Integrand size = 19, antiderivative size = 107 \[ \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{5/4}} \, dx=-\frac {2 c \sqrt {c x}}{b \sqrt [4]{a+b x^2}}+\frac {c^{3/2} \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{b^{5/4}}+\frac {c^{3/2} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{b^{5/4}} \]
c^(3/2)*arctan(b^(1/4)*(c*x)^(1/2)/(b*x^2+a)^(1/4)/c^(1/2))/b^(5/4)+c^(3/2 )*arctanh(b^(1/4)*(c*x)^(1/2)/(b*x^2+a)^(1/4)/c^(1/2))/b^(5/4)-2*c*(c*x)^( 1/2)/b/(b*x^2+a)^(1/4)
Time = 0.35 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.85 \[ \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {c \sqrt {c x} \left (-\frac {2 \sqrt [4]{b}}{\sqrt [4]{a+b x^2}}+\frac {\arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )}{\sqrt {x}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )}{\sqrt {x}}\right )}{b^{5/4}} \]
(c*Sqrt[c*x]*((-2*b^(1/4))/(a + b*x^2)^(1/4) + ArcTan[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)]/Sqrt[x] + ArcTanh[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)]/Sq rt[x]))/b^(5/4)
Time = 0.23 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {252, 266, 770, 756, 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.
\(\displaystyle \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {c^2 \int \frac {1}{\sqrt {c x} \sqrt [4]{b x^2+a}}dx}{b}-\frac {2 c \sqrt {c x}}{b \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 c \int \frac {1}{\sqrt [4]{b x^2+a}}d\sqrt {c x}}{b}-\frac {2 c \sqrt {c x}}{b \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 770 |
\(\displaystyle \frac {2 c \int \frac {1}{1-b x^2}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}}{b}-\frac {2 c \sqrt {c x}}{b \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {2 c \left (\frac {1}{2} c \int \frac {1}{c-\sqrt {b} c x}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}+\frac {1}{2} c \int \frac {1}{\sqrt {b} x c+c}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}\right )}{b}-\frac {2 c \sqrt {c x}}{b \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2 c \left (\frac {1}{2} c \int \frac {1}{c-\sqrt {b} c x}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}+\frac {\sqrt {c} \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{2 \sqrt [4]{b}}\right )}{b}-\frac {2 c \sqrt {c x}}{b \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 c \left (\frac {\sqrt {c} \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{2 \sqrt [4]{b}}+\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{2 \sqrt [4]{b}}\right )}{b}-\frac {2 c \sqrt {c x}}{b \sqrt [4]{a+b x^2}}\) |
(-2*c*Sqrt[c*x])/(b*(a + b*x^2)^(1/4)) + (2*c*((Sqrt[c]*ArcTan[(b^(1/4)*Sq rt[c*x])/(Sqrt[c]*(a + b*x^2)^(1/4))])/(2*b^(1/4)) + (Sqrt[c]*ArcTanh[(b^( 1/4)*Sqrt[c*x])/(Sqrt[c]*(a + b*x^2)^(1/4))])/(2*b^(1/4))))/b
3.10.86.3.1 Defintions of rubi rules used
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] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[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)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n) Subst[In t[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p + 1 /n]
\[\int \frac {\left (c x \right )^{\frac {3}{2}}}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 327, normalized size of antiderivative = 3.06 \[ \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{5/4}} \, dx=-\frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} c - {\left (b^{2} x^{2} + a b\right )} \left (\frac {c^{6}}{b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} c + {\left (b^{2} x^{2} + a b\right )} \left (\frac {c^{6}}{b^{5}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right ) + {\left (b^{2} x^{2} + a b\right )} \left (\frac {c^{6}}{b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} c - {\left (b^{2} x^{2} + a b\right )} \left (\frac {c^{6}}{b^{5}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right ) - {\left (-i \, b^{2} x^{2} - i \, a b\right )} \left (\frac {c^{6}}{b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} c - {\left (i \, b^{2} x^{2} + i \, a b\right )} \left (\frac {c^{6}}{b^{5}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right ) - {\left (i \, b^{2} x^{2} + i \, a b\right )} \left (\frac {c^{6}}{b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} c - {\left (-i \, b^{2} x^{2} - i \, a b\right )} \left (\frac {c^{6}}{b^{5}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right )}{2 \, {\left (b^{2} x^{2} + a b\right )}} \]
-1/2*(4*(b*x^2 + a)^(3/4)*sqrt(c*x)*c - (b^2*x^2 + a*b)*(c^6/b^5)^(1/4)*lo g(((b*x^2 + a)^(3/4)*sqrt(c*x)*c + (b^2*x^2 + a*b)*(c^6/b^5)^(1/4))/(b*x^2 + a)) + (b^2*x^2 + a*b)*(c^6/b^5)^(1/4)*log(((b*x^2 + a)^(3/4)*sqrt(c*x)* c - (b^2*x^2 + a*b)*(c^6/b^5)^(1/4))/(b*x^2 + a)) - (-I*b^2*x^2 - I*a*b)*( c^6/b^5)^(1/4)*log(((b*x^2 + a)^(3/4)*sqrt(c*x)*c - (I*b^2*x^2 + I*a*b)*(c ^6/b^5)^(1/4))/(b*x^2 + a)) - (I*b^2*x^2 + I*a*b)*(c^6/b^5)^(1/4)*log(((b* x^2 + a)^(3/4)*sqrt(c*x)*c - (-I*b^2*x^2 - I*a*b)*(c^6/b^5)^(1/4))/(b*x^2 + a)))/(b^2*x^2 + a*b)
Result contains complex when optimal does not.
Time = 2.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.41 \[ \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {c^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {9}{4}\right )} \]
c**(3/2)*x**(5/2)*gamma(5/4)*hyper((5/4, 5/4), (9/4,), b*x**2*exp_polar(I* pi)/a)/(2*a**(5/4)*gamma(9/4))
\[ \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {\left (c x\right )^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \]
\[ \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {\left (c x\right )^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \]
Timed out. \[ \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\int \frac {{\left (c\,x\right )}^{3/2}}{{\left (b\,x^2+a\right )}^{5/4}} \,d x \]