\(\int \sqrt {a+b x} \sqrt [4]{c+d x} \, dx\) [432]

Optimal result

Integrand size = 19, antiderivative size = 147 \[ \int \sqrt {a+b x} \sqrt [4]{c+d x} \, dx=\frac {4 (b c-a d) \sqrt {a+b x} \sqrt [4]{c+d x}}{21 b d}+\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 b}-\frac {8 (b c-a d)^{9/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{21 b^{5/4} d^2 \sqrt {a+b x}} \] Output:

4/21*(-a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/4)/b/d+4/7*(b*x+a)^(3/2)*(d*x+c)^ 
(1/4)/b-8/21*(-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^(5/4)/d^2/(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 = 73, normalized size of antiderivative = 0.50 \[ \int \sqrt {a+b x} \sqrt [4]{c+d x} \, dx=\frac {2 (a+b x)^{3/2} \sqrt [4]{c+d x} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{2},\frac {5}{2},\frac {d (a+b x)}{-b c+a d}\right )}{3 b \sqrt [4]{\frac {b (c+d x)}{b c-a d}}} \] Input:

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

Output:

(2*(a + b*x)^(3/2)*(c + d*x)^(1/4)*Hypergeometric2F1[-1/4, 3/2, 5/2, (d*(a 
 + b*x))/(-(b*c) + a*d)])/(3*b*((b*(c + d*x))/(b*c - a*d))^(1/4))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.14, 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.

\(\displaystyle \int \sqrt {a+b x} \sqrt [4]{c+d x} \, dx\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(b c-a d) \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}}dx}{7 b}+\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {2 (b c-a d) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}}dx}{3 d}\right )}{7 b}+\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d) \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [4]{c+d x}}{3 d^2}\right )}{7 b}+\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 b}\)

\(\Big \downarrow \) 765

\(\displaystyle \frac {(b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d) \sqrt {1-\frac {b (c+d x)}{b c-a d}} \int \frac {1}{\sqrt {1-\frac {b (c+d x)}{b c-a d}}}d\sqrt [4]{c+d x}}{3 d^2 \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\right )}{7 b}+\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 b}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {(b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d)^{5/4} \sqrt {1-\frac {b (c+d x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{3 \sqrt [4]{b} d^2 \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\right )}{7 b}+\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 b}\)

Input:

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

Output:

(4*(a + b*x)^(3/2)*(c + d*x)^(1/4))/(7*b) + ((b*c - a*d)*((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])))/(7*b)
 

Defintions of rubi rules used

rule 60

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]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

\[ \int \sqrt {a+b x} \sqrt [4]{c+d x} \, dx=\int \sqrt {a + b x} \sqrt [4]{c + d x}\, dx \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt {a+b x} \sqrt [4]{c+d x} \, dx=\frac {12 \left (d x +c \right )^{\frac {1}{4}} \sqrt {b x +a}\, a c +4 \left (d x +c \right )^{\frac {1}{4}} \sqrt {b x +a}\, a d x +8 \left (d x +c \right )^{\frac {1}{4}} \sqrt {b x +a}\, b c x +2 \left (\int \frac {\left (d x +c \right )^{\frac {1}{4}} \sqrt {b x +a}\, x}{a b \,d^{2} x^{2}+2 b^{2} c d \,x^{2}+a^{2} d^{2} x +3 a b c d x +2 b^{2} c^{2} x +a^{2} c d +2 a b \,c^{2}}d x \right ) a^{3} d^{3}-6 \left (\int \frac {\left (d x +c \right )^{\frac {1}{4}} \sqrt {b x +a}\, x}{a b \,d^{2} x^{2}+2 b^{2} c d \,x^{2}+a^{2} d^{2} x +3 a b c d x +2 b^{2} c^{2} x +a^{2} c d +2 a b \,c^{2}}d x \right ) a \,b^{2} c^{2} d +4 \left (\int \frac {\left (d x +c \right )^{\frac {1}{4}} \sqrt {b x +a}\, x}{a b \,d^{2} x^{2}+2 b^{2} c d \,x^{2}+a^{2} d^{2} x +3 a b c d x +2 b^{2} c^{2} x +a^{2} c d +2 a b \,c^{2}}d x \right ) b^{3} c^{3}}{7 a d +14 b c} \] Input:

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

Output:

(2*(6*(c + d*x)**(1/4)*sqrt(a + b*x)*a*c + 2*(c + d*x)**(1/4)*sqrt(a + b*x 
)*a*d*x + 4*(c + d*x)**(1/4)*sqrt(a + b*x)*b*c*x + int(((c + d*x)**(1/4)*s 
qrt(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**3 - 3*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 
*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))/(7*(a*d + 2*b*c))