Integrand size = 19, antiderivative size = 146 \[ \int \frac {(c x)^{7/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {5 a c^3 \sqrt {c x}}{2 b^2 \sqrt [4]{a+b x^2}}+\frac {c (c x)^{5/2}}{2 b \sqrt [4]{a+b x^2}}-\frac {5 a c^{7/2} \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}-\frac {5 a c^{7/2} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}} \]
1/2*c*(c*x)^(5/2)/b/(b*x^2+a)^(1/4)-5/4*a*c^(7/2)*arctan(b^(1/4)*(c*x)^(1/ 2)/(b*x^2+a)^(1/4)/c^(1/2))/b^(9/4)-5/4*a*c^(7/2)*arctanh(b^(1/4)*(c*x)^(1 /2)/(b*x^2+a)^(1/4)/c^(1/2))/b^(9/4)+5/2*a*c^3*(c*x)^(1/2)/b^2/(b*x^2+a)^( 1/4)
Time = 0.46 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.90 \[ \int \frac {(c x)^{7/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {c^3 \sqrt {c x} \left (2 \sqrt [4]{b} \sqrt {x} \left (5 a+b x^2\right )-5 a \sqrt [4]{a+b x^2} \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )-5 a \sqrt [4]{a+b x^2} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{4 b^{9/4} \sqrt {x} \sqrt [4]{a+b x^2}} \]
(c^3*Sqrt[c*x]*(2*b^(1/4)*Sqrt[x]*(5*a + b*x^2) - 5*a*(a + b*x^2)^(1/4)*Ar cTan[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)] - 5*a*(a + b*x^2)^(1/4)*ArcTanh[ (b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)]))/(4*b^(9/4)*Sqrt[x]*(a + b*x^2)^(1/4 ))
Time = 0.26 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {250, 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.
| \(\displaystyle \int \frac {(c x)^{7/2}}{\left (a+b x^2\right )^{5/4}} \, dx\) |
| \(\Big \downarrow \) 250 |
| \(\displaystyle \frac {c (c x)^{5/2}}{2 b \sqrt [4]{a+b x^2}}-\frac {5 a c^2 \int \frac {(c x)^{3/2}}{\left (b x^2+a\right )^{5/4}}dx}{4 b}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {c (c x)^{5/2}}{2 b \sqrt [4]{a+b x^2}}-\frac {5 a c^2 \left (\frac {c^2 \int \frac {1}{\sqrt {c x} \sqrt [4]{b x^2+a}}dx}{b}-\frac {2 c \sqrt {c x}}{b \sqrt [4]{a+b x^2}}\right )}{4 b}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {c (c x)^{5/2}}{2 b \sqrt [4]{a+b x^2}}-\frac {5 a c^2 \left (\frac {2 c \int \frac {1}{\sqrt [4]{b x^2+a}}d\sqrt {c x}}{b}-\frac {2 c \sqrt {c x}}{b \sqrt [4]{a+b x^2}}\right )}{4 b}\) |
| \(\Big \downarrow \) 770 |
| \(\displaystyle \frac {c (c x)^{5/2}}{2 b \sqrt [4]{a+b x^2}}-\frac {5 a c^2 \left (\frac {2 c \int \frac {1}{1-b x^2}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}}{b}-\frac {2 c \sqrt {c x}}{b \sqrt [4]{a+b x^2}}\right )}{4 b}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {c (c x)^{5/2}}{2 b \sqrt [4]{a+b x^2}}-\frac {5 a c^2 \left (\frac {2 c \left (\frac {1}{2} c \int \frac {1}{c-\sqrt {b} c x}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}+\frac {1}{2} c \int \frac {1}{\sqrt {b} x c+c}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}\right )}{b}-\frac {2 c \sqrt {c x}}{b \sqrt [4]{a+b x^2}}\right )}{4 b}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {c (c x)^{5/2}}{2 b \sqrt [4]{a+b x^2}}-\frac {5 a c^2 \left (\frac {2 c \left (\frac {1}{2} c \int \frac {1}{c-\sqrt {b} c x}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}+\frac {\sqrt {c} \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{2 \sqrt [4]{b}}\right )}{b}-\frac {2 c \sqrt {c x}}{b \sqrt [4]{a+b x^2}}\right )}{4 b}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {c (c x)^{5/2}}{2 b \sqrt [4]{a+b x^2}}-\frac {5 a c^2 \left (\frac {2 c \left (\frac {\sqrt {c} \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{2 \sqrt [4]{b}}+\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{2 \sqrt [4]{b}}\right )}{b}-\frac {2 c \sqrt {c x}}{b \sqrt [4]{a+b x^2}}\right )}{4 b}\) |
(c*(c*x)^(5/2))/(2*b*(a + b*x^2)^(1/4)) - (5*a*c^2*((-2*c*Sqrt[c*x])/(b*(a + b*x^2)^(1/4)) + (2*c*((Sqrt[c]*ArcTan[(b^(1/4)*Sqrt[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))/(4*b)
3.10.85.3.1 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)^(5/4), x_Symbol] :> Simp[2*c*(( c*x)^(m - 1)/(b*(2*m - 3)*(a + b*x^2)^(1/4))), x] - Simp[2*a*c^2*((m - 1)/( b*(2*m - 3))) Int[(c*x)^(m - 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a] && IntegerQ[2*m] && GtQ[m, 3/2]
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]
\[\int \frac {\left (c x \right )^{\frac {7}{2}}}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.74 \[ \int \frac {(c x)^{7/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {4 \, {\left (b c^{3} x^{2} + 5 \, a c^{3}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} - 5 \, \left (\frac {a^{4} c^{14}}{b^{9}}\right )^{\frac {1}{4}} {\left (b^{3} x^{2} + a b^{2}\right )} \log \left (\frac {5 \, {\left ({\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} a c^{3} + \left (\frac {a^{4} c^{14}}{b^{9}}\right )^{\frac {1}{4}} {\left (b^{3} x^{2} + a b^{2}\right )}\right )}}{b x^{2} + a}\right ) + 5 \, \left (\frac {a^{4} c^{14}}{b^{9}}\right )^{\frac {1}{4}} {\left (b^{3} x^{2} + a b^{2}\right )} \log \left (\frac {5 \, {\left ({\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} a c^{3} - \left (\frac {a^{4} c^{14}}{b^{9}}\right )^{\frac {1}{4}} {\left (b^{3} x^{2} + a b^{2}\right )}\right )}}{b x^{2} + a}\right ) - 5 \, \left (\frac {a^{4} c^{14}}{b^{9}}\right )^{\frac {1}{4}} {\left (-i \, b^{3} x^{2} - i \, a b^{2}\right )} \log \left (\frac {5 \, {\left ({\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} a c^{3} - \left (\frac {a^{4} c^{14}}{b^{9}}\right )^{\frac {1}{4}} {\left (i \, b^{3} x^{2} + i \, a b^{2}\right )}\right )}}{b x^{2} + a}\right ) - 5 \, \left (\frac {a^{4} c^{14}}{b^{9}}\right )^{\frac {1}{4}} {\left (i \, b^{3} x^{2} + i \, a b^{2}\right )} \log \left (\frac {5 \, {\left ({\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} a c^{3} - \left (\frac {a^{4} c^{14}}{b^{9}}\right )^{\frac {1}{4}} {\left (-i \, b^{3} x^{2} - i \, a b^{2}\right )}\right )}}{b x^{2} + a}\right )}{8 \, {\left (b^{3} x^{2} + a b^{2}\right )}} \]
1/8*(4*(b*c^3*x^2 + 5*a*c^3)*(b*x^2 + a)^(3/4)*sqrt(c*x) - 5*(a^4*c^14/b^9 )^(1/4)*(b^3*x^2 + a*b^2)*log(5*((b*x^2 + a)^(3/4)*sqrt(c*x)*a*c^3 + (a^4* c^14/b^9)^(1/4)*(b^3*x^2 + a*b^2))/(b*x^2 + a)) + 5*(a^4*c^14/b^9)^(1/4)*( b^3*x^2 + a*b^2)*log(5*((b*x^2 + a)^(3/4)*sqrt(c*x)*a*c^3 - (a^4*c^14/b^9) ^(1/4)*(b^3*x^2 + a*b^2))/(b*x^2 + a)) - 5*(a^4*c^14/b^9)^(1/4)*(-I*b^3*x^ 2 - I*a*b^2)*log(5*((b*x^2 + a)^(3/4)*sqrt(c*x)*a*c^3 - (a^4*c^14/b^9)^(1/ 4)*(I*b^3*x^2 + I*a*b^2))/(b*x^2 + a)) - 5*(a^4*c^14/b^9)^(1/4)*(I*b^3*x^2 + I*a*b^2)*log(5*((b*x^2 + a)^(3/4)*sqrt(c*x)*a*c^3 - (a^4*c^14/b^9)^(1/4 )*(-I*b^3*x^2 - I*a*b^2))/(b*x^2 + a)))/(b^3*x^2 + a*b^2)
Result contains complex when optimal does not.
Time = 13.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.30 \[ \int \frac {(c x)^{7/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {c^{\frac {7}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {13}{4}\right )} \]
c**(7/2)*x**(9/2)*gamma(9/4)*hyper((5/4, 9/4), (13/4,), b*x**2*exp_polar(I *pi)/a)/(2*a**(5/4)*gamma(13/4))
\[ \int \frac {(c x)^{7/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {\left (c x\right )^{\frac {7}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \]
\[ \int \frac {(c x)^{7/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {\left (c x\right )^{\frac {7}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \]
Timed out. \[ \int \frac {(c x)^{7/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\int \frac {{\left (c\,x\right )}^{7/2}}{{\left (b\,x^2+a\right )}^{5/4}} \,d x \]