^3)^(5/2))/(9*b*x^3) - (4*a*((-4*a*(a*x^2 + b*x^3)^(5/2))/(35*b^2*x^5) + 
(2*(a*x^2 + b*x^3)^(5/2))/(7*b*x^4)))/(9*b)))/(11*b)
 

Defintions of rubi rules used

rule 1920

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

rule 1922

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

rule 1945

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 3 vs. order 2.

Time = 0.47 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.49

method	result	size
pseudoelliptic	\(\frac {-2 c \,a^{\frac {3}{2}} b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {2 \sqrt {b x +a}\, \left (\left (d \,x^{2}+\frac {5}{3} c x \right ) b^{2}+\frac {20 \left (\frac {3 d x}{10}+c \right ) a b}{3}+a^{2} d \right )}{5}}{b}\)	\(64\)
gosper	\(-\frac {2 \left (b x +a \right ) \left (-315 b^{3} d \,x^{3}+210 a \,b^{2} d \,x^{2}-385 b^{3} c \,x^{2}-120 a^{2} b d x +220 a \,b^{2} c x +48 a^{3} d -88 c \,a^{2} b \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{3465 b^{4} x^{3}}\)	\(85\)
default	\(-\frac {2 \left (b x +a \right ) \left (-315 b^{3} d \,x^{3}+210 a \,b^{2} d \,x^{2}-385 b^{3} c \,x^{2}-120 a^{2} b d x +220 a \,b^{2} c x +48 a^{3} d -88 c \,a^{2} b \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{3465 b^{4} x^{3}}\)	\(85\)
orering	\(-\frac {2 \left (b x +a \right ) \left (-315 b^{3} d \,x^{3}+210 a \,b^{2} d \,x^{2}-385 b^{3} c \,x^{2}-120 a^{2} b d x +220 a \,b^{2} c x +48 a^{3} d -88 c \,a^{2} b \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{3465 b^{4} x^{3}}\)	\(85\)
risch	\(-\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (-315 d \,x^{5} b^{5}-420 a \,b^{4} d \,x^{4}-385 b^{5} c \,x^{4}-15 a^{2} b^{3} d \,x^{3}-550 a \,b^{4} c \,x^{3}+18 a^{3} b^{2} d \,x^{2}-33 a^{2} b^{3} c \,x^{2}-24 a^{4} b d x +44 a^{3} b^{2} c x +48 a^{5} d -88 a^{4} b c \right )}{3465 x \,b^{4}}\)	\(126\)
trager	\(-\frac {2 \left (-315 d \,x^{5} b^{5}-420 a \,b^{4} d \,x^{4}-385 b^{5} c \,x^{4}-15 a^{2} b^{3} d \,x^{3}-550 a \,b^{4} c \,x^{3}+18 a^{3} b^{2} d \,x^{2}-33 a^{2} b^{3} c \,x^{2}-24 a^{4} b d x +44 a^{3} b^{2} c x +48 a^{5} d -88 a^{4} b c \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{3465 b^{4} x}\)	\(128\)

Input:

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

Output:

2/5*(-5*c*a^(3/2)*b*arctanh((b*x+a)^(1/2)/a^(1/2))+(b*x+a)^(1/2)*((d*x^2+5 
/3*c*x)*b^2+20/3*(3/10*d*x+c)*a*b+a^2*d))/b
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.98 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x} \, dx=\frac {2 \, {\left (315 \, b^{5} d x^{5} + 88 \, a^{4} b c - 48 \, a^{5} d + 35 \, {\left (11 \, b^{5} c + 12 \, a b^{4} d\right )} x^{4} + 5 \, {\left (110 \, a b^{4} c + 3 \, a^{2} b^{3} d\right )} x^{3} + 3 \, {\left (11 \, a^{2} b^{3} c - 6 \, a^{3} b^{2} d\right )} x^{2} - 4 \, {\left (11 \, a^{3} b^{2} c - 6 \, a^{4} b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{3465 \, b^{4} x} \] Input:

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

Output:

2/3465*(315*b^5*d*x^5 + 88*a^4*b*c - 48*a^5*d + 35*(11*b^5*c + 12*a*b^4*d) 
*x^4 + 5*(110*a*b^4*c + 3*a^2*b^3*d)*x^3 + 3*(11*a^2*b^3*c - 6*a^3*b^2*d)* 
x^2 - 4*(11*a^3*b^2*c - 6*a^4*b*d)*x)*sqrt(b*x^3 + a*x^2)/(b^4*x)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.92 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x} \, dx=\frac {2 \, {\left (35 \, b^{4} x^{4} + 50 \, a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} - 4 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {b x + a} c}{315 \, b^{3}} + \frac {2 \, {\left (105 \, b^{5} x^{5} + 140 \, a b^{4} x^{4} + 5 \, a^{2} b^{3} x^{3} - 6 \, a^{3} b^{2} x^{2} + 8 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt {b x + a} d}{1155 \, b^{4}} \] Input:

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

Output:

2/315*(35*b^4*x^4 + 50*a*b^3*x^3 + 3*a^2*b^2*x^2 - 4*a^3*b*x + 8*a^4)*sqrt 
(b*x + a)*c/b^3 + 2/1155*(105*b^5*x^5 + 140*a*b^4*x^4 + 5*a^2*b^3*x^3 - 6* 
a^3*b^2*x^2 + 8*a^4*b*x - 16*a^5)*sqrt(b*x + a)*d/b^4
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 384 vs. \(2 (115) = 230\).

Time = 0.12 (sec) , antiderivative size = 384, normalized size of antiderivative = 2.93 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x} \, dx=\frac {2 \, {\left (\frac {231 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a^{2} c \mathrm {sgn}\left (x\right )}{b^{2}} + \frac {198 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a c \mathrm {sgn}\left (x\right )}{b^{2}} + \frac {99 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a^{2} d \mathrm {sgn}\left (x\right )}{b^{3}} + \frac {11 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} c \mathrm {sgn}\left (x\right )}{b^{2}} + \frac {22 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a d \mathrm {sgn}\left (x\right )}{b^{3}} + \frac {5 \, {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )} d \mathrm {sgn}\left (x\right )}{b^{3}}\right )}}{3465 \, b} - \frac {16 \, {\left (11 \, a^{\frac {9}{2}} b c - 6 \, a^{\frac {11}{2}} d\right )} \mathrm {sgn}\left (x\right )}{3465 \, b^{4}} \] Input:

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

Output:

2/3465*(231*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a 
^2)*a^2*c*sgn(x)/b^2 + 198*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35* 
(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*a*c*sgn(x)/b^2 + 99*(5*(b*x + 
a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a 
)*a^3)*a^2*d*sgn(x)/b^3 + 11*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 
 378*(b*x + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4 
)*c*sgn(x)/b^2 + 22*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x 
 + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*a*d*sgn 
(x)/b^3 + 5*(63*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 990*(b*x + a)^( 
7/2)*a^2 - 1386*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4 - 693*sqrt( 
b*x + a)*a^5)*d*sgn(x)/b^3)/b - 16/3465*(11*a^(9/2)*b*c - 6*a^(11/2)*d)*sg 
n(x)/b^4
 

Mupad [B] (verification not implemented)

Time = 9.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.93 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x} \, dx=\frac {\sqrt {b\,x^3+a\,x^2}\,\left (\frac {x^4\,\left (770\,c\,b^5+840\,a\,d\,b^4\right )}{3465\,b^4}-\frac {96\,a^5\,d-176\,a^4\,b\,c}{3465\,b^4}+\frac {2\,b\,d\,x^5}{11}+\frac {8\,a^3\,x\,\left (6\,a\,d-11\,b\,c\right )}{3465\,b^3}+\frac {2\,a\,x^3\,\left (3\,a\,d+110\,b\,c\right )}{693\,b}-\frac {2\,a^2\,x^2\,\left (6\,a\,d-11\,b\,c\right )}{1155\,b^2}\right )}{x} \] Input:

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

Output:

((a*x^2 + b*x^3)^(1/2)*((x^4*(770*b^5*c + 840*a*b^4*d))/(3465*b^4) - (96*a 
^5*d - 176*a^4*b*c)/(3465*b^4) + (2*b*d*x^5)/11 + (8*a^3*x*(6*a*d - 11*b*c 
))/(3465*b^3) + (2*a*x^3*(3*a*d + 110*b*c))/(693*b) - (2*a^2*x^2*(6*a*d - 
11*b*c))/(1155*b^2)))/x
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.89 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x} \, dx=\frac {2 \sqrt {b x +a}\, \left (315 b^{5} d \,x^{5}+420 a \,b^{4} d \,x^{4}+385 b^{5} c \,x^{4}+15 a^{2} b^{3} d \,x^{3}+550 a \,b^{4} c \,x^{3}-18 a^{3} b^{2} d \,x^{2}+33 a^{2} b^{3} c \,x^{2}+24 a^{4} b d x -44 a^{3} b^{2} c x -48 a^{5} d +88 a^{4} b c \right )}{3465 b^{4}} \] Input:

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

Output:

(2*sqrt(a + b*x)*( - 48*a**5*d + 88*a**4*b*c + 24*a**4*b*d*x - 44*a**3*b** 
2*c*x - 18*a**3*b**2*d*x**2 + 33*a**2*b**3*c*x**2 + 15*a**2*b**3*d*x**3 + 
550*a*b**4*c*x**3 + 420*a*b**4*d*x**4 + 385*b**5*c*x**4 + 315*b**5*d*x**5) 
)/(3465*b**4)