Integrand size = 19, antiderivative size = 144 \[ \int \frac {(c+d x)^{5/4}}{\sqrt {a+b x}} \, dx=\frac {20 (b c-a d) \sqrt {a+b x} \sqrt [4]{c+d x}}{21 b^2}+\frac {4 \sqrt {a+b x} (c+d x)^{5/4}}{7 b}+\frac {20 (b c-a d)^{9/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{21 b^{9/4} d \sqrt {a+b x}} \] Output:
20/21*(-a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/4)/b^2+4/7*(b*x+a)^(1/2)*(d*x+c) ^(5/4)/b+20/21*(-a*d+b*c)^(9/4)*(-d*(b*x+a)/(-a*d+b*c))^(1/2)*EllipticF(b^ (1/4)*(d*x+c)^(1/4)/(-a*d+b*c)^(1/4),I)/b^(9/4)/d/(b*x+a)^(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 = 0.49 \[ \int \frac {(c+d x)^{5/4}}{\sqrt {a+b x}} \, dx=\frac {2 \sqrt {a+b x} (c+d x)^{5/4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{2},\frac {3}{2},\frac {d (a+b x)}{-b c+a d}\right )}{b \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \] Input:
Integrate[(c + d*x)^(5/4)/Sqrt[a + b*x],x]
Output:
(2*Sqrt[a + b*x]*(c + d*x)^(5/4)*Hypergeometric2F1[-5/4, 1/2, 3/2, (d*(a + b*x))/(-(b*c) + a*d)])/(b*((b*(c + d*x))/(b*c - a*d))^(5/4))
Time = 0.24 (sec) , antiderivative size = 168, 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 = {60, 60, 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.
\(\displaystyle \int \frac {(c+d x)^{5/4}}{\sqrt {a+b x}} \, dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5 (b c-a d) \int \frac {\sqrt [4]{c+d x}}{\sqrt {a+b x}}dx}{7 b}+\frac {4 \sqrt {a+b x} (c+d x)^{5/4}}{7 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5 (b c-a d) \left (\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}}dx}{3 b}+\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 b}\right )}{7 b}+\frac {4 \sqrt {a+b x} (c+d x)^{5/4}}{7 b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {5 (b c-a d) \left (\frac {4 (b c-a d) \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [4]{c+d x}}{3 b d}+\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 b}\right )}{7 b}+\frac {4 \sqrt {a+b x} (c+d x)^{5/4}}{7 b}\) |
\(\Big \downarrow \) 765 |
\(\displaystyle \frac {5 (b c-a d) \left (\frac {4 (b c-a d) \sqrt {1-\frac {b (c+d x)}{b c-a d}} \int \frac {1}{\sqrt {1-\frac {b (c+d x)}{b c-a d}}}d\sqrt [4]{c+d x}}{3 b d \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}+\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 b}\right )}{7 b}+\frac {4 \sqrt {a+b x} (c+d x)^{5/4}}{7 b}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {5 (b c-a d) \left (\frac {4 (b c-a d)^{5/4} \sqrt {1-\frac {b (c+d x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{3 b^{5/4} d \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}+\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 b}\right )}{7 b}+\frac {4 \sqrt {a+b x} (c+d x)^{5/4}}{7 b}\) |
Input:
Int[(c + d*x)^(5/4)/Sqrt[a + b*x],x]
Output:
(4*Sqrt[a + b*x]*(c + d*x)^(5/4))/(7*b) + (5*(b*c - a*d)*((4*Sqrt[a + b*x] *(c + d*x)^(1/4))/(3*b) + (4*(b*c - a*d)^(5/4)*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 ])/(3*b^(5/4)*d*Sqrt[a - (b*c)/d + (b*(c + d*x))/d])))/(7*b)
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[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]
\[\int \frac {\left (x d +c \right )^{\frac {5}{4}}}{\sqrt {b x +a}}d x\]
Input:
int((d*x+c)^(5/4)/(b*x+a)^(1/2),x)
Output:
int((d*x+c)^(5/4)/(b*x+a)^(1/2),x)
\[ \int \frac {(c+d x)^{5/4}}{\sqrt {a+b x}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{\sqrt {b x + a}} \,d x } \] Input:
integrate((d*x+c)^(5/4)/(b*x+a)^(1/2),x, algorithm="fricas")
Output:
integral((d*x + c)^(5/4)/sqrt(b*x + a), x)
\[ \int \frac {(c+d x)^{5/4}}{\sqrt {a+b x}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{4}}}{\sqrt {a + b x}}\, dx \] Input:
integrate((d*x+c)**(5/4)/(b*x+a)**(1/2),x)
Output:
Integral((c + d*x)**(5/4)/sqrt(a + b*x), x)
\[ \int \frac {(c+d x)^{5/4}}{\sqrt {a+b x}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{\sqrt {b x + a}} \,d x } \] Input:
integrate((d*x+c)^(5/4)/(b*x+a)^(1/2),x, algorithm="maxima")
Output:
integrate((d*x + c)^(5/4)/sqrt(b*x + a), x)
\[ \int \frac {(c+d x)^{5/4}}{\sqrt {a+b x}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{\sqrt {b x + a}} \,d x } \] Input:
integrate((d*x+c)^(5/4)/(b*x+a)^(1/2),x, algorithm="giac")
Output:
integrate((d*x + c)^(5/4)/sqrt(b*x + a), x)
Timed out. \[ \int \frac {(c+d x)^{5/4}}{\sqrt {a+b x}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/4}}{\sqrt {a+b\,x}} \,d x \] Input:
int((c + d*x)^(5/4)/(a + b*x)^(1/2),x)
Output:
int((c + d*x)^(5/4)/(a + b*x)^(1/2), x)
\[ \int \frac {(c+d x)^{5/4}}{\sqrt {a+b x}} \, dx=\frac {-16 \left (d x +c \right )^{\frac {1}{4}} \sqrt {b x +a}\, a c d +4 \left (d x +c \right )^{\frac {1}{4}} \sqrt {b x +a}\, a \,d^{2} x +28 \left (d x +c \right )^{\frac {1}{4}} \sqrt {b x +a}\, b \,c^{2}+8 \left (d x +c \right )^{\frac {1}{4}} \sqrt {b x +a}\, b c d x -5 \left (\int \frac {\left (d x +c \right )^{\frac {1}{4}} \sqrt {b x +a}\, x}{a b \,d^{2} x^{2}+2 b^{2} c d \,x^{2}+a^{2} d^{2} x +3 a b c d x +2 b^{2} c^{2} x +a^{2} c d +2 a b \,c^{2}}d x \right ) a^{3} d^{4}+15 \left (\int \frac {\left (d x +c \right )^{\frac {1}{4}} \sqrt {b x +a}\, x}{a b \,d^{2} x^{2}+2 b^{2} c d \,x^{2}+a^{2} d^{2} x +3 a b c d x +2 b^{2} c^{2} x +a^{2} c d +2 a b \,c^{2}}d x \right ) a \,b^{2} c^{2} d^{2}-10 \left (\int \frac {\left (d x +c \right )^{\frac {1}{4}} \sqrt {b x +a}\, x}{a b \,d^{2} x^{2}+2 b^{2} c d \,x^{2}+a^{2} d^{2} x +3 a b c d x +2 b^{2} c^{2} x +a^{2} c d +2 a b \,c^{2}}d x \right ) b^{3} c^{3} d}{7 b \left (a d +2 b c \right )} \] Input:
int((d*x+c)^(5/4)/(b*x+a)^(1/2),x)
Output:
( - 16*(c + d*x)**(1/4)*sqrt(a + b*x)*a*c*d + 4*(c + d*x)**(1/4)*sqrt(a + b*x)*a*d**2*x + 28*(c + d*x)**(1/4)*sqrt(a + b*x)*b*c**2 + 8*(c + d*x)**(1 /4)*sqrt(a + b*x)*b*c*d*x - 5*int(((c + d*x)**(1/4)*sqrt(a + b*x)*x)/(a**2 *c*d + a**2*d**2*x + 2*a*b*c**2 + 3*a*b*c*d*x + a*b*d**2*x**2 + 2*b**2*c** 2*x + 2*b**2*c*d*x**2),x)*a**3*d**4 + 15*int(((c + d*x)**(1/4)*sqrt(a + b* x)*x)/(a**2*c*d + a**2*d**2*x + 2*a*b*c**2 + 3*a*b*c*d*x + a*b*d**2*x**2 + 2*b**2*c**2*x + 2*b**2*c*d*x**2),x)*a*b**2*c**2*d**2 - 10*int(((c + d*x)* *(1/4)*sqrt(a + b*x)*x)/(a**2*c*d + a**2*d**2*x + 2*a*b*c**2 + 3*a*b*c*d*x + a*b*d**2*x**2 + 2*b**2*c**2*x + 2*b**2*c*d*x**2),x)*b**3*c**3*d)/(7*b*( a*d + 2*b*c))