\(\int \frac {(c+d x) (a x^2+b x^3)^{5/2}}{x^8} \, dx\) [268]

Optimal result

Integrand size = 24, antiderivative size = 155 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^8} \, dx=\frac {5 b (3 b c+4 a d) \sqrt {a x^2+b x^3}}{4 x}-\frac {(5 b c+4 a d) \left (a x^2+b x^3\right )^{3/2}}{4 x^4}+\frac {2 b d \left (a x^2+b x^3\right )^{3/2}}{3 x^3}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{2 x^7}-\frac {5}{4} \sqrt {a} b (3 b c+4 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.74 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^8} \, dx=\frac {\sqrt {x^2 (a+b x)} \left (\sqrt {a+b x} \left (8 b^2 x^2 (3 c+d x)-6 a^2 (c+2 d x)+a b x (-27 c+56 d x)\right )-15 \sqrt {a} b (3 b c+4 a d) x^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{12 x^3 \sqrt {a+b x}} \] Input:

Integrate[((c + d*x)*(a*x^2 + b*x^3)^(5/2))/x^8,x]
 

Output:

(Sqrt[x^2*(a + b*x)]*(Sqrt[a + b*x]*(8*b^2*x^2*(3*c + d*x) - 6*a^2*(c + 2* 
d*x) + a*b*x*(-27*c + 56*d*x)) - 15*Sqrt[a]*b*(3*b*c + 4*a*d)*x^2*ArcTanh[ 
Sqrt[a + b*x]/Sqrt[a]]))/(12*x^3*Sqrt[a + b*x])
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 145, 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, 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.

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

\(\Big \downarrow \) 1944

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

\(\Big \downarrow \) 1926

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

\(\Big \downarrow \) 1927

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

\(\Big \downarrow \) 1927

\(\displaystyle \frac {(4 a d+3 b c) \left (\frac {5}{2} b \left (a \left (a \int \frac {1}{\sqrt {b x^3+a x^2}}dx+\frac {2 \sqrt {a x^2+b x^3}}{x}\right )+\frac {2 \left (a x^2+b x^3\right )^{3/2}}{3 x^3}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{x^6}\right )}{4 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{2 a x^9}\)

\(\Big \downarrow \) 1914

\(\displaystyle \frac {(4 a d+3 b c) \left (\frac {5}{2} b \left (a \left (\frac {2 \sqrt {a x^2+b x^3}}{x}-2 a \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}\right )+\frac {2 \left (a x^2+b x^3\right )^{3/2}}{3 x^3}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{x^6}\right )}{4 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{2 a x^9}\)

\(\Big \downarrow \) 219

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

Input:

Int[((c + d*x)*(a*x^2 + b*x^3)^(5/2))/x^8,x]
 

Output:

-1/2*(c*(a*x^2 + b*x^3)^(7/2))/(a*x^9) + ((3*b*c + 4*a*d)*(-((a*x^2 + b*x^ 
3)^(5/2)/x^6) + (5*b*((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))/( 
4*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 = 0.64 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {a \left (4 a d x +9 c b x +2 a c \right ) \sqrt {x^{2} \left (b x +a \right )}}{4 x^{3}}+\frac {b \left (\frac {16 \left (b x +a \right )^{\frac {3}{2}} d}{3}+32 \sqrt {b x +a}\, a d +16 \sqrt {b x +a}\, b c -10 \sqrt {a}\, \left (4 a d +3 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )\right ) \sqrt {x^{2} \left (b x +a \right )}}{8 x \sqrt {b x +a}}\)	\(117\)
pseudoelliptic	\(\frac {\frac {5 b^{6} x^{7} \left (a d -\frac {b c}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{512}+\frac {\left (-\frac {928 \left (\frac {35 d x}{29}+c \right ) x b \,a^{\frac {11}{2}}}{7}+\left (-64 d x -\frac {384 c}{7}\right ) a^{\frac {13}{2}}+x^{2} b^{2} \left (-\frac {5 b^{3} x^{3} \left (3 d x +c \right ) a^{\frac {3}{2}}}{4}+b^{2} x^{2} \left (\frac {5 d x}{2}+c \right ) a^{\frac {5}{2}}-\frac {6 x b \left (\frac {7 d x}{3}+c \right ) a^{\frac {7}{2}}}{7}+\left (-108 d x -\frac {592 c}{7}\right ) a^{\frac {9}{2}}+\frac {15 \sqrt {a}\, b^{4} c \,x^{4}}{8}\right )\right ) \sqrt {b x +a}}{384}}{a^{\frac {9}{2}} x^{7}}\)	\(150\)
default	\(-\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}} \left (-8 \left (b x +a \right )^{\frac {3}{2}} d \,b^{2} x^{2} \sqrt {a}-48 \sqrt {b x +a}\, a^{\frac {3}{2}} d \,b^{2} x^{2}-24 \sqrt {b x +a}\, b^{3} c \,x^{2} \sqrt {a}+60 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{2} b^{2} d \,x^{2}+45 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \,b^{3} c \,x^{2}+12 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {5}{2}} d +27 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {3}{2}} b c -12 \sqrt {b x +a}\, a^{\frac {7}{2}} d -21 \sqrt {b x +a}\, a^{\frac {5}{2}} b c \right )}{12 b \,x^{7} \left (b x +a \right )^{\frac {5}{2}} \sqrt {a}}\)	\(190\)

Input:

int((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^8,x,method=_RETURNVERBOSE)
 

Output:

-1/4*a*(4*a*d*x+9*b*c*x+2*a*c)/x^3*(x^2*(b*x+a))^(1/2)+1/8*b*(16/3*(b*x+a) 
^(3/2)*d+32*(b*x+a)^(1/2)*a*d+16*(b*x+a)^(1/2)*b*c-10*a^(1/2)*(4*a*d+3*b*c 
)*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 = 0.17 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.57 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^8} \, dx=\left [\frac {15 \, {\left (3 \, b^{2} c + 4 \, a b d\right )} \sqrt {a} x^{3} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (8 \, b^{2} d x^{3} - 6 \, a^{2} c + 8 \, {\left (3 \, b^{2} c + 7 \, a b d\right )} x^{2} - 3 \, {\left (9 \, a b c + 4 \, a^{2} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, x^{3}}, \frac {15 \, {\left (3 \, b^{2} c + 4 \, a b d\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (8 \, b^{2} d x^{3} - 6 \, a^{2} c + 8 \, {\left (3 \, b^{2} c + 7 \, a b d\right )} x^{2} - 3 \, {\left (9 \, a b c + 4 \, a^{2} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{12 \, x^{3}}\right ] \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^8,x, algorithm="fricas")
 

Output:

[1/24*(15*(3*b^2*c + 4*a*b*d)*sqrt(a)*x^3*log((b*x^2 + 2*a*x - 2*sqrt(b*x^ 
3 + a*x^2)*sqrt(a))/x^2) + 2*(8*b^2*d*x^3 - 6*a^2*c + 8*(3*b^2*c + 7*a*b*d 
)*x^2 - 3*(9*a*b*c + 4*a^2*d)*x)*sqrt(b*x^3 + a*x^2))/x^3, 1/12*(15*(3*b^2 
*c + 4*a*b*d)*sqrt(-a)*x^3*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a* 
x)) + (8*b^2*d*x^3 - 6*a^2*c + 8*(3*b^2*c + 7*a*b*d)*x^2 - 3*(9*a*b*c + 4* 
a^2*d)*x)*sqrt(b*x^3 + a*x^2))/x^3]
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^8} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {5}{2}} \left (c + d x\right )}{x^{8}}\, dx \] Input:

integrate((d*x+c)*(b*x**3+a*x**2)**(5/2)/x**8,x)
 

Output:

Integral((x**2*(a + b*x))**(5/2)*(c + d*x)/x**8, x)
 

Maxima [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^8} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {5}{2}} {\left (d x + c\right )}}{x^{8}} \,d x } \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^8,x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a*x^2)^(5/2)*(d*x + c)/x^8, x)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.12 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^8} \, dx=\frac {24 \, \sqrt {b x + a} b^{3} c \mathrm {sgn}\left (x\right ) + 8 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} d \mathrm {sgn}\left (x\right ) + 48 \, \sqrt {b x + a} a b^{2} d \mathrm {sgn}\left (x\right ) + \frac {15 \, {\left (3 \, a b^{3} c \mathrm {sgn}\left (x\right ) + 4 \, a^{2} b^{2} d \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {3 \, {\left (9 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} c \mathrm {sgn}\left (x\right ) - 7 \, \sqrt {b x + a} a^{2} b^{3} c \mathrm {sgn}\left (x\right ) + 4 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{2} d \mathrm {sgn}\left (x\right ) - 4 \, \sqrt {b x + a} a^{3} b^{2} d \mathrm {sgn}\left (x\right )\right )}}{b^{2} x^{2}}}{12 \, b} \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^8,x, algorithm="giac")
 

Output:

1/12*(24*sqrt(b*x + a)*b^3*c*sgn(x) + 8*(b*x + a)^(3/2)*b^2*d*sgn(x) + 48* 
sqrt(b*x + a)*a*b^2*d*sgn(x) + 15*(3*a*b^3*c*sgn(x) + 4*a^2*b^2*d*sgn(x))* 
arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) - 3*(9*(b*x + a)^(3/2)*a*b^3*c*sgn 
(x) - 7*sqrt(b*x + a)*a^2*b^3*c*sgn(x) + 4*(b*x + a)^(3/2)*a^2*b^2*d*sgn(x 
) - 4*sqrt(b*x + a)*a^3*b^2*d*sgn(x))/(b^2*x^2))/b
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^8} \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^{5/2}\,\left (c+d\,x\right )}{x^8} \,d x \] Input:

int(((a*x^2 + b*x^3)^(5/2)*(c + d*x))/x^8,x)
 

Output:

int(((a*x^2 + b*x^3)^(5/2)*(c + d*x))/x^8, x)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.12 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^8} \, dx=\frac {-12 \sqrt {b x +a}\, a^{2} c -24 \sqrt {b x +a}\, a^{2} d x -54 \sqrt {b x +a}\, a b c x +112 \sqrt {b x +a}\, a b d \,x^{2}+48 \sqrt {b x +a}\, b^{2} c \,x^{2}+16 \sqrt {b x +a}\, b^{2} d \,x^{3}+60 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a b d \,x^{2}+45 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{2} c \,x^{2}-60 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a b d \,x^{2}-45 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{2} c \,x^{2}}{24 x^{2}} \] Input:

int((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^8,x)
 

Output:

( - 12*sqrt(a + b*x)*a**2*c - 24*sqrt(a + b*x)*a**2*d*x - 54*sqrt(a + b*x) 
*a*b*c*x + 112*sqrt(a + b*x)*a*b*d*x**2 + 48*sqrt(a + b*x)*b**2*c*x**2 + 1 
6*sqrt(a + b*x)*b**2*d*x**3 + 60*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*a*b* 
d*x**2 + 45*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*b**2*c*x**2 - 60*sqrt(a)* 
log(sqrt(a + b*x) + sqrt(a))*a*b*d*x**2 - 45*sqrt(a)*log(sqrt(a + b*x) + s 
qrt(a))*b**2*c*x**2)/(24*x**2)