Integrand size = 19, antiderivative size = 149 \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/4}} \, dx=-\frac {2 \sqrt [4]{c+d x}}{3 (b c-a d) (a+b x)^{3/2}}+\frac {5 d \sqrt [4]{c+d x}}{3 (b c-a d)^2 \sqrt {a+b x}}+\frac {5 d \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{3 \sqrt [4]{b} (b c-a d)^{7/4} \sqrt {a+b x}} \]
-2/3*(d*x+c)^(1/4)/(-a*d+b*c)/(b*x+a)^(3/2)+5/3*d*(d*x+c)^(1/4)/(-a*d+b*c) ^2/(b*x+a)^(1/2)+5/3*d*EllipticF(b^(1/4)*(d*x+c)^(1/4)/(-a*d+b*c)^(1/4),I) *(-d*(b*x+a)/(-a*d+b*c))^(1/2)/b^(1/4)/(-a*d+b*c)^(7/4)/(b*x+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.49 \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/4}} \, dx=-\frac {2 \left (\frac {b (c+d x)}{b c-a d}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},-\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )}{3 b (a+b x)^{3/2} (c+d x)^{3/4}} \]
(-2*((b*(c + d*x))/(b*c - a*d))^(3/4)*Hypergeometric2F1[-3/2, 3/4, -1/2, ( d*(a + b*x))/(-(b*c) + a*d)])/(3*b*(a + b*x)^(3/2)*(c + d*x)^(3/4))
Time = 0.25 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, 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.
\(\displaystyle \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/4}} \, dx\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {5 d \int \frac {1}{(a+b x)^{3/2} (c+d x)^{3/4}}dx}{6 (b c-a d)}-\frac {2 \sqrt [4]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {5 d \left (-\frac {d \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}}dx}{2 (b c-a d)}-\frac {2 \sqrt [4]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{6 (b c-a d)}-\frac {2 \sqrt [4]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {5 d \left (-\frac {2 \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [4]{c+d x}}{b c-a d}-\frac {2 \sqrt [4]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{6 (b c-a d)}-\frac {2 \sqrt [4]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\) |
\(\Big \downarrow \) 765 |
\(\displaystyle -\frac {5 d \left (-\frac {2 \sqrt {1-\frac {b (c+d x)}{b c-a d}} \int \frac {1}{\sqrt {1-\frac {b (c+d x)}{b c-a d}}}d\sqrt [4]{c+d x}}{(b c-a d) \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}-\frac {2 \sqrt [4]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{6 (b c-a d)}-\frac {2 \sqrt [4]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle -\frac {5 d \left (-\frac {2 \sqrt {1-\frac {b (c+d x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{\sqrt [4]{b} (b c-a d)^{3/4} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}-\frac {2 \sqrt [4]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{6 (b c-a d)}-\frac {2 \sqrt [4]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\) |
(-2*(c + d*x)^(1/4))/(3*(b*c - a*d)*(a + b*x)^(3/2)) - (5*d*((-2*(c + d*x) ^(1/4))/((b*c - a*d)*Sqrt[a + b*x]) - (2*Sqrt[1 - (b*(c + d*x))/(b*c - a*d )]*EllipticF[ArcSin[(b^(1/4)*(c + d*x)^(1/4))/(b*c - a*d)^(1/4)], -1])/(b^ (1/4)*(b*c - a*d)^(3/4)*Sqrt[a - (b*c)/d + (b*(c + d*x))/d])))/(6*(b*c - a *d))
3.17.58.3.1 Defintions of rubi rules used
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]
\[\int \frac {1}{\left (b x +a \right )^{\frac {5}{2}} \left (d x +c \right )^{\frac {3}{4}}}d x\]
\[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {5}{2}} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]
integral(sqrt(b*x + a)*(d*x + c)^(1/4)/(b^3*d*x^4 + a^3*c + (b^3*c + 3*a*b ^2*d)*x^3 + 3*(a*b^2*c + a^2*b*d)*x^2 + (3*a^2*b*c + a^3*d)*x), x)
\[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/4}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {3}{4}}}\, dx \]
\[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {5}{2}} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]
\[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {5}{2}} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/4}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{3/4}} \,d x \]