Integrand size = 19, antiderivative size = 401 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx=-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 d^2 \sqrt [4]{c+d x}}{231 b^2 (b c-a d) (a+b x)^{7/4}}+\frac {8 d^3 \sqrt [4]{c+d x}}{231 b^2 (b c-a d)^2 (a+b x)^{3/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}+\frac {4 \sqrt {2} d^{15/4} ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right ),\frac {1}{2}\right )}{231 b^{9/4} (b c-a d)^{3/2} (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}} \]
-4/33*d*(d*x+c)^(1/4)/b^2/(b*x+a)^(11/4)-4/231*d^2*(d*x+c)^(1/4)/b^2/(-a*d +b*c)/(b*x+a)^(7/4)+8/231*d^3*(d*x+c)^(1/4)/b^2/(-a*d+b*c)^2/(b*x+a)^(3/4) -4/15*(d*x+c)^(5/4)/b/(b*x+a)^(15/4)+4/231*d^(15/4)*((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))*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)/(-a*d+b*c)^(3/2)/(b*x+a)^(3/4)/(d*x+c)^(3/ 4)/(2*b*d*x+a*d+b*c)/((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 = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.18 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx=-\frac {4 (c+d x)^{5/4} \operatorname {Hypergeometric2F1}\left (-\frac {15}{4},-\frac {5}{4},-\frac {11}{4},\frac {d (a+b x)}{-b c+a d}\right )}{15 b (a+b x)^{15/4} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \]
(-4*(c + d*x)^(5/4)*Hypergeometric2F1[-15/4, -5/4, -11/4, (d*(a + b*x))/(- (b*c) + a*d)])/(15*b*(a + b*x)^(15/4)*((b*(c + d*x))/(b*c - a*d))^(5/4))
Time = 0.37 (sec) , antiderivative size = 264, normalized size of antiderivative = 0.66, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {57, 57, 61, 61, 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.
| \(\displaystyle \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx\) |
| \(\Big \downarrow \) 57 |
| \(\displaystyle \frac {d \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{15/4}}dx}{3 b}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}\) |
| \(\Big \downarrow \) 57 |
| \(\displaystyle \frac {d \left (\frac {d \int \frac {1}{(a+b x)^{11/4} (c+d x)^{3/4}}dx}{11 b}-\frac {4 \sqrt [4]{c+d x}}{11 b (a+b x)^{11/4}}\right )}{3 b}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {d \left (\frac {d \left (-\frac {6 d \int \frac {1}{(a+b x)^{7/4} (c+d x)^{3/4}}dx}{7 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{7 (a+b x)^{7/4} (b c-a d)}\right )}{11 b}-\frac {4 \sqrt [4]{c+d x}}{11 b (a+b x)^{11/4}}\right )}{3 b}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {d \left (\frac {d \left (-\frac {6 d \left (-\frac {2 d \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}}dx}{3 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{3 (a+b x)^{3/4} (b c-a d)}\right )}{7 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{7 (a+b x)^{7/4} (b c-a d)}\right )}{11 b}-\frac {4 \sqrt [4]{c+d x}}{11 b (a+b x)^{11/4}}\right )}{3 b}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {d \left (\frac {d \left (-\frac {6 d \left (-\frac {8 d \int \frac {1}{\left (c-\frac {a d}{b}+\frac {d (a+b x)}{b}\right )^{3/4}}d\sqrt [4]{a+b x}}{3 b (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{3 (a+b x)^{3/4} (b c-a d)}\right )}{7 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{7 (a+b x)^{7/4} (b c-a d)}\right )}{11 b}-\frac {4 \sqrt [4]{c+d x}}{11 b (a+b x)^{11/4}}\right )}{3 b}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}\) |
| \(\Big \downarrow \) 768 |
| \(\displaystyle \frac {d \left (\frac {d \left (-\frac {6 d \left (-\frac {8 d (a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{(a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4}}d\sqrt [4]{a+b x}}{3 b (b c-a d) \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}-\frac {4 \sqrt [4]{c+d x}}{3 (a+b x)^{3/4} (b c-a d)}\right )}{7 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{7 (a+b x)^{7/4} (b c-a d)}\right )}{11 b}-\frac {4 \sqrt [4]{c+d x}}{11 b (a+b x)^{11/4}}\right )}{3 b}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}\) |
| \(\Big \downarrow \) 858 |
| \(\displaystyle \frac {d \left (\frac {d \left (-\frac {6 d \left (\frac {8 d (a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{\sqrt [4]{a+b x} \left (\frac {(b c-a d) (a+b x)}{d}+1\right )^{3/4}}d\frac {1}{\sqrt [4]{a+b x}}}{3 b (b c-a d) \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}-\frac {4 \sqrt [4]{c+d x}}{3 (a+b x)^{3/4} (b c-a d)}\right )}{7 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{7 (a+b x)^{7/4} (b c-a d)}\right )}{11 b}-\frac {4 \sqrt [4]{c+d x}}{11 b (a+b x)^{11/4}}\right )}{3 b}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}\) |
| \(\Big \downarrow \) 807 |
| \(\displaystyle \frac {d \left (\frac {d \left (-\frac {6 d \left (\frac {4 d (a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{\left (\frac {\sqrt {a+b x} (b c-a d)}{d}+1\right )^{3/4}}d\sqrt {a+b x}}{3 b (b c-a d) \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}-\frac {4 \sqrt [4]{c+d x}}{3 (a+b x)^{3/4} (b c-a d)}\right )}{7 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{7 (a+b x)^{7/4} (b c-a d)}\right )}{11 b}-\frac {4 \sqrt [4]{c+d x}}{11 b (a+b x)^{11/4}}\right )}{3 b}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}\) |
| \(\Big \downarrow \) 229 |
| \(\displaystyle \frac {d \left (\frac {d \left (-\frac {6 d \left (\frac {8 d^{3/2} (a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d} \sqrt {a+b x}}{\sqrt {d}}\right ),2\right )}{3 b (b c-a d)^{3/2} \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}-\frac {4 \sqrt [4]{c+d x}}{3 (a+b x)^{3/4} (b c-a d)}\right )}{7 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{7 (a+b x)^{7/4} (b c-a d)}\right )}{11 b}-\frac {4 \sqrt [4]{c+d x}}{11 b (a+b x)^{11/4}}\right )}{3 b}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}\) |
(-4*(c + d*x)^(5/4))/(15*b*(a + b*x)^(15/4)) + (d*((-4*(c + d*x)^(1/4))/(1 1*b*(a + b*x)^(11/4)) + (d*((-4*(c + d*x)^(1/4))/(7*(b*c - a*d)*(a + b*x)^ (7/4)) - (6*d*((-4*(c + d*x)^(1/4))/(3*(b*c - a*d)*(a + b*x)^(3/4)) + (8*d ^(3/2)*(a + b*x)^(3/4)*(1 + (b*c - a*d)/(d*(a + b*x)))^(3/4)*EllipticF[Arc Tan[(Sqrt[b*c - a*d]*Sqrt[a + b*x])/Sqrt[d]]/2, 2])/(3*b*(b*c - a*d)^(3/2) *(c - (a*d)/b + (d*(a + b*x))/b)^(3/4))))/(7*(b*c - a*d))))/(11*b)))/(3*b)
3.17.92.3.1 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 + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, x]
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[((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]
\[\int \frac {\left (d x +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {19}{4}}}d x\]
\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {19}{4}}} \,d x } \]
integral((b*x + a)^(1/4)*(d*x + c)^(5/4)/(b^5*x^5 + 5*a*b^4*x^4 + 10*a^2*b ^3*x^3 + 10*a^3*b^2*x^2 + 5*a^4*b*x + a^5), x)
Timed out. \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx=\text {Timed out} \]
\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {19}{4}}} \,d x } \]
\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {19}{4}}} \,d x } \]
Timed out. \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{19/4}} \,d x \]