Integrand size = 19, antiderivative size = 95 \[ \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx=-\frac {4 \sqrt {d} (a+b x)^{3/4} \left (\frac {b (c+d x)}{d (a+b x)}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d}}{\sqrt {d} \sqrt {a+b x}}\right ),2\right )}{b \sqrt {b c-a d} (c+d x)^{3/4}} \] Output:
-4*d^(1/2)*(b*x+a)^(3/4)*(b*(d*x+c)/d/(b*x+a))^(3/4)*InverseJacobiAM(1/2*a rctan(1/d^(1/2)/(b*x+a)^(1/2)*(-a*d+b*c)^(1/2)),2^(1/2))/b/(-a*d+b*c)^(1/2 )/(d*x+c)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.00 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx=\frac {4 \sqrt [4]{a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},\frac {d (a+b x)}{-b c+a d}\right )}{b (c+d x)^{3/4}} \] Input:
Integrate[1/((a + b*x)^(3/4)*(c + d*x)^(3/4)),x]
Output:
(4*(a + b*x)^(1/4)*((b*(c + d*x))/(b*c - a*d))^(3/4)*Hypergeometric2F1[1/4 , 3/4, 5/4, (d*(a + b*x))/(-(b*c) + a*d)])/(b*(c + d*x)^(3/4))
Time = 0.25 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.19, 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, 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.
\(\displaystyle \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {4 \int \frac {1}{\left (c-\frac {a d}{b}+\frac {d (a+b x)}{b}\right )^{3/4}}d\sqrt [4]{a+b x}}{b}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle \frac {4 (a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{(a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4}}d\sqrt [4]{a+b x}}{b \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\frac {4 (a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{\sqrt [4]{a+b x} \left (\frac {(b c-a d) (a+b x)}{d}+1\right )^{3/4}}d\frac {1}{\sqrt [4]{a+b x}}}{b \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {2 (a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{\left (\frac {\sqrt {a+b x} (b c-a d)}{d}+1\right )^{3/4}}d\sqrt {a+b x}}{b \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle -\frac {4 \sqrt {d} (a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d} \sqrt {a+b x}}{\sqrt {d}}\right ),2\right )}{b \sqrt {b c-a d} \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}\) |
Input:
Int[1/((a + b*x)^(3/4)*(c + d*x)^(3/4)),x]
Output:
(-4*Sqrt[d]*(a + b*x)^(3/4)*(1 + (b*c - a*d)/(d*(a + b*x)))^(3/4)*Elliptic F[ArcTan[(Sqrt[b*c - a*d]*Sqrt[a + b*x])/Sqrt[d]]/2, 2])/(b*Sqrt[b*c - a*d ]*(c - (a*d)/b + (d*(a + b*x))/b)^(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)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[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]
\[\int \frac {1}{\left (b x +a \right )^{\frac {3}{4}} \left (x d +c \right )^{\frac {3}{4}}}d x\]
Input:
int(1/(b*x+a)^(3/4)/(d*x+c)^(3/4),x)
Output:
int(1/(b*x+a)^(3/4)/(d*x+c)^(3/4),x)
\[ \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate(1/(b*x+a)^(3/4)/(d*x+c)^(3/4),x, algorithm="fricas")
Output:
integral((b*x + a)^(1/4)*(d*x + c)^(1/4)/(b*d*x^2 + a*c + (b*c + a*d)*x), x)
\[ \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {3}{4}} \left (c + d x\right )^{\frac {3}{4}}}\, dx \] Input:
integrate(1/(b*x+a)**(3/4)/(d*x+c)**(3/4),x)
Output:
Integral(1/((a + b*x)**(3/4)*(c + d*x)**(3/4)), x)
\[ \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate(1/(b*x+a)^(3/4)/(d*x+c)^(3/4),x, algorithm="maxima")
Output:
integrate(1/((b*x + a)^(3/4)*(d*x + c)^(3/4)), x)
\[ \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate(1/(b*x+a)^(3/4)/(d*x+c)^(3/4),x, algorithm="giac")
Output:
integrate(1/((b*x + a)^(3/4)*(d*x + c)^(3/4)), x)
Timed out. \[ \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{3/4}\,{\left (c+d\,x\right )}^{3/4}} \,d x \] Input:
int(1/((a + b*x)^(3/4)*(c + d*x)^(3/4)),x)
Output:
int(1/((a + b*x)^(3/4)*(c + d*x)^(3/4)), x)
\[ \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx=\int \frac {1}{\left (d x +c \right )^{\frac {3}{4}} \left (b x +a \right )^{\frac {3}{4}}}d x \] Input:
int(1/(b*x+a)^(3/4)/(d*x+c)^(3/4),x)
Output:
int(1/((c + d*x)**(3/4)*(a + b*x)**(3/4)),x)