\(\int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx\) [460]

Optimal result

Integrand size = 19, antiderivative size = 83 \[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx=\frac {4 \sqrt [4]{b c-a d} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{\sqrt [4]{b} d \sqrt {a+b x}} \] Output:

4*(-a*d+b*c)^(1/4)*(-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/(b*x+a)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx=\frac {2 \sqrt {a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\frac {d (a+b x)}{-b c+a d}\right )}{b (c+d x)^{3/4}} \] Input:

Integrate[1/(Sqrt[a + b*x]*(c + d*x)^(3/4)),x]
 

Output:

(2*Sqrt[a + b*x]*((b*(c + d*x))/(b*c - a*d))^(3/4)*Hypergeometric2F1[1/2, 
3/4, 3/2, (d*(a + b*x))/(-(b*c) + a*d)])/(b*(c + d*x)^(3/4))
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.19, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {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.

\(\displaystyle \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {4 \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [4]{c+d x}}{d}\)

\(\Big \downarrow \) 765

\(\displaystyle \frac {4 \sqrt {1-\frac {b (c+d x)}{b c-a d}} \int \frac {1}{\sqrt {1-\frac {b (c+d x)}{b c-a d}}}d\sqrt [4]{c+d x}}{d \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {4 \sqrt [4]{b c-a d} \sqrt {1-\frac {b (c+d x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{\sqrt [4]{b} d \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\)

Input:

Int[1/(Sqrt[a + b*x]*(c + d*x)^(3/4)),x]
 

Output:

(4*(b*c - a*d)^(1/4)*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])/(b^(1/4)*d*Sqrt[a - (b* 
c)/d + (b*(c + d*x))/d])
 

Defintions of rubi rules used

rule 73

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]
 

rule 762

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]
 

rule 765

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]
 

Maple [F]

\[\int \frac {1}{\sqrt {b x +a}\, \left (x d +c \right )^{\frac {3}{4}}}d x\]

Input:

int(1/(b*x+a)^(1/2)/(d*x+c)^(3/4),x)
 

Output:

int(1/(b*x+a)^(1/2)/(d*x+c)^(3/4),x)
 

Fricas [F]

\[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx=\int { \frac {1}{\sqrt {b x + a} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate(1/(b*x+a)^(1/2)/(d*x+c)^(3/4),x, algorithm="fricas")
 

Output:

integral(sqrt(b*x + a)*(d*x + c)^(1/4)/(b*d*x^2 + a*c + (b*c + a*d)*x), x)
 

Sympy [F]

\[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx=\int \frac {1}{\sqrt {a + b x} \left (c + d x\right )^{\frac {3}{4}}}\, dx \] Input:

integrate(1/(b*x+a)**(1/2)/(d*x+c)**(3/4),x)
 

Output:

Integral(1/(sqrt(a + b*x)*(c + d*x)**(3/4)), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx=\int { \frac {1}{\sqrt {b x + a} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate(1/(b*x+a)^(1/2)/(d*x+c)^(3/4),x, algorithm="maxima")
 

Output:

integrate(1/(sqrt(b*x + a)*(d*x + c)^(3/4)), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx=\int { \frac {1}{\sqrt {b x + a} {\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate(1/(b*x+a)^(1/2)/(d*x+c)^(3/4),x, algorithm="giac")
 

Output:

integrate(1/(sqrt(b*x + a)*(d*x + c)^(3/4)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx=\int \frac {1}{\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{3/4}} \,d x \] Input:

int(1/((a + b*x)^(1/2)*(c + d*x)^(3/4)),x)
 

Output:

int(1/((a + b*x)^(1/2)*(c + d*x)^(3/4)), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx=\frac {4 \left (d x +c \right )^{\frac {3}{4}} \sqrt {b x +a}-15 \sqrt {d x +c}\, \left (\int \frac {\left (d x +c \right )^{\frac {1}{4}} \sqrt {b x +a}\, x}{3 a b \,d^{2} x^{2}+2 b^{2} c d \,x^{2}+3 a^{2} d^{2} x +5 a b c d x +2 b^{2} c^{2} x +3 a^{2} c d +2 a b \,c^{2}}d x \right ) a b \,d^{2}-10 \sqrt {d x +c}\, \left (\int \frac {\left (d x +c \right )^{\frac {1}{4}} \sqrt {b x +a}\, x}{3 a b \,d^{2} x^{2}+2 b^{2} c d \,x^{2}+3 a^{2} d^{2} x +5 a b c d x +2 b^{2} c^{2} x +3 a^{2} c d +2 a b \,c^{2}}d x \right ) b^{2} c d}{\sqrt {d x +c}\, \left (3 a d +2 b c \right )} \] Input:

int(1/(b*x+a)^(1/2)/(d*x+c)^(3/4),x)
 

Output:

(4*(c + d*x)**(3/4)*sqrt(a + b*x) - 15*sqrt(c + d*x)*int(((c + d*x)**(1/4) 
*sqrt(a + b*x)*x)/(3*a**2*c*d + 3*a**2*d**2*x + 2*a*b*c**2 + 5*a*b*c*d*x + 
 3*a*b*d**2*x**2 + 2*b**2*c**2*x + 2*b**2*c*d*x**2),x)*a*b*d**2 - 10*sqrt( 
c + d*x)*int(((c + d*x)**(1/4)*sqrt(a + b*x)*x)/(3*a**2*c*d + 3*a**2*d**2* 
x + 2*a*b*c**2 + 5*a*b*c*d*x + 3*a*b*d**2*x**2 + 2*b**2*c**2*x + 2*b**2*c* 
d*x**2),x)*b**2*c*d)/(sqrt(c + d*x)*(3*a*d + 2*b*c))