\(\int \frac {(c+d x) (a x^2+b x^3)^{5/2}}{x^7} \, dx\) [267]

Optimal result

Integrand size = 24, antiderivative size = 146 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^7} \, dx=\frac {a (5 b c+2 a d) \sqrt {a x^2+b x^3}}{x}+\frac {(5 b c+2 a d) \left (a x^2+b x^3\right )^{3/2}}{3 x^3}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{x^6}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{5 x^5}-a^{3/2} (5 b c+2 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.74 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^7} \, dx=\frac {(a+b x) \left (2 b^2 x^2 (5 c+3 d x)+2 a b x (35 c+11 d x)+a^2 (-15 c+46 d x)\right )-15 a^{3/2} (5 b c+2 a d) x \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{15 \sqrt {x^2 (a+b x)}} \] Input:

Integrate[((c + d*x)*(a*x^2 + b*x^3)^(5/2))/x^7,x]
 

Output:

((a + b*x)*(2*b^2*x^2*(5*c + 3*d*x) + 2*a*b*x*(35*c + 11*d*x) + a^2*(-15*c 
 + 46*d*x)) - 15*a^(3/2)*(5*b*c + 2*a*d)*x*Sqrt[a + b*x]*ArcTanh[Sqrt[a + 
b*x]/Sqrt[a]])/(15*Sqrt[x^2*(a + b*x)])
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1944, 1927, 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.

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

\(\Big \downarrow \) 1944

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

\(\Big \downarrow \) 1927

\(\displaystyle \frac {(2 a d+5 b c) \left (a \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^4}dx+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{5 x^5}\right )}{2 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{a x^8}\)

\(\Big \downarrow \) 1927

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

\(\Big \downarrow \) 1927

\(\displaystyle \frac {(2 a d+5 b c) \left (a \left (a \left (a \int \frac {1}{\sqrt {b x^3+a x^2}}dx+\frac {2 \sqrt {a x^2+b x^3}}{x}\right )+\frac {2 \left (a x^2+b x^3\right )^{3/2}}{3 x^3}\right )+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{5 x^5}\right )}{2 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{a x^8}\)

\(\Big \downarrow \) 1914

\(\displaystyle \frac {(2 a d+5 b c) \left (a \left (a \left (\frac {2 \sqrt {a x^2+b x^3}}{x}-2 a \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}\right )+\frac {2 \left (a x^2+b x^3\right )^{3/2}}{3 x^3}\right )+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{5 x^5}\right )}{2 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{a x^8}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\left (a \left (a \left (\frac {2 \sqrt {a x^2+b x^3}}{x}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )\right )+\frac {2 \left (a x^2+b x^3\right )^{3/2}}{3 x^3}\right )+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{5 x^5}\right ) (2 a d+5 b c)}{2 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{a x^8}\)

Input:

Int[((c + d*x)*(a*x^2 + b*x^3)^(5/2))/x^7,x]
 

Output:

-((c*(a*x^2 + b*x^3)^(7/2))/(a*x^8)) + ((5*b*c + 2*a*d)*((2*(a*x^2 + b*x^3 
)^(5/2))/(5*x^5) + a*((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*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 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 = 0.62 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.88

method	result	size
risch	\(-\frac {a^{2} c \sqrt {x^{2} \left (b x +a \right )}}{x^{2}}+\frac {\left (\frac {2 d \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 \left (b x +a \right )^{\frac {3}{2}} a d}{3}+\frac {2 \left (b x +a \right )^{\frac {3}{2}} b c}{3}+2 \sqrt {b x +a}\, a^{2} d +4 \sqrt {b x +a}\, a b c -\left (2 a d +5 b c \right ) a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )\right ) \sqrt {x^{2} \left (b x +a \right )}}{x \sqrt {b x +a}}\)	\(128\)
pseudoelliptic	\(-\frac {3 \left (b^{5} x^{6} \left (a d -\frac {5 b c}{12}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {2 \left (\left (\frac {192 d x}{5}+32 c \right ) a^{\frac {11}{2}}+x b \left (-\frac {5 \left (\frac {18 d x}{5}+c \right ) x^{3} b^{3} a^{\frac {3}{2}}}{4}+b^{2} x^{2} \left (3 d x +c \right ) a^{\frac {5}{2}}+54 x b \left (\frac {62 d x}{45}+c \right ) a^{\frac {7}{2}}+\left (\frac {504 d x}{5}+80 c \right ) a^{\frac {9}{2}}+\frac {15 \sqrt {a}\, b^{4} c \,x^{4}}{8}\right )\right ) \sqrt {b x +a}}{9}\right )}{128 a^{\frac {7}{2}} x^{6}}\)	\(133\)
default	\(\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}} \left (6 d \left (b x +a \right )^{\frac {5}{2}} x \sqrt {a}+10 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {3}{2}} d x +10 \left (b x +a \right )^{\frac {3}{2}} b c x \sqrt {a}+30 a^{\frac {5}{2}} d x \sqrt {b x +a}+60 a^{\frac {3}{2}} b c x \sqrt {b x +a}-30 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{3} d x -75 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{2} b c x -15 a^{\frac {5}{2}} c \sqrt {b x +a}\right )}{15 x^{6} \left (b x +a \right )^{\frac {5}{2}} \sqrt {a}}\)	\(154\)

Input:

int((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^7,x,method=_RETURNVERBOSE)
 

Output:

-a^2*c/x^2*(x^2*(b*x+a))^(1/2)+(2/5*d*(b*x+a)^(5/2)+2/3*(b*x+a)^(3/2)*a*d+ 
2/3*(b*x+a)^(3/2)*b*c+2*(b*x+a)^(1/2)*a^2*d+4*(b*x+a)^(1/2)*a*b*c-(2*a*d+5 
*b*c)*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 = 0.16 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.67 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^7} \, dx=\left [\frac {15 \, {\left (5 \, a b c + 2 \, a^{2} d\right )} \sqrt {a} x^{2} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (6 \, b^{2} d x^{3} - 15 \, a^{2} c + 2 \, {\left (5 \, b^{2} c + 11 \, a b d\right )} x^{2} + 2 \, {\left (35 \, a b c + 23 \, a^{2} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{30 \, x^{2}}, \frac {15 \, {\left (5 \, a b c + 2 \, a^{2} d\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (6 \, b^{2} d x^{3} - 15 \, a^{2} c + 2 \, {\left (5 \, b^{2} c + 11 \, a b d\right )} x^{2} + 2 \, {\left (35 \, a b c + 23 \, a^{2} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{15 \, x^{2}}\right ] \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^7,x, algorithm="fricas")
 

Output:

[1/30*(15*(5*a*b*c + 2*a^2*d)*sqrt(a)*x^2*log((b*x^2 + 2*a*x - 2*sqrt(b*x^ 
3 + a*x^2)*sqrt(a))/x^2) + 2*(6*b^2*d*x^3 - 15*a^2*c + 2*(5*b^2*c + 11*a*b 
*d)*x^2 + 2*(35*a*b*c + 23*a^2*d)*x)*sqrt(b*x^3 + a*x^2))/x^2, 1/15*(15*(5 
*a*b*c + 2*a^2*d)*sqrt(-a)*x^2*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 
+ a*x)) + (6*b^2*d*x^3 - 15*a^2*c + 2*(5*b^2*c + 11*a*b*d)*x^2 + 2*(35*a*b 
*c + 23*a^2*d)*x)*sqrt(b*x^3 + a*x^2))/x^2]
 

Sympy [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^7} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {5}{2}} \left (c + d x\right )}{x^{7}}\, dx \] Input:

integrate((d*x+c)*(b*x**3+a*x**2)**(5/2)/x**7,x)
 

Output:

Integral((x**2*(a + b*x))**(5/2)*(c + d*x)/x**7, x)
 

Maxima [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^7} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {5}{2}} {\left (d x + c\right )}}{x^{7}} \,d x } \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^7,x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a*x^2)^(5/2)*(d*x + c)/x^7, x)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.05 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^7} \, dx=-\frac {1}{15} \, {\left (\frac {15 \, \sqrt {b x + a} a^{2} c \mathrm {sgn}\left (x\right )}{b x} - \frac {15 \, {\left (5 \, a^{2} b c \mathrm {sgn}\left (x\right ) + 2 \, a^{3} d \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} b} - \frac {2 \, {\left (5 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{5} c \mathrm {sgn}\left (x\right ) + 30 \, \sqrt {b x + a} a b^{5} c \mathrm {sgn}\left (x\right ) + 3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} d \mathrm {sgn}\left (x\right ) + 5 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} d \mathrm {sgn}\left (x\right ) + 15 \, \sqrt {b x + a} a^{2} b^{4} d \mathrm {sgn}\left (x\right )\right )}}{b^{5}}\right )} b \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^7,x, algorithm="giac")
 

Output:

-1/15*(15*sqrt(b*x + a)*a^2*c*sgn(x)/(b*x) - 15*(5*a^2*b*c*sgn(x) + 2*a^3* 
d*sgn(x))*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*b) - 2*(5*(b*x + a)^(3/ 
2)*b^5*c*sgn(x) + 30*sqrt(b*x + a)*a*b^5*c*sgn(x) + 3*(b*x + a)^(5/2)*b^4* 
d*sgn(x) + 5*(b*x + a)^(3/2)*a*b^4*d*sgn(x) + 15*sqrt(b*x + a)*a^2*b^4*d*s 
gn(x))/b^5)*b
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^7} \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^{5/2}\,\left (c+d\,x\right )}{x^7} \,d x \] Input:

int(((a*x^2 + b*x^3)^(5/2)*(c + d*x))/x^7,x)
 

Output:

int(((a*x^2 + b*x^3)^(5/2)*(c + d*x))/x^7, x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.13 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^7} \, dx=\frac {-30 \sqrt {b x +a}\, a^{2} c +92 \sqrt {b x +a}\, a^{2} d x +140 \sqrt {b x +a}\, a b c x +44 \sqrt {b x +a}\, a b d \,x^{2}+20 \sqrt {b x +a}\, b^{2} c \,x^{2}+12 \sqrt {b x +a}\, b^{2} d \,x^{3}+30 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a^{2} d x +75 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a b c x -30 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a^{2} d x -75 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a b c x}{30 x} \] Input:

int((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^7,x)
 

Output:

( - 30*sqrt(a + b*x)*a**2*c + 92*sqrt(a + b*x)*a**2*d*x + 140*sqrt(a + b*x 
)*a*b*c*x + 44*sqrt(a + b*x)*a*b*d*x**2 + 20*sqrt(a + b*x)*b**2*c*x**2 + 1 
2*sqrt(a + b*x)*b**2*d*x**3 + 30*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*a**2 
*d*x + 75*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*a*b*c*x - 30*sqrt(a)*log(sq 
rt(a + b*x) + sqrt(a))*a**2*d*x - 75*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))* 
a*b*c*x)/(30*x)