Integrand size = 19, antiderivative size = 225 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx=-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 d^2 \sqrt [4]{c+d x}}{231 b^2 (b c-a d) (a+b x)^{7/4}}+\frac {8 d^3 \sqrt [4]{c+d x}}{231 b^2 (b c-a d)^2 (a+b x)^{3/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}-\frac {16 d^{9/2} (a+b x)^{3/4} \left (\frac {b (c+d x)}{d (a+b x)}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d}}{\sqrt {d} \sqrt {a+b x}}\right ),2\right )}{231 b^3 (b c-a d)^{5/2} (c+d x)^{3/4}} \] Output:
-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)-16/231*d^(9/2)*(b*x+a)^(3/4)*(b*(d*x+ c)/d/(b*x+a))^(3/4)*InverseJacobiAM(1/2*arctan(1/d^(1/2)/(b*x+a)^(1/2)*(-a *d+b*c)^(1/2)),2^(1/2))/b^3/(-a*d+b*c)^(5/2)/(d*x+c)^(3/4)
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 = 0.32 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx=-\frac {4 (c+d x)^{5/4} \operatorname {Hypergeometric2F1}\left (-\frac {15}{4},-\frac {5}{4},-\frac {11}{4},\frac {d (a+b x)}{-b c+a d}\right )}{15 b (a+b x)^{15/4} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \] Input:
Integrate[(c + d*x)^(5/4)/(a + b*x)^(19/4),x]
Output:
(-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 = 0.38 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.17, 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 |
| \(\displaystyle \frac {d \left (\frac {d \left (-\frac {6 d \left (\frac {8 d^{3/2} (a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d} \sqrt {a+b x}}{\sqrt {d}}\right ),2\right )}{3 b (b c-a d)^{3/2} \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}-\frac {4 \sqrt [4]{c+d x}}{3 (a+b x)^{3/4} (b c-a d)}\right )}{7 (b c-a d)}-\frac {4 \sqrt [4]{c+d x}}{7 (a+b x)^{7/4} (b c-a d)}\right )}{11 b}-\frac {4 \sqrt [4]{c+d x}}{11 b (a+b x)^{11/4}}\right )}{3 b}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}\) |
Input:
Int[(c + d*x)^(5/4)/(a + b*x)^(19/4),x]
Output:
(-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)
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]
\[\int \frac {\left (x d +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {19}{4}}}d x\]
Input:
int((d*x+c)^(5/4)/(b*x+a)^(19/4),x)
Output:
int((d*x+c)^(5/4)/(b*x+a)^(19/4),x)
\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {19}{4}}} \,d x } \] Input:
integrate((d*x+c)^(5/4)/(b*x+a)^(19/4),x, algorithm="fricas")
Output:
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. \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx=\text {Timed out} \] Input:
integrate((d*x+c)**(5/4)/(b*x+a)**(19/4),x)
Output:
Timed out
\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {19}{4}}} \,d x } \] Input:
integrate((d*x+c)^(5/4)/(b*x+a)^(19/4),x, algorithm="maxima")
Output:
integrate((d*x + c)^(5/4)/(b*x + a)^(19/4), x)
\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {19}{4}}} \,d x } \] Input:
integrate((d*x+c)^(5/4)/(b*x+a)^(19/4),x, algorithm="giac")
Output:
integrate((d*x + c)^(5/4)/(b*x + a)^(19/4), x)
Timed out. \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{19/4}} \,d x \] Input:
int((c + d*x)^(5/4)/(a + b*x)^(19/4),x)
Output:
int((c + d*x)^(5/4)/(a + b*x)^(19/4), x)
\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx=\left (\int \frac {\left (d x +c \right )^{\frac {1}{4}}}{\left (b x +a \right )^{\frac {3}{4}} a^{4}+4 \left (b x +a \right )^{\frac {3}{4}} a^{3} b x +6 \left (b x +a \right )^{\frac {3}{4}} a^{2} b^{2} x^{2}+4 \left (b x +a \right )^{\frac {3}{4}} a \,b^{3} x^{3}+\left (b x +a \right )^{\frac {3}{4}} b^{4} x^{4}}d x \right ) c +\left (\int \frac {\left (d x +c \right )^{\frac {1}{4}} x}{\left (b x +a \right )^{\frac {3}{4}} a^{4}+4 \left (b x +a \right )^{\frac {3}{4}} a^{3} b x +6 \left (b x +a \right )^{\frac {3}{4}} a^{2} b^{2} x^{2}+4 \left (b x +a \right )^{\frac {3}{4}} a \,b^{3} x^{3}+\left (b x +a \right )^{\frac {3}{4}} b^{4} x^{4}}d x \right ) d \] Input:
int((d*x+c)^(5/4)/(b*x+a)^(19/4),x)
Output:
int((c + d*x)**(1/4)/((a + b*x)**(3/4)*a**4 + 4*(a + b*x)**(3/4)*a**3*b*x + 6*(a + b*x)**(3/4)*a**2*b**2*x**2 + 4*(a + b*x)**(3/4)*a*b**3*x**3 + (a + b*x)**(3/4)*b**4*x**4),x)*c + int(((c + d*x)**(1/4)*x)/((a + b*x)**(3/4) *a**4 + 4*(a + b*x)**(3/4)*a**3*b*x + 6*(a + b*x)**(3/4)*a**2*b**2*x**2 + 4*(a + b*x)**(3/4)*a*b**3*x**3 + (a + b*x)**(3/4)*b**4*x**4),x)*d