\(\int \frac {(c+d x) (a x^2+b x^3)^{5/2}}{x^3} \, dx\) [263]

Optimal result

Integrand size = 24, antiderivative size = 131 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^3} \, dx=\frac {2 a^2 (b c-a d) \left (a x^2+b x^3\right )^{7/2}}{7 b^4 x^7}-\frac {2 a (2 b c-3 a d) \left (a x^2+b x^3\right )^{9/2}}{9 b^4 x^9}+\frac {2 (b c-3 a d) \left (a x^2+b x^3\right )^{11/2}}{11 b^4 x^{11}}+\frac {2 d \left (a x^2+b x^3\right )^{13/2}}{13 b^4 x^{13}} \] Output:

2/7*a^2*(-a*d+b*c)*(b*x^3+a*x^2)^(7/2)/b^4/x^7-2/9*a*(-3*a*d+2*b*c)*(b*x^3 
+a*x^2)^(9/2)/b^4/x^9+2/11*(-3*a*d+b*c)*(b*x^3+a*x^2)^(11/2)/b^4/x^11+2/13 
*d*(b*x^3+a*x^2)^(13/2)/b^4/x^13
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.61 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^3} \, dx=\frac {2 x (a+b x)^4 \left (-48 a^3 d+63 b^3 x^2 (13 c+11 d x)+8 a^2 b (13 c+21 d x)-14 a b^2 x (26 c+27 d x)\right )}{9009 b^4 \sqrt {x^2 (a+b x)}} \] Input:

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

Output:

(2*x*(a + b*x)^4*(-48*a^3*d + 63*b^3*x^2*(13*c + 11*d*x) + 8*a^2*b*(13*c + 
 21*d*x) - 14*a*b^2*x*(26*c + 27*d*x)))/(9009*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.

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

\(\Big \downarrow \) 1945

\(\displaystyle \frac {(13 b c-6 a d) \int \frac {\left (b x^3+a x^2\right )^{5/2}}{x^3}dx}{13 b}+\frac {2 d \left (a x^2+b x^3\right )^{7/2}}{13 b x^4}\)

\(\Big \downarrow \) 1922

\(\displaystyle \frac {(13 b c-6 a d) \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{11 b x^5}-\frac {4 a \int \frac {\left (b x^3+a x^2\right )^{5/2}}{x^4}dx}{11 b}\right )}{13 b}+\frac {2 d \left (a x^2+b x^3\right )^{7/2}}{13 b x^4}\)

\(\Big \downarrow \) 1922

\(\displaystyle \frac {(13 b c-6 a d) \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{11 b x^5}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{9 b x^6}-\frac {2 a \int \frac {\left (b x^3+a x^2\right )^{5/2}}{x^5}dx}{9 b}\right )}{11 b}\right )}{13 b}+\frac {2 d \left (a x^2+b x^3\right )^{7/2}}{13 b x^4}\)

\(\Big \downarrow \) 1920

\(\displaystyle \frac {\left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{11 b x^5}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{9 b x^6}-\frac {4 a \left (a x^2+b x^3\right )^{7/2}}{63 b^2 x^7}\right )}{11 b}\right ) (13 b c-6 a d)}{13 b}+\frac {2 d \left (a x^2+b x^3\right )^{7/2}}{13 b x^4}\)

Input:

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

Output:

(2*d*(a*x^2 + b*x^3)^(7/2))/(13*b*x^4) + ((13*b*c - 6*a*d)*((2*(a*x^2 + b* 
x^3)^(7/2))/(11*b*x^5) - (4*a*((-4*a*(a*x^2 + b*x^3)^(7/2))/(63*b^2*x^7) + 
 (2*(a*x^2 + b*x^3)^(7/2))/(9*b*x^6)))/(11*b)))/(13*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 [A] (verified)

Time = 0.52 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.65

method	result	size
gosper	\(-\frac {2 \left (b x +a \right ) \left (-693 b^{3} d \,x^{3}+378 a \,b^{2} d \,x^{2}-819 b^{3} c \,x^{2}-168 a^{2} b d x +364 a \,b^{2} c x +48 a^{3} d -104 c \,a^{2} b \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}{9009 b^{4} x^{5}}\)	\(85\)
default	\(-\frac {2 \left (b x +a \right ) \left (-693 b^{3} d \,x^{3}+378 a \,b^{2} d \,x^{2}-819 b^{3} c \,x^{2}-168 a^{2} b d x +364 a \,b^{2} c x +48 a^{3} d -104 c \,a^{2} b \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}{9009 b^{4} x^{5}}\)	\(85\)
orering	\(-\frac {2 \left (b x +a \right ) \left (-693 b^{3} d \,x^{3}+378 a \,b^{2} d \,x^{2}-819 b^{3} c \,x^{2}-168 a^{2} b d x +364 a \,b^{2} c x +48 a^{3} d -104 c \,a^{2} b \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}{9009 b^{4} x^{5}}\)	\(85\)
pseudoelliptic	\(-\frac {10 x^{2} b \left (a d +\frac {3 b c}{4}\right ) a \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\sqrt {b x +a}\, \left (-4 x^{2} \left (\frac {d x}{3}+c \right ) b^{2} \sqrt {a}+\left (-\frac {28 b d \,x^{2}}{3}+\left (2 a d +\frac {9 b c}{2}\right ) x +a c \right ) a^{\frac {3}{2}}\right )}{2 \sqrt {a}\, x^{2}}\)	\(89\)
risch	\(-\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (-693 d \,x^{6} b^{6}-1701 a \,b^{5} d \,x^{5}-819 b^{6} c \,x^{5}-1113 a^{2} b^{4} d \,x^{4}-2093 a \,b^{5} c \,x^{4}-15 a^{3} b^{3} d \,x^{3}-1469 a^{2} b^{4} c \,x^{3}+18 a^{4} b^{2} d \,x^{2}-39 a^{3} b^{3} c \,x^{2}-24 a^{5} b d x +52 a^{4} b^{2} c x +48 a^{6} d -104 a^{5} b c \right )}{9009 x \,b^{4}}\)	\(150\)
trager	\(-\frac {2 \left (-693 d \,x^{6} b^{6}-1701 a \,b^{5} d \,x^{5}-819 b^{6} c \,x^{5}-1113 a^{2} b^{4} d \,x^{4}-2093 a \,b^{5} c \,x^{4}-15 a^{3} b^{3} d \,x^{3}-1469 a^{2} b^{4} c \,x^{3}+18 a^{4} b^{2} d \,x^{2}-39 a^{3} b^{3} c \,x^{2}-24 a^{5} b d x +52 a^{4} b^{2} c x +48 a^{6} d -104 a^{5} b c \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{9009 b^{4} x}\)	\(152\)

Input:

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

Output:

-2/9009*(b*x+a)*(-693*b^3*d*x^3+378*a*b^2*d*x^2-819*b^3*c*x^2-168*a^2*b*d* 
x+364*a*b^2*c*x+48*a^3*d-104*a^2*b*c)*(b*x^3+a*x^2)^(5/2)/b^4/x^5
 

Fricas [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.16 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^3} \, dx=\frac {2 \, {\left (693 \, b^{6} d x^{6} + 104 \, a^{5} b c - 48 \, a^{6} d + 63 \, {\left (13 \, b^{6} c + 27 \, a b^{5} d\right )} x^{5} + 7 \, {\left (299 \, a b^{5} c + 159 \, a^{2} b^{4} d\right )} x^{4} + {\left (1469 \, a^{2} b^{4} c + 15 \, a^{3} b^{3} d\right )} x^{3} + 3 \, {\left (13 \, a^{3} b^{3} c - 6 \, a^{4} b^{2} d\right )} x^{2} - 4 \, {\left (13 \, a^{4} b^{2} c - 6 \, a^{5} b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{9009 \, b^{4} x} \] Input:

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

Output:

2/9009*(693*b^6*d*x^6 + 104*a^5*b*c - 48*a^6*d + 63*(13*b^6*c + 27*a*b^5*d 
)*x^5 + 7*(299*a*b^5*c + 159*a^2*b^4*d)*x^4 + (1469*a^2*b^4*c + 15*a^3*b^3 
*d)*x^3 + 3*(13*a^3*b^3*c - 6*a^4*b^2*d)*x^2 - 4*(13*a^4*b^2*c - 6*a^5*b*d 
)*x)*sqrt(b*x^3 + a*x^2)/(b^4*x)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.08 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^3} \, dx=\frac {2 \, {\left (63 \, b^{5} x^{5} + 161 \, a b^{4} x^{4} + 113 \, a^{2} b^{3} x^{3} + 3 \, a^{3} b^{2} x^{2} - 4 \, a^{4} b x + 8 \, a^{5}\right )} \sqrt {b x + a} c}{693 \, b^{3}} + \frac {2 \, {\left (231 \, b^{6} x^{6} + 567 \, a b^{5} x^{5} + 371 \, a^{2} b^{4} x^{4} + 5 \, a^{3} b^{3} x^{3} - 6 \, a^{4} b^{2} x^{2} + 8 \, a^{5} b x - 16 \, a^{6}\right )} \sqrt {b x + a} d}{3003 \, b^{4}} \] Input:

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

Output:

2/693*(63*b^5*x^5 + 161*a*b^4*x^4 + 113*a^2*b^3*x^3 + 3*a^3*b^2*x^2 - 4*a^ 
4*b*x + 8*a^5)*sqrt(b*x + a)*c/b^3 + 2/3003*(231*b^6*x^6 + 567*a*b^5*x^5 + 
 371*a^2*b^4*x^4 + 5*a^3*b^3*x^3 - 6*a^4*b^2*x^2 + 8*a^5*b*x - 16*a^6)*sqr 
t(b*x + a)*d/b^4
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 554 vs. \(2 (115) = 230\).

Time = 0.38 (sec) , antiderivative size = 554, normalized size of antiderivative = 4.23 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^3} \, dx =\text {Too large to display} \] Input:

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

Output:

2/45045*(3003*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a) 
*a^2)*a^3*c*sgn(x)/b^2 + 3861*(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*c*sgn(x)/b^2 + 1287*(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^3*d*sgn(x)/b^3 + 429*(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*c*sgn(x)/b^2 + 429*(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^2*d*sgn(x)/b^3 + 65*(63*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 9 
90*(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)*c*sgn(x)/b^2 + 195*(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 + 1 
155*(b*x + a)^(3/2)*a^4 - 693*sqrt(b*x + a)*a^5)*a*d*sgn(x)/b^3 + 15*(231* 
(b*x + a)^(13/2) - 1638*(b*x + a)^(11/2)*a + 5005*(b*x + a)^(9/2)*a^2 - 85 
80*(b*x + a)^(7/2)*a^3 + 9009*(b*x + a)^(5/2)*a^4 - 6006*(b*x + a)^(3/2)*a 
^5 + 3003*sqrt(b*x + a)*a^6)*d*sgn(x)/b^3)/b - 16/9009*(13*a^(11/2)*b*c - 
6*a^(13/2)*d)*sgn(x)/b^4
 

Mupad [B] (verification not implemented)

Time = 8.90 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.08 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^3} \, dx=\frac {\sqrt {b\,x^3+a\,x^2}\,\left (\frac {2\,a\,x^4\,\left (159\,a\,d+299\,b\,c\right )}{1287}-\frac {96\,a^6\,d-208\,a^5\,b\,c}{9009\,b^4}+\frac {2\,b^2\,d\,x^6}{13}+\frac {x^5\,\left (1638\,c\,b^6+3402\,a\,d\,b^5\right )}{9009\,b^4}+\frac {8\,a^4\,x\,\left (6\,a\,d-13\,b\,c\right )}{9009\,b^3}-\frac {2\,a^3\,x^2\,\left (6\,a\,d-13\,b\,c\right )}{3003\,b^2}+\frac {2\,a^2\,x^3\,\left (15\,a\,d+1469\,b\,c\right )}{9009\,b}\right )}{x} \] Input:

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

Output:

((a*x^2 + b*x^3)^(1/2)*((2*a*x^4*(159*a*d + 299*b*c))/1287 - (96*a^6*d - 2 
08*a^5*b*c)/(9009*b^4) + (2*b^2*d*x^6)/13 + (x^5*(1638*b^6*c + 3402*a*b^5* 
d))/(9009*b^4) + (8*a^4*x*(6*a*d - 13*b*c))/(9009*b^3) - (2*a^3*x^2*(6*a*d 
 - 13*b*c))/(3003*b^2) + (2*a^2*x^3*(15*a*d + 1469*b*c))/(9009*b)))/x
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.08 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^3} \, dx=\frac {2 \sqrt {b x +a}\, \left (693 b^{6} d \,x^{6}+1701 a \,b^{5} d \,x^{5}+819 b^{6} c \,x^{5}+1113 a^{2} b^{4} d \,x^{4}+2093 a \,b^{5} c \,x^{4}+15 a^{3} b^{3} d \,x^{3}+1469 a^{2} b^{4} c \,x^{3}-18 a^{4} b^{2} d \,x^{2}+39 a^{3} b^{3} c \,x^{2}+24 a^{5} b d x -52 a^{4} b^{2} c x -48 a^{6} d +104 a^{5} b c \right )}{9009 b^{4}} \] Input:

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

Output:

(2*sqrt(a + b*x)*( - 48*a**6*d + 104*a**5*b*c + 24*a**5*b*d*x - 52*a**4*b* 
*2*c*x - 18*a**4*b**2*d*x**2 + 39*a**3*b**3*c*x**2 + 15*a**3*b**3*d*x**3 + 
 1469*a**2*b**4*c*x**3 + 1113*a**2*b**4*d*x**4 + 2093*a*b**5*c*x**4 + 1701 
*a*b**5*d*x**5 + 819*b**6*c*x**5 + 693*b**6*d*x**6))/(9009*b**4)