Integrand size = 15, antiderivative size = 102 \[ \int \frac {1}{x^{5/2} (a+b x)^{3/4}} \, dx=-\frac {2 \sqrt [4]{a+b x}}{3 a x^{3/2}}+\frac {5 b \sqrt [4]{a+b x}}{3 a^2 \sqrt {x}}+\frac {5 b^{3/2} \left (1+\frac {b x}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right ),2\right )}{3 a^{3/2} (a+b x)^{3/4}} \] Output:
-2/3*(b*x+a)^(1/4)/a/x^(3/2)+5/3*b*(b*x+a)^(1/4)/a^2/x^(1/2)+5/3*b^(3/2)*( 1+b*x/a)^(3/4)*InverseJacobiAM(1/2*arctan(b^(1/2)*x^(1/2)/a^(1/2)),2^(1/2) )/a^(3/2)/(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 = 47, normalized size of antiderivative = 0.46 \[ \int \frac {1}{x^{5/2} (a+b x)^{3/4}} \, dx=-\frac {2 \left (1+\frac {b x}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},-\frac {1}{2},-\frac {b x}{a}\right )}{3 x^{3/2} (a+b x)^{3/4}} \] Input:
Integrate[1/(x^(5/2)*(a + b*x)^(3/4)),x]
Output:
(-2*(1 + (b*x)/a)^(3/4)*Hypergeometric2F1[-3/2, 3/4, -1/2, -((b*x)/a)])/(3 *x^(3/2)*(a + b*x)^(3/4))
Time = 0.20 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {61, 61, 73, 765, 762}
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.
\(\displaystyle \int \frac {1}{x^{5/2} (a+b x)^{3/4}} \, dx\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {5 b \int \frac {1}{x^{3/2} (a+b x)^{3/4}}dx}{6 a}-\frac {2 \sqrt [4]{a+b x}}{3 a x^{3/2}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {5 b \left (-\frac {b \int \frac {1}{\sqrt {x} (a+b x)^{3/4}}dx}{2 a}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )}{6 a}-\frac {2 \sqrt [4]{a+b x}}{3 a x^{3/2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {5 b \left (-\frac {2 \int \frac {1}{\sqrt {\frac {a+b x}{b}-\frac {a}{b}}}d\sqrt [4]{a+b x}}{a}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )}{6 a}-\frac {2 \sqrt [4]{a+b x}}{3 a x^{3/2}}\) |
\(\Big \downarrow \) 765 |
\(\displaystyle -\frac {5 b \left (-\frac {2 \sqrt {1-\frac {a+b x}{a}} \int \frac {1}{\sqrt {1-\frac {a+b x}{a}}}d\sqrt [4]{a+b x}}{a \sqrt {\frac {a+b x}{b}-\frac {a}{b}}}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )}{6 a}-\frac {2 \sqrt [4]{a+b x}}{3 a x^{3/2}}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle -\frac {5 b \left (-\frac {2 \sqrt {1-\frac {a+b x}{a}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{a+b x}}{\sqrt [4]{a}}\right ),-1\right )}{a^{3/4} \sqrt {\frac {a+b x}{b}-\frac {a}{b}}}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )}{6 a}-\frac {2 \sqrt [4]{a+b x}}{3 a x^{3/2}}\) |
Input:
Int[1/(x^(5/2)*(a + b*x)^(3/4)),x]
Output:
(-2*(a + b*x)^(1/4))/(3*a*x^(3/2)) - (5*b*((-2*(a + b*x)^(1/4))/(a*Sqrt[x] ) - (2*Sqrt[1 - (a + b*x)/a]*EllipticF[ArcSin[(a + b*x)^(1/4)/a^(1/4)], -1 ])/(a^(3/4)*Sqrt[-(a/b) + (a + b*x)/b])))/(6*a)
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]
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[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) )*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[Sqrt[1 + b*(x^4/a)]/Sqrt [a + b*x^4] Int[1/Sqrt[1 + b*(x^4/a)], x], x] /; FreeQ[{a, b}, x] && NegQ [b/a] && !GtQ[a, 0]
\[\int \frac {1}{x^{\frac {5}{2}} \left (b x +a \right )^{\frac {3}{4}}}d x\]
Input:
int(1/x^(5/2)/(b*x+a)^(3/4),x)
Output:
int(1/x^(5/2)/(b*x+a)^(3/4),x)
\[ \int \frac {1}{x^{5/2} (a+b x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} x^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/x^(5/2)/(b*x+a)^(3/4),x, algorithm="fricas")
Output:
integral((b*x + a)^(1/4)*sqrt(x)/(b*x^4 + a*x^3), x)
Result contains complex when optimal does not.
Time = 1.56 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.33 \[ \int \frac {1}{x^{5/2} (a+b x)^{3/4}} \, dx=- \frac {2 {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {3}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{3 a^{\frac {3}{4}} x^{\frac {3}{2}}} \] Input:
integrate(1/x**(5/2)/(b*x+a)**(3/4),x)
Output:
-2*hyper((-3/2, 3/4), (-1/2,), b*x*exp_polar(I*pi)/a)/(3*a**(3/4)*x**(3/2) )
\[ \int \frac {1}{x^{5/2} (a+b x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} x^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/x^(5/2)/(b*x+a)^(3/4),x, algorithm="maxima")
Output:
integrate(1/((b*x + a)^(3/4)*x^(5/2)), x)
\[ \int \frac {1}{x^{5/2} (a+b x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} x^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/x^(5/2)/(b*x+a)^(3/4),x, algorithm="giac")
Output:
integrate(1/((b*x + a)^(3/4)*x^(5/2)), x)
Timed out. \[ \int \frac {1}{x^{5/2} (a+b x)^{3/4}} \, dx=\int \frac {1}{x^{5/2}\,{\left (a+b\,x\right )}^{3/4}} \,d x \] Input:
int(1/(x^(5/2)*(a + b*x)^(3/4)),x)
Output:
int(1/(x^(5/2)*(a + b*x)^(3/4)), x)
\[ \int \frac {1}{x^{5/2} (a+b x)^{3/4}} \, dx=\int \frac {\sqrt {x}\, \left (b x +a \right )^{\frac {3}{4}}}{\sqrt {b x +a}\, a \,x^{3}+\sqrt {b x +a}\, b \,x^{4}}d x \] Input:
int(1/x^(5/2)/(b*x+a)^(3/4),x)
Output:
int((sqrt(x)*(a + b*x)**(3/4))/(sqrt(a + b*x)*a*x**3 + sqrt(a + b*x)*b*x** 4),x)