Integrand size = 18, antiderivative size = 67 \[ \int \frac {1}{(-c x)^{3/4} (a+b x)^{3/4}} \, dx=\frac {2 \sqrt {2} a \left (-\frac {b x}{a}-\frac {b^2 x^2}{a^2}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (1+\frac {2 b x}{a}\right ),2\right )}{b (-c x)^{3/4} (a+b x)^{3/4}} \] Output:
2*2^(1/2)*a*(-b*x/a-b^2*x^2/a^2)^(3/4)*InverseJacobiAM(1/2*arcsin(1+2*b*x/ a),2^(1/2))/b/(-c*x)^(3/4)/(b*x+a)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(-c x)^{3/4} (a+b x)^{3/4}} \, dx=\frac {4 x \left (1+\frac {b x}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\frac {b x}{a}\right )}{(-c x)^{3/4} (a+b x)^{3/4}} \] Input:
Integrate[1/((-(c*x))^(3/4)*(a + b*x)^(3/4)),x]
Output:
(4*x*(1 + (b*x)/a)^(3/4)*Hypergeometric2F1[1/4, 3/4, 5/4, -((b*x)/a)])/((- (c*x))^(3/4)*(a + b*x)^(3/4))
Time = 0.22 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.18, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, 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.
\(\displaystyle \int \frac {1}{(-c x)^{3/4} (a+b x)^{3/4}} \, dx\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {4 \int \frac {1}{(a+b x)^{3/4}}d\sqrt [4]{-c x}}{c}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle -\frac {4 (-c x)^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x}+1\right )^{3/4} (-c x)^{3/4}}d\sqrt [4]{-c x}}{c (a+b x)^{3/4}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {4 (-c x)^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \int \frac {1}{\sqrt [4]{-c x} \left (\frac {a x c^2}{b}+1\right )^{3/4}}d\frac {1}{\sqrt [4]{-c x}}}{c (a+b x)^{3/4}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {2 (-c x)^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \int \frac {1}{\left (1-\frac {a c \sqrt {-c x}}{b}\right )^{3/4}}d\sqrt {-c x}}{c (a+b x)^{3/4}}\) |
\(\Big \downarrow \) 230 |
\(\displaystyle \frac {4 \sqrt {b} (-c x)^{3/4} \left (\frac {a}{b x}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {a} \sqrt {c} \sqrt {-c x}}{\sqrt {b}}\right ),2\right )}{\sqrt {a} c^{3/2} (a+b x)^{3/4}}\) |
Input:
Int[1/((-(c*x))^(3/4)*(a + b*x)^(3/4)),x]
Output:
(4*Sqrt[b]*(1 + a/(b*x))^(3/4)*(-(c*x))^(3/4)*EllipticF[ArcSin[(Sqrt[a]*Sq rt[c]*Sqrt[-(c*x)])/Sqrt[b]]/2, 2])/(Sqrt[a]*c^(3/2)*(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]
\[\int \frac {1}{\left (-c x \right )^{\frac {3}{4}} \left (b x +a \right )^{\frac {3}{4}}}d x\]
Input:
int(1/(-c*x)^(3/4)/(b*x+a)^(3/4),x)
Output:
int(1/(-c*x)^(3/4)/(b*x+a)^(3/4),x)
\[ \int \frac {1}{(-c x)^{3/4} (a+b x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} \left (-c x\right )^{\frac {3}{4}}} \,d x } \] Input:
integrate(1/(-c*x)^(3/4)/(b*x+a)^(3/4),x, algorithm="fricas")
Output:
integral(-(b*x + a)^(1/4)*(-c*x)^(1/4)/(b*c*x^2 + a*c*x), x)
Result contains complex when optimal does not.
Time = 1.33 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(-c x)^{3/4} (a+b x)^{3/4}} \, dx=\frac {\sqrt [4]{x} e^{- \frac {3 i \pi }{4}} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{a^{\frac {3}{4}} c^{\frac {3}{4}} \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate(1/(-c*x)**(3/4)/(b*x+a)**(3/4),x)
Output:
x**(1/4)*exp(-3*I*pi/4)*gamma(1/4)*hyper((1/4, 3/4), (5/4,), b*x*exp_polar (I*pi)/a)/(a**(3/4)*c**(3/4)*gamma(5/4))
\[ \int \frac {1}{(-c x)^{3/4} (a+b x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} \left (-c x\right )^{\frac {3}{4}}} \,d x } \] Input:
integrate(1/(-c*x)^(3/4)/(b*x+a)^(3/4),x, algorithm="maxima")
Output:
integrate(1/((b*x + a)^(3/4)*(-c*x)^(3/4)), x)
\[ \int \frac {1}{(-c x)^{3/4} (a+b x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} \left (-c x\right )^{\frac {3}{4}}} \,d x } \] Input:
integrate(1/(-c*x)^(3/4)/(b*x+a)^(3/4),x, algorithm="giac")
Output:
integrate(1/((b*x + a)^(3/4)*(-c*x)^(3/4)), x)
Timed out. \[ \int \frac {1}{(-c x)^{3/4} (a+b x)^{3/4}} \, dx=\int \frac {1}{{\left (-c\,x\right )}^{3/4}\,{\left (a+b\,x\right )}^{3/4}} \,d x \] Input:
int(1/((-c*x)^(3/4)*(a + b*x)^(3/4)),x)
Output:
int(1/((-c*x)^(3/4)*(a + b*x)^(3/4)), x)
\[ \int \frac {1}{(-c x)^{3/4} (a+b x)^{3/4}} \, dx=-\frac {\left (\int \frac {1}{x^{\frac {3}{4}} \left (b x +a \right )^{\frac {3}{4}}}d x \right ) \left (-1\right )^{\frac {1}{4}}}{c^{\frac {3}{4}}} \] Input:
int(1/(-c*x)^(3/4)/(b*x+a)^(3/4),x)
Output:
int(1/(x**(3/4)*(a + b*x)**(3/4)),x)/(c**(3/4)*( - 1)**(3/4))