\(\int \frac {1}{(2+b x)^{5/4} \sqrt [4]{a+b x}} \, dx\) [649]

Optimal result
Mathematica [C] (verified)
Rubi [A] (warning: unable to verify)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 19, antiderivative size = 75 \[ \int \frac {1}{(2+b x)^{5/4} \sqrt [4]{a+b x}} \, dx=-\frac {4 \sqrt [4]{\frac {2+b x}{a+b x}} \sqrt [4]{a+b x} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {2-a}}{\sqrt {a+b x}}\right )\right |2\right )}{\sqrt {2-a} b \sqrt [4]{2+b x}} \] Output:

-4*((b*x+2)/(b*x+a))^(1/4)*(b*x+a)^(1/4)*EllipticE(sin(1/2*arctan((2-a)^(1 
/2)/(b*x+a)^(1/2))),2^(1/2))/(2-a)^(1/2)/b/(b*x+2)^(1/4)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(2+b x)^{5/4} \sqrt [4]{a+b x}} \, dx=-\frac {4 \sqrt [4]{\frac {a+b x}{-2+a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4},\frac {3}{4},\frac {2+b x}{2-a}\right )}{b \sqrt [4]{2+b x} \sqrt [4]{a+b x}} \] Input:

Integrate[1/((2 + b*x)^(5/4)*(a + b*x)^(1/4)),x]
 

Output:

(-4*((a + b*x)/(-2 + a))^(1/4)*Hypergeometric2F1[-1/4, 1/4, 3/4, (2 + b*x) 
/(2 - a)])/(b*(2 + b*x)^(1/4)*(a + b*x)^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.27 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.91, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {61, 73, 840, 842, 858, 807, 226}

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.

\(\displaystyle \int \frac {1}{(b x+2)^{5/4} \sqrt [4]{a+b x}} \, dx\)

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {4 (a+b x)^{3/4}}{(2-a) b \sqrt [4]{b x+2}}-\frac {8 \int \frac {\sqrt {b x+2}}{\sqrt [4]{a+b x}}d\sqrt [4]{b x+2}}{(2-a) b}\)

\(\Big \downarrow \) 840

\(\displaystyle \frac {4 (a+b x)^{3/4}}{(2-a) b \sqrt [4]{b x+2}}-\frac {8 \left (\frac {(a+b x)^{3/4}}{2 \sqrt [4]{b x+2}}-\frac {1}{2} (2-a) \int \frac {1}{\sqrt {b x+2} \sqrt [4]{a+b x}}d\sqrt [4]{b x+2}\right )}{(2-a) b}\)

\(\Big \downarrow \) 842

\(\displaystyle \frac {4 (a+b x)^{3/4}}{(2-a) b \sqrt [4]{b x+2}}-\frac {8 \left (\frac {(a+b x)^{3/4}}{2 \sqrt [4]{b x+2}}-\frac {(2-a) \sqrt [4]{b x+2} \sqrt [4]{1-\frac {2-a}{b x+2}} \int \frac {1}{(b x+2)^{3/4} \sqrt [4]{1-\frac {2-a}{b x+2}}}d\sqrt [4]{b x+2}}{2 \sqrt [4]{a+b x}}\right )}{(2-a) b}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {4 (a+b x)^{3/4}}{(2-a) b \sqrt [4]{b x+2}}-\frac {8 \left (\frac {(2-a) \sqrt [4]{b x+2} \sqrt [4]{1-\frac {2-a}{b x+2}} \int \frac {1}{\sqrt [4]{b x+2} \sqrt [4]{(a-2) (b x+2)+1}}d\frac {1}{\sqrt [4]{b x+2}}}{2 \sqrt [4]{a+b x}}+\frac {(a+b x)^{3/4}}{2 \sqrt [4]{b x+2}}\right )}{(2-a) b}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {4 (a+b x)^{3/4}}{(2-a) b \sqrt [4]{b x+2}}-\frac {8 \left (\frac {(2-a) \sqrt [4]{b x+2} \sqrt [4]{1-\frac {2-a}{b x+2}} \int \frac {1}{\sqrt [4]{1-(2-a) \sqrt {b x+2}}}d\sqrt {b x+2}}{4 \sqrt [4]{a+b x}}+\frac {(a+b x)^{3/4}}{2 \sqrt [4]{b x+2}}\right )}{(2-a) b}\)

\(\Big \downarrow \) 226

\(\displaystyle \frac {4 (a+b x)^{3/4}}{(2-a) b \sqrt [4]{b x+2}}-\frac {8 \left (\frac {\sqrt {2-a} \sqrt [4]{b x+2} \sqrt [4]{1-\frac {2-a}{b x+2}} E\left (\left .\frac {1}{2} \arcsin \left (\sqrt {2-a} \sqrt {b x+2}\right )\right |2\right )}{2 \sqrt [4]{a+b x}}+\frac {(a+b x)^{3/4}}{2 \sqrt [4]{b x+2}}\right )}{(2-a) b}\)

Input:

Int[1/((2 + b*x)^(5/4)*(a + b*x)^(1/4)),x]
 

Output:

(4*(a + b*x)^(3/4))/((2 - a)*b*(2 + b*x)^(1/4)) - (8*((a + b*x)^(3/4)/(2*( 
2 + b*x)^(1/4)) + (Sqrt[2 - a]*(2 + b*x)^(1/4)*(1 - (2 - a)/(2 + b*x))^(1/ 
4)*EllipticE[ArcSin[Sqrt[2 - a]*Sqrt[2 + b*x]]/2, 2])/(2*(a + b*x)^(1/4))) 
)/((2 - a)*b)
 

Defintions of rubi rules used

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 226
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2/(a^(1/4)*Rt[-b/a, 2] 
))*EllipticE[(1/2)*ArcSin[Rt[-b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ 
[a, 0] && NegQ[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 840
Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(1/4), x_Symbol] :> Simp[(a + b*x^4)^(3/4) 
/(2*b*x), x] + Simp[a/(2*b)   Int[1/(x^2*(a + b*x^4)^(1/4)), x], x] /; Free 
Q[{a, b}, x] && NegQ[b/a]
 

rule 842
Int[1/((x_)^2*((a_) + (b_.)*(x_)^4)^(1/4)), x_Symbol] :> Simp[x*((1 + a/(b* 
x^4))^(1/4)/(a + b*x^4)^(1/4))   Int[1/(x^3*(1 + a/(b*x^4))^(1/4)), x], x] 
/; FreeQ[{a, b}, x] && NegQ[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]

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

Input:

int(1/(b*x+2)^(5/4)/(b*x+a)^(1/4),x)
 

Output:

int(1/(b*x+2)^(5/4)/(b*x+a)^(1/4),x)
 

Fricas [F]

\[ \int \frac {1}{(2+b x)^{5/4} \sqrt [4]{a+b x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {1}{4}} {\left (b x + 2\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(1/(b*x+2)^(5/4)/(b*x+a)^(1/4),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {1}{(2+b x)^{5/4} \sqrt [4]{a+b x}} \, dx=\int \frac {1}{\sqrt [4]{a + b x} \left (b x + 2\right )^{\frac {5}{4}}}\, dx \] Input:

integrate(1/(b*x+2)**(5/4)/(b*x+a)**(1/4),x)
 

Output:

Integral(1/((a + b*x)**(1/4)*(b*x + 2)**(5/4)), x)
 

Maxima [F]

\[ \int \frac {1}{(2+b x)^{5/4} \sqrt [4]{a+b x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {1}{4}} {\left (b x + 2\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(1/(b*x+2)^(5/4)/(b*x+a)^(1/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x + a)^(1/4)*(b*x + 2)^(5/4)), x)
 

Giac [F]

\[ \int \frac {1}{(2+b x)^{5/4} \sqrt [4]{a+b x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {1}{4}} {\left (b x + 2\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(1/(b*x+2)^(5/4)/(b*x+a)^(1/4),x, algorithm="giac")
 

Output:

integrate(1/((b*x + a)^(1/4)*(b*x + 2)^(5/4)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(2+b x)^{5/4} \sqrt [4]{a+b x}} \, dx=\int \frac {1}{{\left (b\,x+2\right )}^{5/4}\,{\left (a+b\,x\right )}^{1/4}} \,d x \] Input:

int(1/((b*x + 2)^(5/4)*(a + b*x)^(1/4)),x)
 

Output:

int(1/((b*x + 2)^(5/4)*(a + b*x)^(1/4)), x)
 

Reduce [F]

\[ \int \frac {1}{(2+b x)^{5/4} \sqrt [4]{a+b x}} \, dx=\int \frac {1}{\left (b x +a \right )^{\frac {1}{4}} \left (b x +2\right )^{\frac {1}{4}} b x +2 \left (b x +a \right )^{\frac {1}{4}} \left (b x +2\right )^{\frac {1}{4}}}d x \] Input:

int(1/(b*x+2)^(5/4)/(b*x+a)^(1/4),x)
 

Output:

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