\(\int \frac {\sqrt [4]{a+b x}}{x^{15/4}} \, dx\) [697]

Optimal result

Integrand size = 15, antiderivative size = 127 \[ \int \frac {\sqrt [4]{a+b x}}{x^{15/4}} \, dx=-\frac {4 \sqrt [4]{a+b x}}{11 x^{11/4}}-\frac {4 b \sqrt [4]{a+b x}}{77 a x^{7/4}}+\frac {8 b^2 \sqrt [4]{a+b x}}{77 a^2 x^{3/4}}-\frac {16 b^2 \left (\frac {b x}{a+b x}\right )^{3/4} (a+b x)^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {a}}{\sqrt {a+b x}}\right ),2\right )}{77 a^{5/2} x^{3/4}} \] Output:

-4/11*(b*x+a)^(1/4)/x^(11/4)-4/77*b*(b*x+a)^(1/4)/a/x^(7/4)+8/77*b^2*(b*x+ 
a)^(1/4)/a^2/x^(3/4)-16/77*b^2*(b*x/(b*x+a))^(3/4)*(b*x+a)^(3/4)*InverseJa 
cobiAM(1/2*arcsin(a^(1/2)/(b*x+a)^(1/2)),2^(1/2))/a^(5/2)/x^(3/4)
 

Mathematica [C] (verified)

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

Time = 10.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.37 \[ \int \frac {\sqrt [4]{a+b x}}{x^{15/4}} \, dx=-\frac {4 \sqrt [4]{a+b x} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {1}{4},-\frac {7}{4},-\frac {b x}{a}\right )}{11 x^{11/4} \sqrt [4]{1+\frac {b x}{a}}} \] Input:

Integrate[(a + b*x)^(1/4)/x^(15/4),x]
 

Output:

(-4*(a + b*x)^(1/4)*Hypergeometric2F1[-11/4, -1/4, -7/4, -((b*x)/a)])/(11* 
x^(11/4)*(1 + (b*x)/a)^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.25 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {57, 61, 61, 73, 768, 858, 807, 229}

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.

\(\displaystyle \int \frac {\sqrt [4]{a+b x}}{x^{15/4}} \, dx\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {1}{11} b \int \frac {1}{x^{11/4} (a+b x)^{3/4}}dx-\frac {4 \sqrt [4]{a+b x}}{11 x^{11/4}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {1}{11} b \left (-\frac {6 b \int \frac {1}{x^{7/4} (a+b x)^{3/4}}dx}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )-\frac {4 \sqrt [4]{a+b x}}{11 x^{11/4}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {1}{11} b \left (-\frac {6 b \left (-\frac {2 b \int \frac {1}{x^{3/4} (a+b x)^{3/4}}dx}{3 a}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )-\frac {4 \sqrt [4]{a+b x}}{11 x^{11/4}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{11} b \left (-\frac {6 b \left (-\frac {8 b \int \frac {1}{(a+b x)^{3/4}}d\sqrt [4]{x}}{3 a}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )-\frac {4 \sqrt [4]{a+b x}}{11 x^{11/4}}\)

\(\Big \downarrow \) 768

\(\displaystyle \frac {1}{11} b \left (-\frac {6 b \left (-\frac {8 b x^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x}+1\right )^{3/4} x^{3/4}}d\sqrt [4]{x}}{3 a (a+b x)^{3/4}}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )-\frac {4 \sqrt [4]{a+b x}}{11 x^{11/4}}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {1}{11} b \left (-\frac {6 b \left (\frac {8 b x^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \int \frac {1}{\sqrt [4]{x} \left (\frac {a x}{b}+1\right )^{3/4}}d\frac {1}{\sqrt [4]{x}}}{3 a (a+b x)^{3/4}}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )-\frac {4 \sqrt [4]{a+b x}}{11 x^{11/4}}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {1}{11} b \left (-\frac {6 b \left (\frac {4 b x^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \int \frac {1}{\left (\frac {\sqrt {x} a}{b}+1\right )^{3/4}}d\sqrt {x}}{3 a (a+b x)^{3/4}}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )-\frac {4 \sqrt [4]{a+b x}}{11 x^{11/4}}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {1}{11} b \left (-\frac {6 b \left (\frac {8 b^{3/2} x^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right ),2\right )}{3 a^{3/2} (a+b x)^{3/4}}-\frac {4 \sqrt [4]{a+b x}}{3 a x^{3/4}}\right )}{7 a}-\frac {4 \sqrt [4]{a+b x}}{7 a x^{7/4}}\right )-\frac {4 \sqrt [4]{a+b x}}{11 x^{11/4}}\)

Input:

Int[(a + b*x)^(1/4)/x^(15/4),x]
 

Output:

(-4*(a + b*x)^(1/4))/(11*x^(11/4)) + (b*((-4*(a + b*x)^(1/4))/(7*a*x^(7/4) 
) - (6*b*((-4*(a + b*x)^(1/4))/(3*a*x^(3/4)) + (8*b^(3/2)*(1 + a/(b*x))^(3 
/4)*x^(3/4)*EllipticF[ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]]/2, 2])/(3*a^(3/2)* 
(a + b*x)^(3/4))))/(7*a)))/11
 

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 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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 229

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

rule 768

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

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 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]

\[\int \frac {\left (b x +a \right )^{\frac {1}{4}}}{x^{\frac {15}{4}}}d x\]

Input:

int((b*x+a)^(1/4)/x^(15/4),x)
 

Output:

int((b*x+a)^(1/4)/x^(15/4),x)
 

Fricas [F]

\[ \int \frac {\sqrt [4]{a+b x}}{x^{15/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{4}}}{x^{\frac {15}{4}}} \,d x } \] Input:

integrate((b*x+a)^(1/4)/x^(15/4),x, algorithm="fricas")
 

Output:

integral((b*x + a)^(1/4)/x^(15/4), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 66.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.35 \[ \int \frac {\sqrt [4]{a+b x}}{x^{15/4}} \, dx=\frac {\sqrt [4]{a} \Gamma \left (- \frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {11}{4}, - \frac {1}{4} \\ - \frac {7}{4} \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{x^{\frac {11}{4}} \Gamma \left (- \frac {7}{4}\right )} \] Input:

integrate((b*x+a)**(1/4)/x**(15/4),x)
 

Output:

a**(1/4)*gamma(-11/4)*hyper((-11/4, -1/4), (-7/4,), b*x*exp_polar(I*pi)/a) 
/(x**(11/4)*gamma(-7/4))
 

Maxima [F]

\[ \int \frac {\sqrt [4]{a+b x}}{x^{15/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{4}}}{x^{\frac {15}{4}}} \,d x } \] Input:

integrate((b*x+a)^(1/4)/x^(15/4),x, algorithm="maxima")
 

Output:

integrate((b*x + a)^(1/4)/x^(15/4), x)
 

Giac [F]

\[ \int \frac {\sqrt [4]{a+b x}}{x^{15/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{4}}}{x^{\frac {15}{4}}} \,d x } \] Input:

integrate((b*x+a)^(1/4)/x^(15/4),x, algorithm="giac")
 

Output:

integrate((b*x + a)^(1/4)/x^(15/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{a+b x}}{x^{15/4}} \, dx=\int \frac {{\left (a+b\,x\right )}^{1/4}}{x^{15/4}} \,d x \] Input:

int((a + b*x)^(1/4)/x^(15/4),x)
 

Output:

int((a + b*x)^(1/4)/x^(15/4), x)
 

Reduce [F]

\[ \int \frac {\sqrt [4]{a+b x}}{x^{15/4}} \, dx=\frac {-4 \left (b x +a \right )^{\frac {1}{4}}-x^{\frac {11}{4}} \left (\int \frac {\left (b x +a \right )^{\frac {1}{4}}}{x^{\frac {15}{4}} a +x^{\frac {19}{4}} b}d x \right ) a}{10 x^{\frac {11}{4}}} \] Input:

int((b*x+a)^(1/4)/x^(15/4),x)
 

Output:

( - 4*(a + b*x)**(1/4) - x**(3/4)*int((a + b*x)**(1/4)/(x**(3/4)*a*x**3 + 
x**(3/4)*b*x**4),x)*a*x**2)/(10*x**(3/4)*x**2)