\(\int \frac {1}{(2+b x)^{5/4} \sqrt [4]{3+b x}} \, dx\) [647]

Optimal result

Integrand size = 19, antiderivative size = 45 \[ \int \frac {1}{(2+b x)^{5/4} \sqrt [4]{3+b x}} \, dx=-\frac {4}{b \sqrt [4]{2+b x} \sqrt [4]{3+b x}}+\frac {4 E\left (\left .\frac {1}{2} \arcsin \left (\frac {1}{\sqrt {3+b x}}\right )\right |2\right )}{b} \] Output:

-4/b/(b*x+2)^(1/4)/(b*x+3)^(1/4)+4*EllipticE(sin(1/2*arcsin(1/(b*x+3)^(1/2 
))),2^(1/2))/b
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.67 \[ \int \frac {1}{(2+b x)^{5/4} \sqrt [4]{3+b x}} \, dx=-\frac {4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4},\frac {3}{4},-2-b x\right )}{b \sqrt [4]{2+b x}} \] Input:

Integrate[1/((2 + b*x)^(5/4)*(3 + b*x)^(1/4)),x]
 

Output:

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

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(103\) vs. \(2(45)=90\).

Time = 0.22 (sec) , antiderivative size = 103, normalized size of antiderivative = 2.29, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {61, 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.

\(\displaystyle \int \frac {1}{(b x+2)^{5/4} \sqrt [4]{b x+3}} \, dx\)

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 839

\(\displaystyle \frac {8 \left (\frac {(b x+2)^{3/4}}{2 \sqrt [4]{b x+3}}-\frac {1}{2} \int \frac {\sqrt {b x+2}}{(b x+3)^{5/4}}d\sqrt [4]{b x+2}\right )}{b}-\frac {4 (b x+3)^{3/4}}{b \sqrt [4]{b x+2}}\)

\(\Big \downarrow \) 813

\(\displaystyle \frac {8 \left (\frac {(b x+2)^{3/4}}{2 \sqrt [4]{b x+3}}-\frac {\sqrt [4]{b x+2} \sqrt [4]{\frac {1}{b x+2}+1} \int \frac {1}{(b x+2)^{3/4} \left (1+\frac {1}{b x+2}\right )^{5/4}}d\sqrt [4]{b x+2}}{2 \sqrt [4]{b x+3}}\right )}{b}-\frac {4 (b x+3)^{3/4}}{b \sqrt [4]{b x+2}}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {8 \left (\frac {\sqrt [4]{b x+2} \sqrt [4]{\frac {1}{b x+2}+1} \int \frac {1}{\sqrt [4]{b x+2} (b x+3)^{5/4}}d\frac {1}{\sqrt [4]{b x+2}}}{2 \sqrt [4]{b x+3}}+\frac {(b x+2)^{3/4}}{2 \sqrt [4]{b x+3}}\right )}{b}-\frac {4 (b x+3)^{3/4}}{b \sqrt [4]{b x+2}}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {8 \left (\frac {\sqrt [4]{b x+2} \sqrt [4]{\frac {1}{b x+2}+1} \int \frac {1}{\left (\sqrt {b x+2}+1\right )^{5/4}}d\sqrt {b x+2}}{4 \sqrt [4]{b x+3}}+\frac {(b x+2)^{3/4}}{2 \sqrt [4]{b x+3}}\right )}{b}-\frac {4 (b x+3)^{3/4}}{b \sqrt [4]{b x+2}}\)

\(\Big \downarrow \) 212

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

Input:

Int[1/((2 + b*x)^(5/4)*(3 + b*x)^(1/4)),x]
 

Output:

(-4*(3 + b*x)^(3/4))/(b*(2 + b*x)^(1/4)) + (8*((2 + b*x)^(3/4)/(2*(3 + b*x 
)^(1/4)) + ((2 + b*x)^(1/4)*(1 + (2 + b*x)^(-1))^(1/4)*EllipticE[ArcTan[Sq 
rt[2 + b*x]]/2, 2])/(2*(3 + b*x)^(1/4))))/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 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]

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

Input:

int(1/(b*x+2)^(5/4)/(b*x+3)^(1/4),x)
 

Output:

int(1/(b*x+2)^(5/4)/(b*x+3)^(1/4),x)
 

Fricas [F]

\[ \int \frac {1}{(2+b x)^{5/4} \sqrt [4]{3+b x}} \, dx=\int { \frac {1}{{\left (b x + 3\right )}^{\frac {1}{4}} {\left (b x + 2\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(1/(b*x+2)^(5/4)/(b*x+3)^(1/4),x, algorithm="fricas")
 

Output:

integral((b*x + 3)^(3/4)*(b*x + 2)^(3/4)/(b^3*x^3 + 7*b^2*x^2 + 16*b*x + 1 
2), x)
 

Sympy [F]

\[ \int \frac {1}{(2+b x)^{5/4} \sqrt [4]{3+b x}} \, dx=\int \frac {1}{\left (b x + 2\right )^{\frac {5}{4}} \sqrt [4]{b x + 3}}\, dx \] Input:

integrate(1/(b*x+2)**(5/4)/(b*x+3)**(1/4),x)
 

Output:

Integral(1/((b*x + 2)**(5/4)*(b*x + 3)**(1/4)), x)
 

Maxima [F]

\[ \int \frac {1}{(2+b x)^{5/4} \sqrt [4]{3+b x}} \, dx=\int { \frac {1}{{\left (b x + 3\right )}^{\frac {1}{4}} {\left (b x + 2\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(1/(b*x+2)^(5/4)/(b*x+3)^(1/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x + 3)^(1/4)*(b*x + 2)^(5/4)), x)
 

Giac [F]

\[ \int \frac {1}{(2+b x)^{5/4} \sqrt [4]{3+b x}} \, dx=\int { \frac {1}{{\left (b x + 3\right )}^{\frac {1}{4}} {\left (b x + 2\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(1/(b*x+2)^(5/4)/(b*x+3)^(1/4),x, algorithm="giac")
 

Output:

integrate(1/((b*x + 3)^(1/4)*(b*x + 2)^(5/4)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(2+b x)^{5/4} \sqrt [4]{3+b x}} \, dx=\int \frac {1}{{\left (b\,x+2\right )}^{5/4}\,{\left (b\,x+3\right )}^{1/4}} \,d x \] Input:

int(1/((b*x + 2)^(5/4)*(b*x + 3)^(1/4)),x)
 

Output:

int(1/((b*x + 2)^(5/4)*(b*x + 3)^(1/4)), x)
 

Reduce [F]

\[ \int \frac {1}{(2+b x)^{5/4} \sqrt [4]{3+b x}} \, dx=\int \frac {1}{\left (b x +2\right )^{\frac {1}{4}} \left (b x +3\right )^{\frac {1}{4}} b x +2 \left (b x +2\right )^{\frac {1}{4}} \left (b x +3\right )^{\frac {1}{4}}}d x \] Input:

int(1/(b*x+2)^(5/4)/(b*x+3)^(1/4),x)
 

Output:

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