Integrand size = 19, antiderivative size = 111 \[ \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}} \, dx=\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d)^{5/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{3 \sqrt [4]{b} d^2 \sqrt {a+b x}} \] Output:
4/3*(b*x+a)^(1/2)*(d*x+c)^(1/4)/d-8/3*(-a*d+b*c)^(5/4)*(-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^(1/4)/d^2 /(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 = 0.66 \[ \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}} \, dx=\frac {2 (a+b x)^{3/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {5}{2},\frac {d (a+b x)}{-b c+a d}\right )}{3 b (c+d x)^{3/4}} \] Input:
Integrate[Sqrt[a + b*x]/(c + d*x)^(3/4),x]
Output:
(2*(a + b*x)^(3/2)*((b*(c + d*x))/(b*c - a*d))^(3/4)*Hypergeometric2F1[3/4 , 3/2, 5/2, (d*(a + b*x))/(-(b*c) + a*d)])/(3*b*(c + d*x)^(3/4))
Time = 0.22 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {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.
\(\displaystyle \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}} \, dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {2 (b c-a d) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}}dx}{3 d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d) \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [4]{c+d x}}{3 d^2}\) |
\(\Big \downarrow \) 765 |
\(\displaystyle \frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d) \sqrt {1-\frac {b (c+d x)}{b c-a d}} \int \frac {1}{\sqrt {1-\frac {b (c+d x)}{b c-a d}}}d\sqrt [4]{c+d x}}{3 d^2 \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d)^{5/4} \sqrt {1-\frac {b (c+d x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{3 \sqrt [4]{b} d^2 \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\) |
Input:
Int[Sqrt[a + b*x]/(c + d*x)^(3/4),x]
Output:
(4*Sqrt[a + b*x]*(c + d*x)^(1/4))/(3*d) - (8*(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^(1/4)*d^2*Sqrt[a - (b*c)/d + (b*(c + d*x))/d])
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]
\[\int \frac {\sqrt {b x +a}}{\left (x d +c \right )^{\frac {3}{4}}}d x\]
Input:
int((b*x+a)^(1/2)/(d*x+c)^(3/4),x)
Output:
int((b*x+a)^(1/2)/(d*x+c)^(3/4),x)
\[ \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}} \, dx=\int { \frac {\sqrt {b x + a}}{{\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate((b*x+a)^(1/2)/(d*x+c)^(3/4),x, algorithm="fricas")
Output:
integral(sqrt(b*x + a)/(d*x + c)^(3/4), x)
\[ \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}} \, dx=\int \frac {\sqrt {a + b x}}{\left (c + d x\right )^{\frac {3}{4}}}\, dx \] Input:
integrate((b*x+a)**(1/2)/(d*x+c)**(3/4),x)
Output:
Integral(sqrt(a + b*x)/(c + d*x)**(3/4), x)
\[ \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}} \, dx=\int { \frac {\sqrt {b x + a}}{{\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate((b*x+a)^(1/2)/(d*x+c)^(3/4),x, algorithm="maxima")
Output:
integrate(sqrt(b*x + a)/(d*x + c)^(3/4), x)
\[ \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}} \, dx=\int { \frac {\sqrt {b x + a}}{{\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate((b*x+a)^(1/2)/(d*x+c)^(3/4),x, algorithm="giac")
Output:
integrate(sqrt(b*x + a)/(d*x + c)^(3/4), x)
Timed out. \[ \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}} \, dx=\int \frac {\sqrt {a+b\,x}}{{\left (c+d\,x\right )}^{3/4}} \,d x \] Input:
int((a + b*x)^(1/2)/(c + d*x)^(3/4),x)
Output:
int((a + b*x)^(1/2)/(c + d*x)^(3/4), x)
\[ \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}} \, dx=\int \frac {\sqrt {b x +a}}{\left (d x +c \right )^{\frac {3}{4}}}d x \] Input:
int((b*x+a)^(1/2)/(d*x+c)^(3/4),x)
Output:
int(sqrt(a + b*x)/(c + d*x)**(3/4),x)