Integrand size = 19, antiderivative size = 75 \[ \int \frac {1}{(2+b x)^{3/4} (a+b x)^{3/4}} \, dx=-\frac {4 \left (\frac {2+b x}{a+b x}\right )^{3/4} (a+b x)^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {2-a}}{\sqrt {a+b x}}\right ),2\right )}{\sqrt {2-a} b (2+b x)^{3/4}} \] Output:
-4*((b*x+2)/(b*x+a))^(3/4)*(b*x+a)^(3/4)*InverseJacobiAM(1/2*arctan((2-a)^ (1/2)/(b*x+a)^(1/2)),2^(1/2))/(2-a)^(1/2)/b/(b*x+2)^(3/4)
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 \[ \int \frac {1}{(2+b x)^{3/4} (a+b x)^{3/4}} \, dx=\frac {4 \sqrt [4]{2+b x} \left (\frac {a+b x}{-2+a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},\frac {2+b x}{2-a}\right )}{b (a+b x)^{3/4}} \] Input:
Integrate[1/((2 + b*x)^(3/4)*(a + b*x)^(3/4)),x]
Output:
(4*(2 + b*x)^(1/4)*((a + b*x)/(-2 + a))^(3/4)*Hypergeometric2F1[1/4, 3/4, 5/4, (2 + b*x)/(2 - a)])/(b*(a + b*x)^(3/4))
Time = 0.21 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {73, 768, 858, 807, 230}
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.
\(\displaystyle \int \frac {1}{(b x+2)^{3/4} (a+b x)^{3/4}} \, dx\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {4 \int \frac {1}{(a+b x)^{3/4}}d\sqrt [4]{b x+2}}{b}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle \frac {4 (b x+2)^{3/4} \left (1-\frac {2-a}{b x+2}\right )^{3/4} \int \frac {1}{(b x+2)^{3/4} \left (1-\frac {2-a}{b x+2}\right )^{3/4}}d\sqrt [4]{b x+2}}{b (a+b x)^{3/4}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\frac {4 (b x+2)^{3/4} \left (1-\frac {2-a}{b x+2}\right )^{3/4} \int \frac {1}{\sqrt [4]{b x+2} ((a-2) (b x+2)+1)^{3/4}}d\frac {1}{\sqrt [4]{b x+2}}}{b (a+b x)^{3/4}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {2 (b x+2)^{3/4} \left (1-\frac {2-a}{b x+2}\right )^{3/4} \int \frac {1}{\left (1-(2-a) \sqrt {b x+2}\right )^{3/4}}d\sqrt {b x+2}}{b (a+b x)^{3/4}}\) |
\(\Big \downarrow \) 230 |
\(\displaystyle -\frac {4 (b x+2)^{3/4} \left (1-\frac {2-a}{b x+2}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\sqrt {2-a} \sqrt {b x+2}\right ),2\right )}{\sqrt {2-a} b (a+b x)^{3/4}}\) |
Input:
Int[1/((2 + b*x)^(3/4)*(a + b*x)^(3/4)),x]
Output:
(-4*(2 + b*x)^(3/4)*(1 - (2 - a)/(2 + b*x))^(3/4)*EllipticF[ArcSin[Sqrt[2 - a]*Sqrt[2 + b*x]]/2, 2])/(Sqrt[2 - a]*b*(a + b*x)^(3/4))
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]
Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[-b/a, 2] ))*EllipticF[(1/2)*ArcSin[Rt[-b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ [a, 0] && NegQ[b/a]
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]
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]
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]
\[\int \frac {1}{\left (b x +2\right )^{\frac {3}{4}} \left (b x +a \right )^{\frac {3}{4}}}d x\]
Input:
int(1/(b*x+2)^(3/4)/(b*x+a)^(3/4),x)
Output:
int(1/(b*x+2)^(3/4)/(b*x+a)^(3/4),x)
\[ \int \frac {1}{(2+b x)^{3/4} (a+b x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} {\left (b x + 2\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate(1/(b*x+2)^(3/4)/(b*x+a)^(3/4),x, algorithm="fricas")
Output:
integral((b*x + a)^(1/4)*(b*x + 2)^(1/4)/(b^2*x^2 + (a + 2)*b*x + 2*a), x)
\[ \int \frac {1}{(2+b x)^{3/4} (a+b x)^{3/4}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {3}{4}} \left (b x + 2\right )^{\frac {3}{4}}}\, dx \] Input:
integrate(1/(b*x+2)**(3/4)/(b*x+a)**(3/4),x)
Output:
Integral(1/((a + b*x)**(3/4)*(b*x + 2)**(3/4)), x)
\[ \int \frac {1}{(2+b x)^{3/4} (a+b x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} {\left (b x + 2\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate(1/(b*x+2)^(3/4)/(b*x+a)^(3/4),x, algorithm="maxima")
Output:
integrate(1/((b*x + a)^(3/4)*(b*x + 2)^(3/4)), x)
\[ \int \frac {1}{(2+b x)^{3/4} (a+b x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} {\left (b x + 2\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate(1/(b*x+2)^(3/4)/(b*x+a)^(3/4),x, algorithm="giac")
Output:
integrate(1/((b*x + a)^(3/4)*(b*x + 2)^(3/4)), x)
Timed out. \[ \int \frac {1}{(2+b x)^{3/4} (a+b x)^{3/4}} \, dx=\int \frac {1}{{\left (b\,x+2\right )}^{3/4}\,{\left (a+b\,x\right )}^{3/4}} \,d x \] Input:
int(1/((b*x + 2)^(3/4)*(a + b*x)^(3/4)),x)
Output:
int(1/((b*x + 2)^(3/4)*(a + b*x)^(3/4)), x)
\[ \int \frac {1}{(2+b x)^{3/4} (a+b x)^{3/4}} \, dx=\int \frac {1}{\left (b x +a \right )^{\frac {3}{4}} \left (b x +2\right )^{\frac {3}{4}}}d x \] Input:
int(1/(b*x+2)^(3/4)/(b*x+a)^(3/4),x)
Output:
int(1/((a + b*x)**(3/4)*(b*x + 2)**(3/4)),x)