\(\int \frac {(c+d x) (a x^2+b x^3)^{3/2}}{x^7} \, dx\) [255]

Optimal result

Integrand size = 24, antiderivative size = 133 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^7} \, dx=-\frac {(b c+2 a d) \sqrt {a x^2+b x^3}}{4 x^3}-\frac {b (b c+10 a d) \sqrt {a x^2+b x^3}}{8 a x^2}-\frac {c \left (a x^2+b x^3\right )^{3/2}}{3 x^6}+\frac {b^2 (b c-6 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{8 a^{3/2}} \] Output:

-1/4*(2*a*d+b*c)*(b*x^3+a*x^2)^(1/2)/x^3-1/8*b*(10*a*d+b*c)*(b*x^3+a*x^2)^ 
(1/2)/a/x^2-1/3*c*(b*x^3+a*x^2)^(3/2)/x^6+1/8*b^2*(-6*a*d+b*c)*arctanh((b* 
x^3+a*x^2)^(1/2)/a^(1/2)/x)/a^(3/2)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.89 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^7} \, dx=\frac {\sqrt {x^2 (a+b x)} \left (-\sqrt {a} \sqrt {a+b x} \left (3 b^2 c x^2+4 a^2 (2 c+3 d x)+2 a b x (7 c+15 d x)\right )+3 b^2 (b c-6 a d) x^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{24 a^{3/2} x^4 \sqrt {a+b x}} \] Input:

Integrate[((c + d*x)*(a*x^2 + b*x^3)^(3/2))/x^7,x]
 

Output:

(Sqrt[x^2*(a + b*x)]*(-(Sqrt[a]*Sqrt[a + b*x]*(3*b^2*c*x^2 + 4*a^2*(2*c + 
3*d*x) + 2*a*b*x*(7*c + 15*d*x))) + 3*b^2*(b*c - 6*a*d)*x^3*ArcTanh[Sqrt[a 
 + b*x]/Sqrt[a]]))/(24*a^(3/2)*x^4*Sqrt[a + b*x])
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1944, 1926, 1926, 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.

\(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{3/2} (c+d x)}{x^7} \, dx\)

\(\Big \downarrow \) 1944

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

\(\Big \downarrow \) 1926

\(\displaystyle -\frac {(b c-6 a d) \left (\frac {3}{4} b \int \frac {\sqrt {b x^3+a x^2}}{x^3}dx-\frac {\left (a x^2+b x^3\right )^{3/2}}{2 x^5}\right )}{6 a}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{3 a x^8}\)

\(\Big \downarrow \) 1926

\(\displaystyle -\frac {(b c-6 a d) \left (\frac {3}{4} b \left (\frac {1}{2} b \int \frac {1}{\sqrt {b x^3+a x^2}}dx-\frac {\sqrt {a x^2+b x^3}}{x^2}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{2 x^5}\right )}{6 a}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{3 a x^8}\)

\(\Big \downarrow \) 1914

\(\displaystyle -\frac {(b c-6 a d) \left (\frac {3}{4} b \left (-b \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}-\frac {\sqrt {a x^2+b x^3}}{x^2}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{2 x^5}\right )}{6 a}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{3 a x^8}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\left (\frac {3}{4} b \left (-\frac {b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{\sqrt {a}}-\frac {\sqrt {a x^2+b x^3}}{x^2}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{2 x^5}\right ) (b c-6 a d)}{6 a}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{3 a x^8}\)

Input:

Int[((c + d*x)*(a*x^2 + b*x^3)^(3/2))/x^7,x]
 

Output:

-1/3*(c*(a*x^2 + b*x^3)^(5/2))/(a*x^8) - ((b*c - 6*a*d)*(-1/2*(a*x^2 + b*x 
^3)^(3/2)/x^5 + (3*b*(-(Sqrt[a*x^2 + b*x^3]/x^2) - (b*ArcTanh[(Sqrt[a]*x)/ 
Sqrt[a*x^2 + b*x^3]])/Sqrt[a]))/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 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 = 0.61 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.81

method	result	size
risch	\(-\frac {\left (30 a b d \,x^{2}+3 b^{2} c \,x^{2}+12 a^{2} d x +14 a b c x +8 a^{2} c \right ) \sqrt {x^{2} \left (b x +a \right )}}{24 x^{4} a}-\frac {\left (6 a d -b c \right ) b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{8 a^{\frac {3}{2}} x \sqrt {b x +a}}\)	\(108\)
pseudoelliptic	\(\frac {\frac {3 b^{5} x^{6} \left (a d -\frac {7 b c}{12}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128}+\frac {7 \left (\frac {32 \left (-6 d x -5 c \right ) a^{\frac {11}{2}}}{7}+x b \left (-\frac {5 \left (\frac {18 d x}{7}+c \right ) x^{3} b^{3} a^{\frac {3}{2}}}{4}+b^{2} x^{2} \left (\frac {15 d x}{7}+c \right ) a^{\frac {5}{2}}-\frac {6 b x \left (2 d x +c \right ) a^{\frac {7}{2}}}{7}+\frac {8 \left (-33 d x -26 c \right ) a^{\frac {9}{2}}}{7}+\frac {15 \sqrt {a}\, b^{4} c \,x^{4}}{8}\right )\right ) \sqrt {b x +a}}{960}}{a^{\frac {9}{2}} x^{6}}\)	\(135\)
default	\(-\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (30 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {5}{2}} d +3 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {3}{2}} b c +18 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{2} b^{3} d \,x^{3}-3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \,b^{4} c \,x^{3}-48 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {7}{2}} d +8 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {5}{2}} b c +18 \sqrt {b x +a}\, a^{\frac {9}{2}} d -3 \sqrt {b x +a}\, a^{\frac {7}{2}} b c \right )}{24 b \,x^{6} \left (b x +a \right )^{\frac {3}{2}} a^{\frac {5}{2}}}\)	\(160\)

Input:

int((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^7,x,method=_RETURNVERBOSE)
 

Output:

-1/24*(30*a*b*d*x^2+3*b^2*c*x^2+12*a^2*d*x+14*a*b*c*x+8*a^2*c)/x^4/a*(x^2* 
(b*x+a))^(1/2)-1/8*(6*a*d-b*c)*b^2/a^(3/2)*arctanh((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 = 0.10 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.82 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^7} \, dx=\left [-\frac {3 \, {\left (b^{3} c - 6 \, a b^{2} d\right )} \sqrt {a} x^{4} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (8 \, a^{3} c + 3 \, {\left (a b^{2} c + 10 \, a^{2} b d\right )} x^{2} + 2 \, {\left (7 \, a^{2} b c + 6 \, a^{3} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{48 \, a^{2} x^{4}}, -\frac {3 \, {\left (b^{3} c - 6 \, a b^{2} d\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (8 \, a^{3} c + 3 \, {\left (a b^{2} c + 10 \, a^{2} b d\right )} x^{2} + 2 \, {\left (7 \, a^{2} b c + 6 \, a^{3} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, a^{2} x^{4}}\right ] \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^7,x, algorithm="fricas")
 

Output:

[-1/48*(3*(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*(8*a^3*c + 3*(a*b^2*c + 10*a^2*b*d)*x^2 + 2*( 
7*a^2*b*c + 6*a^3*d)*x)*sqrt(b*x^3 + a*x^2))/(a^2*x^4), -1/24*(3*(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)) 
 + (8*a^3*c + 3*(a*b^2*c + 10*a^2*b*d)*x^2 + 2*(7*a^2*b*c + 6*a^3*d)*x)*sq 
rt(b*x^3 + a*x^2))/(a^2*x^4)]
 

Sympy [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^7} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}} \left (c + d x\right )}{x^{7}}\, dx \] Input:

integrate((d*x+c)*(b*x**3+a*x**2)**(3/2)/x**7,x)
 

Output:

Integral((x**2*(a + b*x))**(3/2)*(c + d*x)/x**7, x)
 

Maxima [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^7} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}} {\left (d x + c\right )}}{x^{7}} \,d x } \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^7,x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a*x^2)^(3/2)*(d*x + c)/x^7, x)
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.07 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^7} \, dx=-\frac {1}{24} \, b^{3} {\left (\frac {3 \, {\left (b c \mathrm {sgn}\left (x\right ) - 6 \, a d \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a b} + \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} b c \mathrm {sgn}\left (x\right ) + 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a b c \mathrm {sgn}\left (x\right ) - 3 \, \sqrt {b x + a} a^{2} b c \mathrm {sgn}\left (x\right ) + 30 \, {\left (b x + a\right )}^{\frac {5}{2}} a d \mathrm {sgn}\left (x\right ) - 48 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} d \mathrm {sgn}\left (x\right ) + 18 \, \sqrt {b x + a} a^{3} d \mathrm {sgn}\left (x\right )}{a b^{4} x^{3}}\right )} \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^7,x, algorithm="giac")
 

Output:

-1/24*b^3*(3*(b*c*sgn(x) - 6*a*d*sgn(x))*arctan(sqrt(b*x + a)/sqrt(-a))/(s 
qrt(-a)*a*b) + (3*(b*x + a)^(5/2)*b*c*sgn(x) + 8*(b*x + a)^(3/2)*a*b*c*sgn 
(x) - 3*sqrt(b*x + a)*a^2*b*c*sgn(x) + 30*(b*x + a)^(5/2)*a*d*sgn(x) - 48* 
(b*x + a)^(3/2)*a^2*d*sgn(x) + 18*sqrt(b*x + a)*a^3*d*sgn(x))/(a*b^4*x^3))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^7} \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^{3/2}\,\left (c+d\,x\right )}{x^7} \,d x \] Input:

int(((a*x^2 + b*x^3)^(3/2)*(c + d*x))/x^7,x)
 

Output:

int(((a*x^2 + b*x^3)^(3/2)*(c + d*x))/x^7, x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.28 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^7} \, dx=\frac {-16 \sqrt {b x +a}\, a^{3} c -24 \sqrt {b x +a}\, a^{3} d x -28 \sqrt {b x +a}\, a^{2} b c x -60 \sqrt {b x +a}\, a^{2} b d \,x^{2}-6 \sqrt {b x +a}\, a \,b^{2} c \,x^{2}+18 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{2} d \,x^{3}-3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{3} c \,x^{3}-18 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{2} d \,x^{3}+3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{3} c \,x^{3}}{48 a^{2} x^{3}} \] Input:

int((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^7,x)
 

Output:

( - 16*sqrt(a + b*x)*a**3*c - 24*sqrt(a + b*x)*a**3*d*x - 28*sqrt(a + b*x) 
*a**2*b*c*x - 60*sqrt(a + b*x)*a**2*b*d*x**2 - 6*sqrt(a + b*x)*a*b**2*c*x* 
*2 + 18*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*a*b**2*d*x**3 - 3*sqrt(a)*log 
(sqrt(a + b*x) - sqrt(a))*b**3*c*x**3 - 18*sqrt(a)*log(sqrt(a + b*x) + sqr 
t(a))*a*b**2*d*x**3 + 3*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b**3*c*x**3)/ 
(48*a**2*x**3)