Integrand size = 19, antiderivative size = 126 \[ \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{3/4}} \, dx=-\frac {2 \sqrt [4]{a+b x^2}}{7 a c (c x)^{7/2}}+\frac {4 b \sqrt [4]{a+b x^2}}{7 a^2 c^3 (c x)^{3/2}}-\frac {8 b^{5/2} \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{7 a^{5/2} c^6 \left (a+b x^2\right )^{3/4}} \] Output:
-2/7*(b*x^2+a)^(1/4)/a/c/(c*x)^(7/2)+4/7*b*(b*x^2+a)^(1/4)/a^2/c^3/(c*x)^( 3/2)-8/7*b^(5/2)*(1+a/b/x^2)^(3/4)*(c*x)^(3/2)*InverseJacobiAM(1/2*arccot( b^(1/2)*x/a^(1/2)),2^(1/2))/a^(5/2)/c^6/(b*x^2+a)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.44 \[ \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{3/4}} \, dx=-\frac {2 x \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {3}{4},-\frac {3}{4},-\frac {b x^2}{a}\right )}{7 (c x)^{9/2} \left (a+b x^2\right )^{3/4}} \] Input:
Integrate[1/((c*x)^(9/2)*(a + b*x^2)^(3/4)),x]
Output:
(-2*x*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[-7/4, 3/4, -3/4, -((b*x^2)/a )])/(7*(c*x)^(9/2)*(a + b*x^2)^(3/4))
Time = 0.30 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {264, 264, 266, 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.
\(\displaystyle \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {6 b \int \frac {1}{(c x)^{5/2} \left (b x^2+a\right )^{3/4}}dx}{7 a c^2}-\frac {2 \sqrt [4]{a+b x^2}}{7 a c (c x)^{7/2}}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {6 b \left (-\frac {2 b \int \frac {1}{\sqrt {c x} \left (b x^2+a\right )^{3/4}}dx}{3 a c^2}-\frac {2 \sqrt [4]{a+b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a+b x^2}}{7 a c (c x)^{7/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {6 b \left (-\frac {4 b \int \frac {1}{\left (b x^2+a\right )^{3/4}}d\sqrt {c x}}{3 a c^3}-\frac {2 \sqrt [4]{a+b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a+b x^2}}{7 a c (c x)^{7/2}}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle -\frac {6 b \left (-\frac {4 b (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4} (c x)^{3/2}}d\sqrt {c x}}{3 a c^3 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a+b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a+b x^2}}{7 a c (c x)^{7/2}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\frac {6 b \left (\frac {4 b (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\sqrt {c x} \left (\frac {a x^2 c^4}{b}+1\right )^{3/4}}d\frac {1}{\sqrt {c x}}}{3 a c^3 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a+b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a+b x^2}}{7 a c (c x)^{7/2}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {6 b \left (\frac {2 b (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a x c^3}{b}+1\right )^{3/4}}d(c x)}{3 a c^3 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a+b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a+b x^2}}{7 a c (c x)^{7/2}}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle -\frac {6 b \left (\frac {4 b^{3/2} (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a} c^2 x}{\sqrt {b}}\right ),2\right )}{3 a^{3/2} c^4 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a+b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a+b x^2}}{7 a c (c x)^{7/2}}\) |
Input:
Int[1/((c*x)^(9/2)*(a + b*x^2)^(3/4)),x]
Output:
(-2*(a + b*x^2)^(1/4))/(7*a*c*(c*x)^(7/2)) - (6*b*((-2*(a + b*x^2)^(1/4))/ (3*a*c*(c*x)^(3/2)) + (4*b^(3/2)*(1 + a/(b*x^2))^(3/4)*(c*x)^(3/2)*Ellipti cF[ArcTan[(Sqrt[a]*c^2*x)/Sqrt[b]]/2, 2])/(3*a^(3/2)*c^4*(a + b*x^2)^(3/4) )))/(7*a*c^2)
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
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]
\[\int \frac {1}{\left (c x \right )^{\frac {9}{2}} \left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x\]
Input:
int(1/(c*x)^(9/2)/(b*x^2+a)^(3/4),x)
Output:
int(1/(c*x)^(9/2)/(b*x^2+a)^(3/4),x)
\[ \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (c x\right )^{\frac {9}{2}}} \,d x } \] Input:
integrate(1/(c*x)^(9/2)/(b*x^2+a)^(3/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(1/4)*sqrt(c*x)/(b*c^5*x^7 + a*c^5*x^5), x)
Result contains complex when optimal does not.
Time = 38.54 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.27 \[ \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{3/4}} \, dx=- \frac {{{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{5 b^{\frac {3}{4}} c^{\frac {9}{2}} x^{5}} \] Input:
integrate(1/(c*x)**(9/2)/(b*x**2+a)**(3/4),x)
Output:
-hyper((3/4, 5/2), (7/2,), a*exp_polar(I*pi)/(b*x**2))/(5*b**(3/4)*c**(9/2 )*x**5)
\[ \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (c x\right )^{\frac {9}{2}}} \,d x } \] Input:
integrate(1/(c*x)^(9/2)/(b*x^2+a)^(3/4),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a)^(3/4)*(c*x)^(9/2)), x)
\[ \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (c x\right )^{\frac {9}{2}}} \,d x } \] Input:
integrate(1/(c*x)^(9/2)/(b*x^2+a)^(3/4),x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a)^(3/4)*(c*x)^(9/2)), x)
Timed out. \[ \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{3/4}} \, dx=\int \frac {1}{{\left (c\,x\right )}^{9/2}\,{\left (b\,x^2+a\right )}^{3/4}} \,d x \] Input:
int(1/((c*x)^(9/2)*(a + b*x^2)^(3/4)),x)
Output:
int(1/((c*x)^(9/2)*(a + b*x^2)^(3/4)), x)
Time = 0.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.37 \[ \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{3/4}} \, dx=\frac {2 \sqrt {c}\, \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (4 b \,x^{2}-3 a \right )}{21 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, a^{2} c^{5} x^{3}} \] Input:
int(1/(c*x)^(9/2)/(b*x^2+a)^(3/4),x)
Output:
(2*sqrt(c)*(a + b*x**2)**(3/4)*( - 3*a + 4*b*x**2))/(21*sqrt(x)*sqrt(a + b *x**2)*a**2*c**5*x**3)