Integrand size = 19, antiderivative size = 332 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/4}} \, dx=\frac {5 (b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{3 b^2}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}+\frac {5 (b c-a d)^{5/2} ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right ),\frac {1}{2}\right )}{6 \sqrt {2} b^{9/4} \sqrt [4]{d} (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}} \]
5/3*(-a*d+b*c)*(b*x+a)^(1/4)*(d*x+c)^(1/4)/b^2+2/3*(b*x+a)^(1/4)*(d*x+c)^( 5/4)/b+5/12*(-a*d+b*c)^(5/2)*((b*x+a)*(d*x+c))^(3/4)*(cos(2*arctan(b^(1/4) *d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2)))^2)^(1/2)/cos(2 *arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2))) *EllipticF(sin(2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(- a*d+b*c)^(1/2))),1/2*2^(1/2))*(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1/2) /(-a*d+b*c))*((2*b*d*x+a*d+b*c)^2)^(1/2)*((a*d+b*(2*d*x+c))^2/(-a*d+b*c)^2 /(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1/2)/(-a*d+b*c))^2)^(1/2)/b^(9/4) /d^(1/4)/(b*x+a)^(3/4)/(d*x+c)^(3/4)/(2*b*d*x+a*d+b*c)*2^(1/2)/((a*d+b*(2* d*x+c))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.21 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/4}} \, dx=\frac {4 \sqrt [4]{a+b x} (c+d x)^{5/4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{4},\frac {5}{4},\frac {d (a+b x)}{-b c+a d}\right )}{b \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \]
(4*(a + b*x)^(1/4)*(c + d*x)^(5/4)*Hypergeometric2F1[-5/4, 1/4, 5/4, (d*(a + b*x))/(-(b*c) + a*d)])/(b*((b*(c + d*x))/(b*c - a*d))^(5/4))
Time = 0.31 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.54, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {60, 60, 73, 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.
\(\displaystyle \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/4}} \, dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5 (b c-a d) \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{3/4}}dx}{6 b}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5 (b c-a d) \left (\frac {(b c-a d) \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}}dx}{2 b}+\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{b}\right )}{6 b}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {5 (b c-a d) \left (\frac {2 (b c-a d) \int \frac {1}{\left (c-\frac {a d}{b}+\frac {d (a+b x)}{b}\right )^{3/4}}d\sqrt [4]{a+b x}}{b^2}+\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{b}\right )}{6 b}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle \frac {5 (b c-a d) \left (\frac {2 (a+b x)^{3/4} (b c-a d) \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{(a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4}}d\sqrt [4]{a+b x}}{b^2 \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}+\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{b}\right )}{6 b}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {5 (b c-a d) \left (\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{b}-\frac {2 (a+b x)^{3/4} (b c-a d) \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{\sqrt [4]{a+b x} \left (\frac {(b c-a d) (a+b x)}{d}+1\right )^{3/4}}d\frac {1}{\sqrt [4]{a+b x}}}{b^2 \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}\right )}{6 b}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {5 (b c-a d) \left (\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{b}-\frac {(a+b x)^{3/4} (b c-a d) \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{\left (\frac {\sqrt {a+b x} (b c-a d)}{d}+1\right )^{3/4}}d\sqrt {a+b x}}{b^2 \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}\right )}{6 b}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle \frac {5 (b c-a d) \left (\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{b}-\frac {2 \sqrt {d} (a+b x)^{3/4} \sqrt {b c-a d} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d} \sqrt {a+b x}}{\sqrt {d}}\right ),2\right )}{b^2 \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}\right )}{6 b}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}\) |
(2*(a + b*x)^(1/4)*(c + d*x)^(5/4))/(3*b) + (5*(b*c - a*d)*((2*(a + b*x)^( 1/4)*(c + d*x)^(1/4))/b - (2*Sqrt[d]*Sqrt[b*c - a*d]*(a + b*x)^(3/4)*(1 + (b*c - a*d)/(d*(a + b*x)))^(3/4)*EllipticF[ArcTan[(Sqrt[b*c - a*d]*Sqrt[a + b*x])/Sqrt[d]]/2, 2])/(b^2*(c - (a*d)/b + (d*(a + b*x))/b)^(3/4))))/(6*b )
3.17.88.3.1 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/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[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[((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[((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]
\[\int \frac {\left (d x +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {3}{4}}}d x\]
\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {3}{4}}} \,d x } \]
\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/4}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{4}}}{\left (a + b x\right )^{\frac {3}{4}}}\, dx \]
\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {3}{4}}} \,d x } \]
\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {3}{4}}} \,d x } \]
Timed out. \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/4}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{3/4}} \,d x \]