\(\int \frac {x^{11/4}}{(a+b x)^{5/4}} \, dx\) [734]

Optimal result

Integrand size = 15, antiderivative size = 134 \[ \int \frac {x^{11/4}}{(a+b x)^{5/4}} \, dx=-\frac {4 a^2 x^{3/4}}{b^3 \sqrt [4]{a+b x}}-\frac {17 a x^{3/4} (a+b x)^{3/4}}{15 b^3}+\frac {2 x^{7/4} (a+b x)^{3/4}}{5 b^2}+\frac {77 a^3 \sqrt [4]{-\frac {b x}{a}-\frac {b^2 x^2}{a^2}} E\left (\left .\frac {1}{2} \arcsin \left (1+\frac {2 b x}{a}\right )\right |2\right )}{10 \sqrt {2} b^4 \sqrt [4]{x} \sqrt [4]{a+b x}} \] Output:

-4*a^2*x^(3/4)/b^3/(b*x+a)^(1/4)-17/15*a*x^(3/4)*(b*x+a)^(3/4)/b^3+2/5*x^( 
7/4)*(b*x+a)^(3/4)/b^2+77/20*a^3*(-b*x/a-b^2*x^2/a^2)^(1/4)*EllipticE(sin( 
1/2*arcsin(1+2*b*x/a)),2^(1/2))*2^(1/2)/b^4/x^(1/4)/(b*x+a)^(1/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.37 \[ \int \frac {x^{11/4}}{(a+b x)^{5/4}} \, dx=\frac {4 x^{15/4} \sqrt [4]{1+\frac {b x}{a}} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {15}{4},\frac {19}{4},-\frac {b x}{a}\right )}{15 a \sqrt [4]{a+b x}} \] Input:

Integrate[x^(11/4)/(a + b*x)^(5/4),x]
 

Output:

(4*x^(15/4)*(1 + (b*x)/a)^(1/4)*Hypergeometric2F1[5/4, 15/4, 19/4, -((b*x) 
/a)])/(15*a*(a + b*x)^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.28 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.25, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {57, 60, 60, 73, 839, 813, 858, 807, 212}

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.

\(\displaystyle \int \frac {x^{11/4}}{(a+b x)^{5/4}} \, dx\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {11 \int \frac {x^{7/4}}{\sqrt [4]{a+b x}}dx}{b}-\frac {4 x^{11/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {11 \left (\frac {2 x^{7/4} (a+b x)^{3/4}}{5 b}-\frac {7 a \int \frac {x^{3/4}}{\sqrt [4]{a+b x}}dx}{10 b}\right )}{b}-\frac {4 x^{11/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {11 \left (\frac {2 x^{7/4} (a+b x)^{3/4}}{5 b}-\frac {7 a \left (\frac {2 x^{3/4} (a+b x)^{3/4}}{3 b}-\frac {a \int \frac {1}{\sqrt [4]{x} \sqrt [4]{a+b x}}dx}{2 b}\right )}{10 b}\right )}{b}-\frac {4 x^{11/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {11 \left (\frac {2 x^{7/4} (a+b x)^{3/4}}{5 b}-\frac {7 a \left (\frac {2 x^{3/4} (a+b x)^{3/4}}{3 b}-\frac {2 a \int \frac {\sqrt {x}}{\sqrt [4]{a+b x}}d\sqrt [4]{x}}{b}\right )}{10 b}\right )}{b}-\frac {4 x^{11/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 839

\(\displaystyle \frac {11 \left (\frac {2 x^{7/4} (a+b x)^{3/4}}{5 b}-\frac {7 a \left (\frac {2 x^{3/4} (a+b x)^{3/4}}{3 b}-\frac {2 a \left (\frac {x^{3/4}}{2 \sqrt [4]{a+b x}}-\frac {1}{2} a \int \frac {\sqrt {x}}{(a+b x)^{5/4}}d\sqrt [4]{x}\right )}{b}\right )}{10 b}\right )}{b}-\frac {4 x^{11/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 813

\(\displaystyle \frac {11 \left (\frac {2 x^{7/4} (a+b x)^{3/4}}{5 b}-\frac {7 a \left (\frac {2 x^{3/4} (a+b x)^{3/4}}{3 b}-\frac {2 a \left (\frac {x^{3/4}}{2 \sqrt [4]{a+b x}}-\frac {a \sqrt [4]{x} \sqrt [4]{\frac {a}{b x}+1} \int \frac {1}{\left (\frac {a}{b x}+1\right )^{5/4} x^{3/4}}d\sqrt [4]{x}}{2 b \sqrt [4]{a+b x}}\right )}{b}\right )}{10 b}\right )}{b}-\frac {4 x^{11/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {11 \left (\frac {2 x^{7/4} (a+b x)^{3/4}}{5 b}-\frac {7 a \left (\frac {2 x^{3/4} (a+b x)^{3/4}}{3 b}-\frac {2 a \left (\frac {a \sqrt [4]{x} \sqrt [4]{\frac {a}{b x}+1} \int \frac {1}{\sqrt [4]{x} \left (\frac {a x}{b}+1\right )^{5/4}}d\frac {1}{\sqrt [4]{x}}}{2 b \sqrt [4]{a+b x}}+\frac {x^{3/4}}{2 \sqrt [4]{a+b x}}\right )}{b}\right )}{10 b}\right )}{b}-\frac {4 x^{11/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {11 \left (\frac {2 x^{7/4} (a+b x)^{3/4}}{5 b}-\frac {7 a \left (\frac {2 x^{3/4} (a+b x)^{3/4}}{3 b}-\frac {2 a \left (\frac {a \sqrt [4]{x} \sqrt [4]{\frac {a}{b x}+1} \int \frac {1}{\left (\frac {\sqrt {x} a}{b}+1\right )^{5/4}}d\sqrt {x}}{4 b \sqrt [4]{a+b x}}+\frac {x^{3/4}}{2 \sqrt [4]{a+b x}}\right )}{b}\right )}{10 b}\right )}{b}-\frac {4 x^{11/4}}{b \sqrt [4]{a+b x}}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {11 \left (\frac {2 x^{7/4} (a+b x)^{3/4}}{5 b}-\frac {7 a \left (\frac {2 x^{3/4} (a+b x)^{3/4}}{3 b}-\frac {2 a \left (\frac {\sqrt {a} \sqrt [4]{x} \sqrt [4]{\frac {a}{b x}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )\right |2\right )}{2 \sqrt {b} \sqrt [4]{a+b x}}+\frac {x^{3/4}}{2 \sqrt [4]{a+b x}}\right )}{b}\right )}{10 b}\right )}{b}-\frac {4 x^{11/4}}{b \sqrt [4]{a+b x}}\)

Input:

Int[x^(11/4)/(a + b*x)^(5/4),x]
 

Output:

(-4*x^(11/4))/(b*(a + b*x)^(1/4)) + (11*((2*x^(7/4)*(a + b*x)^(3/4))/(5*b) 
 - (7*a*((2*x^(3/4)*(a + b*x)^(3/4))/(3*b) - (2*a*(x^(3/4)/(2*(a + b*x)^(1 
/4)) + (Sqrt[a]*(1 + a/(b*x))^(1/4)*x^(1/4)*EllipticE[ArcTan[(Sqrt[a]*Sqrt 
[x])/Sqrt[b]]/2, 2])/(2*Sqrt[b]*(a + b*x)^(1/4))))/b))/(10*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 212

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) 
)*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 813

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x*((1 + a/(b*x^4) 
)^(1/4)/(b*(a + b*x^4)^(1/4)))   Int[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] 
/; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 839

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(1/4), x_Symbol] :> Simp[x^3/(2*(a + b*x^4 
)^(1/4)), x] - Simp[a/2   Int[x^2/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b} 
, x] && PosQ[b/a]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

Maple [F]

\[\int \frac {x^{\frac {11}{4}}}{\left (b x +a \right )^{\frac {5}{4}}}d x\]

Input:

int(x^(11/4)/(b*x+a)^(5/4),x)
 

Output:

int(x^(11/4)/(b*x+a)^(5/4),x)
 

Fricas [F]

\[ \int \frac {x^{11/4}}{(a+b x)^{5/4}} \, dx=\int { \frac {x^{\frac {11}{4}}}{{\left (b x + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(x^(11/4)/(b*x+a)^(5/4),x, algorithm="fricas")
 

Output:

integral((b*x + a)^(3/4)*x^(11/4)/(b^2*x^2 + 2*a*b*x + a^2), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 58.91 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.27 \[ \int \frac {x^{11/4}}{(a+b x)^{5/4}} \, dx=\frac {x^{\frac {15}{4}} \Gamma \left (\frac {15}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {15}{4} \\ \frac {19}{4} \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{a^{\frac {5}{4}} \Gamma \left (\frac {19}{4}\right )} \] Input:

integrate(x**(11/4)/(b*x+a)**(5/4),x)
 

Output:

x**(15/4)*gamma(15/4)*hyper((5/4, 15/4), (19/4,), b*x*exp_polar(I*pi)/a)/( 
a**(5/4)*gamma(19/4))
 

Maxima [F]

\[ \int \frac {x^{11/4}}{(a+b x)^{5/4}} \, dx=\int { \frac {x^{\frac {11}{4}}}{{\left (b x + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(x^(11/4)/(b*x+a)^(5/4),x, algorithm="maxima")
 

Output:

integrate(x^(11/4)/(b*x + a)^(5/4), x)
 

Giac [F]

\[ \int \frac {x^{11/4}}{(a+b x)^{5/4}} \, dx=\int { \frac {x^{\frac {11}{4}}}{{\left (b x + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(x^(11/4)/(b*x+a)^(5/4),x, algorithm="giac")
 

Output:

integrate(x^(11/4)/(b*x + a)^(5/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{11/4}}{(a+b x)^{5/4}} \, dx=\int \frac {x^{11/4}}{{\left (a+b\,x\right )}^{5/4}} \,d x \] Input:

int(x^(11/4)/(a + b*x)^(5/4),x)
 

Output:

int(x^(11/4)/(a + b*x)^(5/4), x)
 

Reduce [F]

\[ \int \frac {x^{11/4}}{(a+b x)^{5/4}} \, dx=\int \frac {x^{\frac {11}{4}}}{\left (b x +a \right )^{\frac {1}{4}} a +\left (b x +a \right )^{\frac {1}{4}} b x}d x \] Input:

int(x^(11/4)/(b*x+a)^(5/4),x)
 

Output:

int((x**(3/4)*x**2)/((a + b*x)**(1/4)*a + (a + b*x)**(1/4)*b*x),x)