\(\int \frac {(c+d x) (a x^2+b x^3)^{5/2}}{x^9} \, dx\) [269]

Optimal result

Integrand size = 24, antiderivative size = 156 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^9} \, dx=-\frac {b (5 b c+14 a d) \sqrt {a x^2+b x^3}}{8 x^2}+\frac {2 b^2 d \sqrt {a x^2+b x^3}}{x}-\frac {(5 b c+6 a d) \left (a x^2+b x^3\right )^{3/2}}{12 x^5}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{3 x^8}-\frac {5 b^2 (b c+6 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{8 \sqrt {a}} \] Output:

-1/8*b*(14*a*d+5*b*c)*(b*x^3+a*x^2)^(1/2)/x^2+2*b^2*d*(b*x^3+a*x^2)^(1/2)/ 
x-1/12*(6*a*d+5*b*c)*(b*x^3+a*x^2)^(3/2)/x^5-1/3*c*(b*x^3+a*x^2)^(5/2)/x^8 
-5/8*b^2*(6*a*d+b*c)*arctanh((b*x^3+a*x^2)^(1/2)/a^(1/2)/x)/a^(1/2)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.80 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^9} \, dx=-\frac {\sqrt {x^2 (a+b x)} \left (\sqrt {a} \sqrt {a+b x} \left (3 b^2 x^2 (11 c-16 d x)+4 a^2 (2 c+3 d x)+2 a b x (13 c+27 d x)\right )+15 b^2 (b c+6 a d) x^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{24 \sqrt {a} x^4 \sqrt {a+b x}} \] Input:

Integrate[((c + d*x)*(a*x^2 + b*x^3)^(5/2))/x^9,x]
 

Output:

-1/24*(Sqrt[x^2*(a + b*x)]*(Sqrt[a]*Sqrt[a + b*x]*(3*b^2*x^2*(11*c - 16*d* 
x) + 4*a^2*(2*c + 3*d*x) + 2*a*b*x*(13*c + 27*d*x)) + 15*b^2*(b*c + 6*a*d) 
*x^3*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(Sqrt[a]*x^4*Sqrt[a + b*x])
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1944, 1926, 1926, 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.

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

\(\Big \downarrow \) 1944

\(\displaystyle \frac {(6 a d+b c) \int \frac {\left (b x^3+a x^2\right )^{5/2}}{x^8}dx}{6 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{3 a x^{10}}\)

\(\Big \downarrow \) 1926

\(\displaystyle \frac {(6 a d+b c) \left (\frac {5}{4} b \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^5}dx-\frac {\left (a x^2+b x^3\right )^{5/2}}{2 x^7}\right )}{6 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{3 a x^{10}}\)

\(\Big \downarrow \) 1926

\(\displaystyle \frac {(6 a d+b c) \left (\frac {5}{4} b \left (\frac {3}{2} b \int \frac {\sqrt {b x^3+a x^2}}{x^2}dx-\frac {\left (a x^2+b x^3\right )^{3/2}}{x^4}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{2 x^7}\right )}{6 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{3 a x^{10}}\)

\(\Big \downarrow \) 1927

\(\displaystyle \frac {(6 a d+b c) \left (\frac {5}{4} b \left (\frac {3}{2} b \left (a \int \frac {1}{\sqrt {b x^3+a x^2}}dx+\frac {2 \sqrt {a x^2+b x^3}}{x}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{x^4}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{2 x^7}\right )}{6 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{3 a x^{10}}\)

\(\Big \downarrow \) 1914

\(\displaystyle \frac {(6 a d+b c) \left (\frac {5}{4} b \left (\frac {3}{2} b \left (\frac {2 \sqrt {a x^2+b x^3}}{x}-2 a \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{x^4}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{2 x^7}\right )}{6 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{3 a x^{10}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\left (\frac {5}{4} b \left (\frac {3}{2} b \left (\frac {2 \sqrt {a x^2+b x^3}}{x}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{x^4}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{2 x^7}\right ) (6 a d+b c)}{6 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{3 a x^{10}}\)

Input:

Int[((c + d*x)*(a*x^2 + b*x^3)^(5/2))/x^9,x]
 

Output:

-1/3*(c*(a*x^2 + b*x^3)^(7/2))/(a*x^10) + ((b*c + 6*a*d)*(-1/2*(a*x^2 + b* 
x^3)^(5/2)/x^7 + (5*b*(-((a*x^2 + b*x^3)^(3/2)/x^4) + (3*b*((2*Sqrt[a*x^2 
+ b*x^3])/x - 2*Sqrt[a]*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]]))/2))/4)) 
/(6*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 1926

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 + j*p + 1))), x] - Simp[b*p 
*((n - j)/(c^n*(m + j*p + 1)))   Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p - 1), 
 x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (Integer 
sQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && LtQ[m + j*p + 1, 0]
 

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 = 0.72 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.76

method	result	size
risch	\(-\frac {\left (54 a b d \,x^{2}+33 b^{2} c \,x^{2}+12 a^{2} d x +26 a b c x +8 a^{2} c \right ) \sqrt {x^{2} \left (b x +a \right )}}{24 x^{4}}+\frac {b^{2} \left (32 \sqrt {b x +a}\, d -\frac {2 \left (30 a d +5 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{16 x \sqrt {b x +a}}\)	\(118\)
pseudoelliptic	\(-\frac {5 \left (b^{7} x^{8} \left (a d -\frac {9 b c}{16}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-\frac {9 \left (-144 \left (\frac {296 d x}{243}+c \right ) x^{2} b^{2} a^{\frac {11}{2}}-\frac {704 x b \left (\frac {116 d x}{99}+c \right ) a^{\frac {13}{2}}}{3}+\frac {128 \left (-8 d x -7 c \right ) a^{\frac {15}{2}}}{9}+x^{3} b^{3} \left (\frac {35 x^{3} b^{3} \left (\frac {8 d x}{3}+c \right ) a^{\frac {3}{2}}}{24}-\frac {7 x^{2} b^{2} \left (\frac {20 d x}{9}+c \right ) a^{\frac {5}{2}}}{6}+b x \left (\frac {56 d x}{27}+c \right ) a^{\frac {7}{2}}+\frac {8 \left (-2 d x -c \right ) a^{\frac {9}{2}}}{9}-\frac {35 \sqrt {a}\, b^{4} c \,x^{4}}{16}\right )\right ) \sqrt {b x +a}}{35}\right )}{1024 a^{\frac {11}{2}} x^{8}}\)	\(169\)
default	\(-\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}} \left (-48 \sqrt {b x +a}\, d \,b^{3} x^{3} \sqrt {a}+90 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \,b^{3} d \,x^{3}+15 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{4} c \,x^{3}+54 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {3}{2}} d +33 \left (b x +a \right )^{\frac {5}{2}} \sqrt {a}\, b c -96 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {5}{2}} d -40 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {3}{2}} b c +42 \sqrt {b x +a}\, a^{\frac {7}{2}} d +15 \sqrt {b x +a}\, a^{\frac {5}{2}} b c \right )}{24 b \,x^{8} \left (b x +a \right )^{\frac {5}{2}} \sqrt {a}}\)	\(176\)

Input:

int((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^9,x,method=_RETURNVERBOSE)
 

Output:

-1/24*(54*a*b*d*x^2+33*b^2*c*x^2+12*a^2*d*x+26*a*b*c*x+8*a^2*c)/x^4*(x^2*( 
b*x+a))^(1/2)+1/16*b^2*(32*(b*x+a)^(1/2)*d-2*(30*a*d+5*b*c)/a^(1/2)*arctan 
h((b*x+a)^(1/2)/a^(1/2)))*(x^2*(b*x+a))^(1/2)/x/(b*x+a)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.69 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^9} \, dx=\left [\frac {15 \, {\left (b^{3} c + 6 \, a b^{2} d\right )} \sqrt {a} x^{4} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (48 \, a b^{2} d x^{3} - 8 \, a^{3} c - 3 \, {\left (11 \, a b^{2} c + 18 \, a^{2} b d\right )} x^{2} - 2 \, {\left (13 \, a^{2} b c + 6 \, a^{3} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{48 \, a x^{4}}, \frac {15 \, {\left (b^{3} c + 6 \, a b^{2} d\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (48 \, a b^{2} d x^{3} - 8 \, a^{3} c - 3 \, {\left (11 \, a b^{2} c + 18 \, a^{2} b d\right )} x^{2} - 2 \, {\left (13 \, a^{2} b c + 6 \, a^{3} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, a x^{4}}\right ] \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^9,x, algorithm="fricas")
 

Output:

[1/48*(15*(b^3*c + 6*a*b^2*d)*sqrt(a)*x^4*log((b*x^2 + 2*a*x - 2*sqrt(b*x^ 
3 + a*x^2)*sqrt(a))/x^2) + 2*(48*a*b^2*d*x^3 - 8*a^3*c - 3*(11*a*b^2*c + 1 
8*a^2*b*d)*x^2 - 2*(13*a^2*b*c + 6*a^3*d)*x)*sqrt(b*x^3 + a*x^2))/(a*x^4), 
 1/24*(15*(b^3*c + 6*a*b^2*d)*sqrt(-a)*x^4*arctan(sqrt(b*x^3 + a*x^2)*sqrt 
(-a)/(b*x^2 + a*x)) + (48*a*b^2*d*x^3 - 8*a^3*c - 3*(11*a*b^2*c + 18*a^2*b 
*d)*x^2 - 2*(13*a^2*b*c + 6*a^3*d)*x)*sqrt(b*x^3 + a*x^2))/(a*x^4)]
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^9} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {5}{2}} \left (c + d x\right )}{x^{9}}\, dx \] Input:

integrate((d*x+c)*(b*x**3+a*x**2)**(5/2)/x**9,x)
 

Output:

Integral((x**2*(a + b*x))**(5/2)*(c + d*x)/x**9, x)
 

Maxima [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^9} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {5}{2}} {\left (d x + c\right )}}{x^{9}} \,d x } \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^9,x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a*x^2)^(5/2)*(d*x + c)/x^9, x)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.97 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^9} \, dx=\frac {1}{24} \, {\left (\frac {48 \, \sqrt {b x + a} d \mathrm {sgn}\left (x\right )}{b} + \frac {15 \, {\left (b c \mathrm {sgn}\left (x\right ) + 6 \, a d \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} b} - \frac {33 \, {\left (b x + a\right )}^{\frac {5}{2}} b c \mathrm {sgn}\left (x\right ) - 40 \, {\left (b x + a\right )}^{\frac {3}{2}} a b c \mathrm {sgn}\left (x\right ) + 15 \, \sqrt {b x + a} a^{2} b c \mathrm {sgn}\left (x\right ) + 54 \, {\left (b x + a\right )}^{\frac {5}{2}} a d \mathrm {sgn}\left (x\right ) - 96 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} d \mathrm {sgn}\left (x\right ) + 42 \, \sqrt {b x + a} a^{3} d \mathrm {sgn}\left (x\right )}{b^{4} x^{3}}\right )} b^{3} \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^9,x, algorithm="giac")
 

Output:

1/24*(48*sqrt(b*x + a)*d*sgn(x)/b + 15*(b*c*sgn(x) + 6*a*d*sgn(x))*arctan( 
sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*b) - (33*(b*x + a)^(5/2)*b*c*sgn(x) - 40 
*(b*x + a)^(3/2)*a*b*c*sgn(x) + 15*sqrt(b*x + a)*a^2*b*c*sgn(x) + 54*(b*x 
+ a)^(5/2)*a*d*sgn(x) - 96*(b*x + a)^(3/2)*a^2*d*sgn(x) + 42*sqrt(b*x + a) 
*a^3*d*sgn(x))/(b^4*x^3))*b^3
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^9} \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^{5/2}\,\left (c+d\,x\right )}{x^9} \,d x \] Input:

int(((a*x^2 + b*x^3)^(5/2)*(c + d*x))/x^9,x)
 

Output:

int(((a*x^2 + b*x^3)^(5/2)*(c + d*x))/x^9, x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.19 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^9} \, dx=\frac {-16 \sqrt {b x +a}\, a^{3} c -24 \sqrt {b x +a}\, a^{3} d x -52 \sqrt {b x +a}\, a^{2} b c x -108 \sqrt {b x +a}\, a^{2} b d \,x^{2}-66 \sqrt {b x +a}\, a \,b^{2} c \,x^{2}+96 \sqrt {b x +a}\, a \,b^{2} d \,x^{3}+90 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{2} d \,x^{3}+15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{3} c \,x^{3}-90 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{2} d \,x^{3}-15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{3} c \,x^{3}}{48 a \,x^{3}} \] Input:

int((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^9,x)
 

Output:

( - 16*sqrt(a + b*x)*a**3*c - 24*sqrt(a + b*x)*a**3*d*x - 52*sqrt(a + b*x) 
*a**2*b*c*x - 108*sqrt(a + b*x)*a**2*b*d*x**2 - 66*sqrt(a + b*x)*a*b**2*c* 
x**2 + 96*sqrt(a + b*x)*a*b**2*d*x**3 + 90*sqrt(a)*log(sqrt(a + b*x) - sqr 
t(a))*a*b**2*d*x**3 + 15*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*b**3*c*x**3 
- 90*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*a*b**2*d*x**3 - 15*sqrt(a)*log(s 
qrt(a + b*x) + sqrt(a))*b**3*c*x**3)/(48*a*x**3)