\(\int \frac {x^{5/4}}{(a+b x)^{5/4}} \, dx\) [728]

Optimal result

Integrand size = 15, antiderivative size = 101 \[ \int \frac {x^{5/4}}{(a+b x)^{5/4}} \, dx=\frac {4 a \sqrt [4]{x}}{b^2 \sqrt [4]{a+b x}}+\frac {\sqrt [4]{x} (a+b x)^{3/4}}{b^2}-\frac {5 a \arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{x}}{\sqrt [4]{a+b x}}\right )}{2 b^{9/4}}-\frac {5 a \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt [4]{x}}{\sqrt [4]{a+b x}}\right )}{2 b^{9/4}} \] Output:

4*a*x^(1/4)/b^2/(b*x+a)^(1/4)+x^(1/4)*(b*x+a)^(3/4)/b^2-5/2*a*arctan(b^(1/ 
4)*x^(1/4)/(b*x+a)^(1/4))/b^(9/4)-5/2*a*arctanh(b^(1/4)*x^(1/4)/(b*x+a)^(1 
/4))/b^(9/4)
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.85 \[ \int \frac {x^{5/4}}{(a+b x)^{5/4}} \, dx=\frac {\frac {2 \sqrt [4]{b} \sqrt [4]{x} (5 a+b x)}{\sqrt [4]{a+b x}}-5 a \arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{x}}{\sqrt [4]{a+b x}}\right )-5 a \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt [4]{x}}{\sqrt [4]{a+b x}}\right )}{2 b^{9/4}} \] Input:

Integrate[x^(5/4)/(a + b*x)^(5/4),x]
 

Output:

((2*b^(1/4)*x^(1/4)*(5*a + b*x))/(a + b*x)^(1/4) - 5*a*ArcTan[(b^(1/4)*x^( 
1/4))/(a + b*x)^(1/4)] - 5*a*ArcTanh[(b^(1/4)*x^(1/4))/(a + b*x)^(1/4)])/( 
2*b^(9/4))
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {57, 60, 73, 770, 756, 216, 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.

\(\displaystyle \int \frac {x^{5/4}}{(a+b x)^{5/4}} \, dx\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {5 \int \frac {\sqrt [4]{x}}{\sqrt [4]{a+b x}}dx}{b}-\frac {4 x^{5/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {5 \left (\frac {\sqrt [4]{x} (a+b x)^{3/4}}{b}-\frac {a \int \frac {1}{x^{3/4} \sqrt [4]{a+b x}}dx}{4 b}\right )}{b}-\frac {4 x^{5/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {5 \left (\frac {\sqrt [4]{x} (a+b x)^{3/4}}{b}-\frac {a \int \frac {1}{\sqrt [4]{a+b x}}d\sqrt [4]{x}}{b}\right )}{b}-\frac {4 x^{5/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 770

\(\displaystyle \frac {5 \left (\frac {\sqrt [4]{x} (a+b x)^{3/4}}{b}-\frac {a \int \frac {1}{1-b x}d\frac {\sqrt [4]{x}}{\sqrt [4]{a+b x}}}{b}\right )}{b}-\frac {4 x^{5/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {5 \left (\frac {\sqrt [4]{x} (a+b x)^{3/4}}{b}-\frac {a \left (\frac {\arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{x}}{\sqrt [4]{a+b x}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt [4]{x}}{\sqrt [4]{a+b x}}\right )}{2 \sqrt [4]{b}}\right )}{b}\right )}{b}-\frac {4 x^{5/4}}{b \sqrt [4]{a+b x}}\)

Input:

Int[x^(5/4)/(a + b*x)^(5/4),x]
 

Output:

(-4*x^(5/4))/(b*(a + b*x)^(1/4)) + (5*((x^(1/4)*(a + b*x)^(3/4))/b - (a*(A 
rcTan[(b^(1/4)*x^(1/4))/(a + b*x)^(1/4)]/(2*b^(1/4)) + ArcTanh[(b^(1/4)*x^ 
(1/4))/(a + b*x)^(1/4)]/(2*b^(1/4))))/b))/b
 

Defintions of rubi rules used

rule 57

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c 
, d, m, n, x]
 

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, 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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 770

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n)   Subst[In 
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, 
 b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 
/n]
 

Maple [F]

\[\int \frac {x^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {5}{4}}}d x\]

Input:

int(x^(5/4)/(b*x+a)^(5/4),x)
 

Output:

int(x^(5/4)/(b*x+a)^(5/4),x)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 310, normalized size of antiderivative = 3.07 \[ \int \frac {x^{5/4}}{(a+b x)^{5/4}} \, dx=-\frac {5 \, {\left (b^{3} x + a b^{2}\right )} \left (\frac {a^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {5 \, {\left ({\left (b x + a\right )}^{\frac {3}{4}} a x^{\frac {1}{4}} + {\left (b^{3} x + a b^{2}\right )} \left (\frac {a^{4}}{b^{9}}\right )^{\frac {1}{4}}\right )}}{b x + a}\right ) - 5 \, {\left (b^{3} x + a b^{2}\right )} \left (\frac {a^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {5 \, {\left ({\left (b x + a\right )}^{\frac {3}{4}} a x^{\frac {1}{4}} - {\left (b^{3} x + a b^{2}\right )} \left (\frac {a^{4}}{b^{9}}\right )^{\frac {1}{4}}\right )}}{b x + a}\right ) - 5 \, {\left (i \, b^{3} x + i \, a b^{2}\right )} \left (\frac {a^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {5 \, {\left ({\left (b x + a\right )}^{\frac {3}{4}} a x^{\frac {1}{4}} - {\left (i \, b^{3} x + i \, a b^{2}\right )} \left (\frac {a^{4}}{b^{9}}\right )^{\frac {1}{4}}\right )}}{b x + a}\right ) - 5 \, {\left (-i \, b^{3} x - i \, a b^{2}\right )} \left (\frac {a^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {5 \, {\left ({\left (b x + a\right )}^{\frac {3}{4}} a x^{\frac {1}{4}} - {\left (-i \, b^{3} x - i \, a b^{2}\right )} \left (\frac {a^{4}}{b^{9}}\right )^{\frac {1}{4}}\right )}}{b x + a}\right ) - 4 \, {\left (b x + 5 \, a\right )} {\left (b x + a\right )}^{\frac {3}{4}} x^{\frac {1}{4}}}{4 \, {\left (b^{3} x + a b^{2}\right )}} \] Input:

integrate(x^(5/4)/(b*x+a)^(5/4),x, algorithm="fricas")
 

Output:

-1/4*(5*(b^3*x + a*b^2)*(a^4/b^9)^(1/4)*log(5*((b*x + a)^(3/4)*a*x^(1/4) + 
 (b^3*x + a*b^2)*(a^4/b^9)^(1/4))/(b*x + a)) - 5*(b^3*x + a*b^2)*(a^4/b^9) 
^(1/4)*log(5*((b*x + a)^(3/4)*a*x^(1/4) - (b^3*x + a*b^2)*(a^4/b^9)^(1/4)) 
/(b*x + a)) - 5*(I*b^3*x + I*a*b^2)*(a^4/b^9)^(1/4)*log(5*((b*x + a)^(3/4) 
*a*x^(1/4) - (I*b^3*x + I*a*b^2)*(a^4/b^9)^(1/4))/(b*x + a)) - 5*(-I*b^3*x 
 - I*a*b^2)*(a^4/b^9)^(1/4)*log(5*((b*x + a)^(3/4)*a*x^(1/4) - (-I*b^3*x - 
 I*a*b^2)*(a^4/b^9)^(1/4))/(b*x + a)) - 4*(b*x + 5*a)*(b*x + a)^(3/4)*x^(1 
/4))/(b^3*x + a*b^2)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.36 \[ \int \frac {x^{5/4}}{(a+b x)^{5/4}} \, dx=\frac {x^{\frac {9}{4}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{a^{\frac {5}{4}} \Gamma \left (\frac {13}{4}\right )} \] Input:

integrate(x**(5/4)/(b*x+a)**(5/4),x)
 

Output:

x**(9/4)*gamma(9/4)*hyper((5/4, 9/4), (13/4,), b*x*exp_polar(I*pi)/a)/(a** 
(5/4)*gamma(13/4))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.16 \[ \int \frac {x^{5/4}}{(a+b x)^{5/4}} \, dx=\frac {4 \, a b - \frac {5 \, {\left (b x + a\right )} a}{x}}{\frac {{\left (b x + a\right )}^{\frac {1}{4}} b^{3}}{x^{\frac {1}{4}}} - \frac {{\left (b x + a\right )}^{\frac {5}{4}} b^{2}}{x^{\frac {5}{4}}}} + \frac {5 \, a {\left (\frac {2 \, \arctan \left (\frac {{\left (b x + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x + a\right )}^{\frac {1}{4}}}{x^{\frac {1}{4}}}}{b^{\frac {1}{4}} + \frac {{\left (b x + a\right )}^{\frac {1}{4}}}{x^{\frac {1}{4}}}}\right )}{b^{\frac {1}{4}}}\right )}}{4 \, b^{2}} \] Input:

integrate(x^(5/4)/(b*x+a)^(5/4),x, algorithm="maxima")
 

Output:

(4*a*b - 5*(b*x + a)*a/x)/((b*x + a)^(1/4)*b^3/x^(1/4) - (b*x + a)^(5/4)*b 
^2/x^(5/4)) + 5/4*a*(2*arctan((b*x + a)^(1/4)/(b^(1/4)*x^(1/4)))/b^(1/4) + 
 log(-(b^(1/4) - (b*x + a)^(1/4)/x^(1/4))/(b^(1/4) + (b*x + a)^(1/4)/x^(1/ 
4)))/b^(1/4))/b^2
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^{5/4}}{(a+b x)^{5/4}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^(5/4)/(b*x+a)^(5/4),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{-1,[0,1,1,0]%%%} / %%%{1,[0,0,0,1]%%%} Error: Bad Argument 
 Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{5/4}}{(a+b x)^{5/4}} \, dx=\int \frac {x^{5/4}}{{\left (a+b\,x\right )}^{5/4}} \,d x \] Input:

int(x^(5/4)/(a + b*x)^(5/4),x)
 

Output:

int(x^(5/4)/(a + b*x)^(5/4), x)
 

Reduce [F]

\[ \int \frac {x^{5/4}}{(a+b x)^{5/4}} \, dx=\int \frac {x^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {1}{4}} a +\left (b x +a \right )^{\frac {1}{4}} b x}d x \] Input:

int(x^(5/4)/(b*x+a)^(5/4),x)
 

Output:

int((x**(1/4)*x)/((a + b*x)**(1/4)*a + (a + b*x)**(1/4)*b*x),x)