Integrand size = 26, antiderivative size = 125 \[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\frac {2 (b c-a d) (e x)^{3/2}}{3 a b e \left (a+b x^2\right )^{3/4}}-\frac {d \sqrt {e} \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{7/4}}+\frac {d \sqrt {e} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{7/4}} \]
2/3*(-a*d+b*c)*(e*x)^(3/2)/a/b/e/(b*x^2+a)^(3/4)-d*arctan(b^(1/4)*(e*x)^(1 /2)/(b*x^2+a)^(1/4)/e^(1/2))*e^(1/2)/b^(7/4)+d*arctanh(b^(1/4)*(e*x)^(1/2) /(b*x^2+a)^(1/4)/e^(1/2))*e^(1/2)/b^(7/4)
Time = 0.75 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\frac {\sqrt {e x} \left (\frac {2 b^{3/4} (b c-a d) x^{3/2}}{a \left (a+b x^2\right )^{3/4}}-3 d \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )+3 d \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{3 b^{7/4} \sqrt {x}} \]
(Sqrt[e*x]*((2*b^(3/4)*(b*c - a*d)*x^(3/2))/(a*(a + b*x^2)^(3/4)) - 3*d*Ar cTan[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)] + 3*d*ArcTanh[(b^(1/4)*Sqrt[x])/ (a + b*x^2)^(1/4)]))/(3*b^(7/4)*Sqrt[x])
Time = 0.27 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {357, 266, 854, 27, 827, 218, 221}
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.
\(\displaystyle \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx\) |
\(\Big \downarrow \) 357 |
\(\displaystyle \frac {d \int \frac {\sqrt {e x}}{\left (b x^2+a\right )^{3/4}}dx}{b}+\frac {2 (e x)^{3/2} (b c-a d)}{3 a b e \left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 d \int \frac {e x}{\left (b x^2+a\right )^{3/4}}d\sqrt {e x}}{b e}+\frac {2 (e x)^{3/2} (b c-a d)}{3 a b e \left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {2 d \int \frac {e^3 x}{e^2-b e^2 x^2}d\frac {\sqrt {e x}}{\sqrt [4]{b x^2+a}}}{b e}+\frac {2 (e x)^{3/2} (b c-a d)}{3 a b e \left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 d e \int \frac {e x}{e^2-b e^2 x^2}d\frac {\sqrt {e x}}{\sqrt [4]{b x^2+a}}}{b}+\frac {2 (e x)^{3/2} (b c-a d)}{3 a b e \left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {2 d e \left (\frac {\int \frac {1}{e-\sqrt {b} e x}d\frac {\sqrt {e x}}{\sqrt [4]{b x^2+a}}}{2 \sqrt {b}}-\frac {\int \frac {1}{\sqrt {b} x e+e}d\frac {\sqrt {e x}}{\sqrt [4]{b x^2+a}}}{2 \sqrt {b}}\right )}{b}+\frac {2 (e x)^{3/2} (b c-a d)}{3 a b e \left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2 d e \left (\frac {\int \frac {1}{e-\sqrt {b} e x}d\frac {\sqrt {e x}}{\sqrt [4]{b x^2+a}}}{2 \sqrt {b}}-\frac {\arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{2 b^{3/4} \sqrt {e}}\right )}{b}+\frac {2 (e x)^{3/2} (b c-a d)}{3 a b e \left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 d e \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{2 b^{3/4} \sqrt {e}}-\frac {\arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{2 b^{3/4} \sqrt {e}}\right )}{b}+\frac {2 (e x)^{3/2} (b c-a d)}{3 a b e \left (a+b x^2\right )^{3/4}}\) |
(2*(b*c - a*d)*(e*x)^(3/2))/(3*a*b*e*(a + b*x^2)^(3/4)) + (2*d*e*(-1/2*Arc Tan[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))]/(b^(3/4)*Sqrt[e]) + A rcTanh[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))]/(2*b^(3/4)*Sqrt[e] )))/b
3.12.17.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_ Symbol] :> Simp[(b*c - a*d)*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*b*e*(m + 1))), x] + Simp[d/b Int[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && EqQ[m + 2*p + 3, 0] && LtQ[p, -1]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 1)/n) Subst[Int[x^m/(1 - b*x^n)^(p + (m + 1)/n + 1), x], x, x/(a + b*x^n )^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, - 2^(-1)] && IntegersQ[m, p + (m + 1)/n]
\[\int \frac {\sqrt {e x}\, \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {7}{4}}}d x\]
Timed out. \[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\text {Timed out} \]
Result contains complex when optimal does not.
Time = 12.73 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\frac {c \sqrt {e} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right )}{2 a^{\frac {7}{4}} \left (1 + \frac {b x^{2}}{a}\right )^{\frac {3}{4}} \Gamma \left (\frac {7}{4}\right )} + \frac {d \sqrt {e} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {7}{4}} \Gamma \left (\frac {11}{4}\right )} \]
c*sqrt(e)*x**(3/2)*gamma(3/4)/(2*a**(7/4)*(1 + b*x**2/a)**(3/4)*gamma(7/4) ) + d*sqrt(e)*x**(7/2)*gamma(7/4)*hyper((7/4, 7/4), (11/4,), b*x**2*exp_po lar(I*pi)/a)/(2*a**(7/4)*gamma(11/4))
\[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \sqrt {e x}}{{\left (b x^{2} + a\right )}^{\frac {7}{4}}} \,d x } \]
\[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \sqrt {e x}}{{\left (b x^{2} + a\right )}^{\frac {7}{4}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\int \frac {\sqrt {e\,x}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{7/4}} \,d x \]