\(\int \frac {(e x)^{1+\frac {3 n}{2}}}{(c+d x) (a x^n+b x^{1+n})^{3/2}} \, dx\) [259]

Optimal result

Integrand size = 36, antiderivative size = 138 \[ \int \frac {(e x)^{1+\frac {3 n}{2}}}{(c+d x) \left (a x^n+b x^{1+n}\right )^{3/2}} \, dx=\frac {2 a e x^{-n} (e x)^{3 n/2}}{b (b c-a d) \sqrt {a x^n+b x^{1+n}}}+\frac {2 c e x^{-n} (e x)^{3 n/2} \sqrt {a+b x} \arctan \left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {d} (b c-a d)^{3/2} \sqrt {a x^n+b x^{1+n}}} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.80 \[ \int \frac {(e x)^{1+\frac {3 n}{2}}}{(c+d x) \left (a x^n+b x^{1+n}\right )^{3/2}} \, dx=\frac {2 e (e x)^{3 n/2} (a+b x) \left (a \sqrt {d} \sqrt {b c-a d}+b c \sqrt {a+b x} \arctan \left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )\right )}{b \sqrt {d} (b c-a d)^{3/2} \left (x^n (a+b x)\right )^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.84, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1948, 87, 73, 218}

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.

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

\(\Big \downarrow \) 1948

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

Defintions of rubi rules used

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 218

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

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 439, normalized size of antiderivative = 3.18 \[ \int \frac {(e x)^{1+\frac {3 n}{2}}}{(c+d x) \left (a x^n+b x^{1+n}\right )^{3/2}} \, dx=\left [\frac {{\left (b^{2} c x + a b c\right )} \sqrt {-b c d + a d^{2}} e^{\frac {3}{2} \, n + 1} x^{\frac {1}{2} \, n + \frac {1}{2}} \log \left (\frac {{\left (b d x - b c + 2 \, a d\right )} x^{\frac {1}{2} \, n + \frac {1}{2}} + 2 \, \sqrt {-b c d + a d^{2}} \sqrt {x} \sqrt {\frac {{\left (b x + a\right )} x^{n + 1}}{x}}}{{\left (d x + c\right )} x^{\frac {1}{2} \, n + \frac {1}{2}}}\right ) + 2 \, {\left (a b c d - a^{2} d^{2}\right )} e^{\frac {3}{2} \, n + 1} \sqrt {x} \sqrt {\frac {{\left (b x + a\right )} x^{n + 1}}{x}}}{{\left (a b^{3} c^{2} d - 2 \, a^{2} b^{2} c d^{2} + a^{3} b d^{3} + {\left (b^{4} c^{2} d - 2 \, a b^{3} c d^{2} + a^{2} b^{2} d^{3}\right )} x\right )} x^{\frac {1}{2} \, n + \frac {1}{2}}}, -\frac {2 \, {\left ({\left (b^{2} c x + a b c\right )} \sqrt {b c d - a d^{2}} e^{\frac {3}{2} \, n + 1} x^{\frac {1}{2} \, n + \frac {1}{2}} \arctan \left (\frac {\sqrt {b c d - a d^{2}} \sqrt {x} \sqrt {\frac {{\left (b x + a\right )} x^{n + 1}}{x}}}{{\left (b d x + a d\right )} x^{\frac {1}{2} \, n + \frac {1}{2}}}\right ) - {\left (a b c d - a^{2} d^{2}\right )} e^{\frac {3}{2} \, n + 1} \sqrt {x} \sqrt {\frac {{\left (b x + a\right )} x^{n + 1}}{x}}\right )}}{{\left (a b^{3} c^{2} d - 2 \, a^{2} b^{2} c d^{2} + a^{3} b d^{3} + {\left (b^{4} c^{2} d - 2 \, a b^{3} c d^{2} + a^{2} b^{2} d^{3}\right )} x\right )} x^{\frac {1}{2} \, n + \frac {1}{2}}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.75 \[ \int \frac {(e x)^{1+\frac {3 n}{2}}}{(c+d x) \left (a x^n+b x^{1+n}\right )^{3/2}} \, dx=\frac {2 e^{\frac {3 n}{2}} e \left (\sqrt {d}\, \sqrt {b x +a}\, \sqrt {-a d +b c}\, \mathit {atan} \left (\frac {\sqrt {b x +a}\, d}{\sqrt {d}\, \sqrt {-a d +b c}}\right ) b c -a^{2} d^{2}+a b c d \right )}{\sqrt {b x +a}\, b d \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )} \] Input:

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

Output:

(2*e**((3*n)/2)*e*(sqrt(d)*sqrt(a + b*x)*sqrt( - a*d + b*c)*atan((sqrt(a + 
 b*x)*d)/(sqrt(d)*sqrt( - a*d + b*c)))*b*c - a**2*d**2 + a*b*c*d))/(sqrt(a 
 + b*x)*b*d*(a**2*d**2 - 2*a*b*c*d + b**2*c**2))