\(\int \frac {(e x)^{-2+\frac {3 n}{2}} (c+d x)^2}{(a x^n+b x^{1+n})^{3/2}} \, dx\) [256]

Optimal result

Integrand size = 36, antiderivative size = 172 \[ \int \frac {(e x)^{-2+\frac {3 n}{2}} (c+d x)^2}{\left (a x^n+b x^{1+n}\right )^{3/2}} \, dx=-\frac {2 (b c-a d)^2 x^{-n} (e x)^{3 n/2}}{a^2 b e^2 \sqrt {a x^n+b x^{1+n}}}-\frac {c^2 x^{-1-2 n} (e x)^{3 n/2} \sqrt {a x^n+b x^{1+n}}}{a^2 e^2}+\frac {c (3 b c-4 a d) x^{-n} (e x)^{3 n/2} \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2} e^2 \sqrt {a x^n+b x^{1+n}}} \] Output:

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.67 \[ \int \frac {(e x)^{-2+\frac {3 n}{2}} (c+d x)^2}{\left (a x^n+b x^{1+n}\right )^{3/2}} \, dx=\frac {x^{-1-n} (e x)^{3 n/2} \left (-\sqrt {a} \left (3 b^2 c^2 x+2 a^2 d^2 x+a b c (c-4 d x)\right )+b c (3 b c-4 a d) x \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{a^{5/2} b e^2 \sqrt {x^n (a+b x)}} \] Input:

Integrate[((e*x)^(-2 + (3*n)/2)*(c + d*x)^2)/(a*x^n + b*x^(1 + n))^(3/2),x 
]
 

Output:

(x^(-1 - n)*(e*x)^((3*n)/2)*(-(Sqrt[a]*(3*b^2*c^2*x + 2*a^2*d^2*x + a*b*c* 
(c - 4*d*x))) + b*c*(3*b*c - 4*a*d)*x*Sqrt[a + b*x]*ArcTanh[Sqrt[a + b*x]/ 
Sqrt[a]]))/(a^(5/2)*b*e^2*Sqrt[x^n*(a + b*x)])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.82, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1948, 100, 27, 87, 73, 221}

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.

\(\displaystyle \int \frac {(c+d x)^2 (e x)^{\frac {3 n}{2}-2}}{\left (a x^n+b x^{n+1}\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1948

\(\displaystyle \frac {x^{-n} \sqrt {a+b x} (e x)^{3 n/2} \int \frac {(c+d x)^2}{x^2 (a+b x)^{3/2}}dx}{e^2 \sqrt {a x^n+b x^{n+1}}}\)

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {x^{-n} \sqrt {a+b x} (e x)^{3 n/2} \left (-\frac {\frac {2 \left (\frac {c (3 b c-4 a d)}{a}+\frac {2 a d^2}{b}\right )}{\sqrt {a+b x}}-\frac {2 c \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) (3 b c-4 a d)}{a^{3/2}}}{2 a}-\frac {c^2}{a x \sqrt {a+b x}}\right )}{e^2 \sqrt {a x^n+b x^{n+1}}}\)

Input:

Int[((e*x)^(-2 + (3*n)/2)*(c + d*x)^2)/(a*x^n + b*x^(1 + n))^(3/2),x]
 

Output:

((e*x)^((3*n)/2)*Sqrt[a + b*x]*(-(c^2/(a*x*Sqrt[a + b*x])) - ((2*((2*a*d^2 
)/b + (c*(3*b*c - 4*a*d))/a))/Sqrt[a + b*x] - (2*c*(3*b*c - 4*a*d)*ArcTanh 
[Sqrt[a + b*x]/Sqrt[a]])/a^(3/2))/(2*a)))/(e^2*x^n*Sqrt[a*x^n + b*x^(1 + n 
)])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 1948

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Simp[e^IntPart[m]*(e*x)^FracPart[m]*( 
(a*x^j + b*x^(j + n))^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x 
^n)^FracPart[p]))   Int[x^(m + j*p)*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; 
FreeQ[{a, b, c, d, e, j, m, n, p, q}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] 
 && NeQ[b*c - a*d, 0] &&  !(EqQ[n, 1] && EqQ[j, 1])
 

Maple [F]

\[\int \frac {\left (e x \right )^{-2+\frac {3 n}{2}} \left (d x +c \right )^{2}}{\left (a \,x^{n}+b \,x^{1+n}\right )^{\frac {3}{2}}}d x\]

Input:

int((e*x)^(-2+3/2*n)*(d*x+c)^2/(a*x^n+b*x^(1+n))^(3/2),x)
 

Output:

int((e*x)^(-2+3/2*n)*(d*x+c)^2/(a*x^n+b*x^(1+n))^(3/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 410, normalized size of antiderivative = 2.38 \[ \int \frac {(e x)^{-2+\frac {3 n}{2}} (c+d x)^2}{\left (a x^n+b x^{1+n}\right )^{3/2}} \, dx=\left [-\frac {{\left ({\left (3 \, b^{3} c^{2} - 4 \, a b^{2} c d\right )} x^{2} + {\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x\right )} \sqrt {a} e^{\frac {3}{2} \, n - 2} x^{\frac {1}{2} \, n + \frac {1}{2}} \log \left (\frac {{\left (b x + 2 \, a\right )} x^{\frac {1}{2} \, n + \frac {1}{2}} - 2 \, \sqrt {a} \sqrt {x} \sqrt {\frac {{\left (b x + a\right )} x^{n + 1}}{x}}}{x x^{\frac {1}{2} \, n + \frac {1}{2}}}\right ) + 2 \, {\left (a^{2} b c^{2} + {\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d + 2 \, a^{3} d^{2}\right )} x\right )} e^{\frac {3}{2} \, n - 2} \sqrt {x} \sqrt {\frac {{\left (b x + a\right )} x^{n + 1}}{x}}}{2 \, {\left (a^{3} b^{2} x^{2} + a^{4} b x\right )} x^{\frac {1}{2} \, n + \frac {1}{2}}}, -\frac {{\left ({\left (3 \, b^{3} c^{2} - 4 \, a b^{2} c d\right )} x^{2} + {\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x\right )} \sqrt {-a} e^{\frac {3}{2} \, n - 2} x^{\frac {1}{2} \, n + \frac {1}{2}} \arctan \left (\frac {\sqrt {-a} \sqrt {x} \sqrt {\frac {{\left (b x + a\right )} x^{n + 1}}{x}}}{{\left (b x + a\right )} x^{\frac {1}{2} \, n + \frac {1}{2}}}\right ) + {\left (a^{2} b c^{2} + {\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d + 2 \, a^{3} d^{2}\right )} x\right )} e^{\frac {3}{2} \, n - 2} \sqrt {x} \sqrt {\frac {{\left (b x + a\right )} x^{n + 1}}{x}}}{{\left (a^{3} b^{2} x^{2} + a^{4} b x\right )} x^{\frac {1}{2} \, n + \frac {1}{2}}}\right ] \] Input:

integrate((e*x)^(-2+3/2*n)*(d*x+c)^2/(a*x^n+b*x^(1+n))^(3/2),x, algorithm= 
"fricas")
 

Output:

[-1/2*(((3*b^3*c^2 - 4*a*b^2*c*d)*x^2 + (3*a*b^2*c^2 - 4*a^2*b*c*d)*x)*sqr 
t(a)*e^(3/2*n - 2)*x^(1/2*n + 1/2)*log(((b*x + 2*a)*x^(1/2*n + 1/2) - 2*sq 
rt(a)*sqrt(x)*sqrt((b*x + a)*x^(n + 1)/x))/(x*x^(1/2*n + 1/2))) + 2*(a^2*b 
*c^2 + (3*a*b^2*c^2 - 4*a^2*b*c*d + 2*a^3*d^2)*x)*e^(3/2*n - 2)*sqrt(x)*sq 
rt((b*x + a)*x^(n + 1)/x))/((a^3*b^2*x^2 + a^4*b*x)*x^(1/2*n + 1/2)), -((( 
3*b^3*c^2 - 4*a*b^2*c*d)*x^2 + (3*a*b^2*c^2 - 4*a^2*b*c*d)*x)*sqrt(-a)*e^( 
3/2*n - 2)*x^(1/2*n + 1/2)*arctan(sqrt(-a)*sqrt(x)*sqrt((b*x + a)*x^(n + 1 
)/x)/((b*x + a)*x^(1/2*n + 1/2))) + (a^2*b*c^2 + (3*a*b^2*c^2 - 4*a^2*b*c* 
d + 2*a^3*d^2)*x)*e^(3/2*n - 2)*sqrt(x)*sqrt((b*x + a)*x^(n + 1)/x))/((a^3 
*b^2*x^2 + a^4*b*x)*x^(1/2*n + 1/2))]
 

Sympy [F]

\[ \int \frac {(e x)^{-2+\frac {3 n}{2}} (c+d x)^2}{\left (a x^n+b x^{1+n}\right )^{3/2}} \, dx=\int \frac {\left (e x\right )^{\frac {3 n}{2} - 2} \left (c + d x\right )^{2}}{\left (a x^{n} + b x^{n + 1}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((e*x)**(-2+3/2*n)*(d*x+c)**2/(a*x**n+b*x**(1+n))**(3/2),x)
 

Output:

Integral((e*x)**(3*n/2 - 2)*(c + d*x)**2/(a*x**n + b*x**(n + 1))**(3/2), x 
)
 

Maxima [F]

\[ \int \frac {(e x)^{-2+\frac {3 n}{2}} (c+d x)^2}{\left (a x^n+b x^{1+n}\right )^{3/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{2} \left (e x\right )^{\frac {3}{2} \, n - 2}}{{\left (b x^{n + 1} + a x^{n}\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((e*x)^(-2+3/2*n)*(d*x+c)^2/(a*x^n+b*x^(1+n))^(3/2),x, algorithm= 
"maxima")
 

Output:

integrate((d*x + c)^2*(e*x)^(3/2*n - 2)/(b*x^(n + 1) + a*x^n)^(3/2), x)
 

Giac [F]

\[ \int \frac {(e x)^{-2+\frac {3 n}{2}} (c+d x)^2}{\left (a x^n+b x^{1+n}\right )^{3/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{2} \left (e x\right )^{\frac {3}{2} \, n - 2}}{{\left (b x^{n + 1} + a x^{n}\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((e*x)^(-2+3/2*n)*(d*x+c)^2/(a*x^n+b*x^(1+n))^(3/2),x, algorithm= 
"giac")
 

Output:

integrate((d*x + c)^2*(e*x)^(3/2*n - 2)/(b*x^(n + 1) + a*x^n)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{-2+\frac {3 n}{2}} (c+d x)^2}{\left (a x^n+b x^{1+n}\right )^{3/2}} \, dx=\int \frac {{\left (e\,x\right )}^{\frac {3\,n}{2}-2}\,{\left (c+d\,x\right )}^2}{{\left (a\,x^n+b\,x^{n+1}\right )}^{3/2}} \,d x \] Input:

int(((e*x)^((3*n)/2 - 2)*(c + d*x)^2)/(a*x^n + b*x^(n + 1))^(3/2),x)
 

Output:

int(((e*x)^((3*n)/2 - 2)*(c + d*x)^2)/(a*x^n + b*x^(n + 1))^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.01 \[ \int \frac {(e x)^{-2+\frac {3 n}{2}} (c+d x)^2}{\left (a x^n+b x^{1+n}\right )^{3/2}} \, dx=\frac {e^{\frac {3 n}{2}} \left (4 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a b c d x -3 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{2} c^{2} x -4 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a b c d x +3 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{2} c^{2} x -4 a^{3} d^{2} x -2 a^{2} b \,c^{2}+8 a^{2} b c d x -6 a \,b^{2} c^{2} x \right )}{2 \sqrt {b x +a}\, a^{3} b \,e^{2} x} \] Input:

int((e*x)^(-2+3/2*n)*(d*x+c)^2/(a*x^n+b*x^(1+n))^(3/2),x)
 

Output:

(e**((3*n)/2)*(4*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*a*b*c* 
d*x - 3*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*b**2*c**2*x - 4 
*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*a*b*c*d*x + 3*sqrt(a)* 
sqrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*b**2*c**2*x - 4*a**3*d**2*x - 2 
*a**2*b*c**2 + 8*a**2*b*c*d*x - 6*a*b**2*c**2*x))/(2*sqrt(a + b*x)*a**3*b* 
e**2*x)