\(\int \frac {x^2 \sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx\) [494]

Optimal result

Integrand size = 22, antiderivative size = 267 \[ \int \frac {x^2 \sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx=\frac {\left (15 b^2 c^2+10 a b c d+7 a^2 d^2\right ) \sqrt [4]{a+b x} (c+d x)^{3/4}}{32 b^2 d^3}-\frac {(3 b c+5 a d) (a+b x)^{5/4} (c+d x)^{3/4}}{8 b^2 d^2}+\frac {(a+b x)^{9/4} (c+d x)^{3/4}}{3 b^2 d}-\frac {(b c-a d) \left (15 b^2 c^2+10 a b c d+7 a^2 d^2\right ) \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{11/4} d^{13/4}}-\frac {(b c-a d) \left (15 b^2 c^2+10 a b c d+7 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{11/4} d^{13/4}} \] Output:

1/32*(7*a^2*d^2+10*a*b*c*d+15*b^2*c^2)*(b*x+a)^(1/4)*(d*x+c)^(3/4)/b^2/d^3 
-1/8*(5*a*d+3*b*c)*(b*x+a)^(5/4)*(d*x+c)^(3/4)/b^2/d^2+1/3*(b*x+a)^(9/4)*( 
d*x+c)^(3/4)/b^2/d-1/64*(-a*d+b*c)*(7*a^2*d^2+10*a*b*c*d+15*b^2*c^2)*arcta 
n(d^(1/4)*(b*x+a)^(1/4)/b^(1/4)/(d*x+c)^(1/4))/b^(11/4)/d^(13/4)-1/64*(-a* 
d+b*c)*(7*a^2*d^2+10*a*b*c*d+15*b^2*c^2)*arctanh(d^(1/4)*(b*x+a)^(1/4)/b^( 
1/4)/(d*x+c)^(1/4))/b^(11/4)/d^(13/4)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 10.10 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.46 \[ \int \frac {x^2 \sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx=\frac {(a+b x)^{5/4} \left (-5 b (c+d x) (9 b c+7 a d-8 b d x)+3 \left (15 b^2 c^2+10 a b c d+7 a^2 d^2\right ) \sqrt [4]{\frac {b (c+d x)}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {5}{4},\frac {9}{4},\frac {d (a+b x)}{-b c+a d}\right )\right )}{120 b^3 d^2 \sqrt [4]{c+d x}} \] Input:

Integrate[(x^2*(a + b*x)^(1/4))/(c + d*x)^(1/4),x]
 

Output:

((a + b*x)^(5/4)*(-5*b*(c + d*x)*(9*b*c + 7*a*d - 8*b*d*x) + 3*(15*b^2*c^2 
 + 10*a*b*c*d + 7*a^2*d^2)*((b*(c + d*x))/(b*c - a*d))^(1/4)*Hypergeometri 
c2F1[1/4, 5/4, 9/4, (d*(a + b*x))/(-(b*c) + a*d)]))/(120*b^3*d^2*(c + d*x) 
^(1/4))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {101, 27, 90, 60, 73, 770, 756, 218, 221}

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.

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

\(\Big \downarrow \) 101

\(\displaystyle \frac {\int -\frac {\sqrt [4]{a+b x} (4 a c+(9 b c+7 a d) x)}{4 \sqrt [4]{c+d x}}dx}{3 b d}+\frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d}-\frac {\int \frac {\sqrt [4]{a+b x} (4 a c+(9 b c+7 a d) x)}{\sqrt [4]{c+d x}}dx}{12 b d}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d}-\frac {\frac {(a+b x)^{5/4} (c+d x)^{3/4} (7 a d+9 b c)}{2 b d}-\frac {3 \left (7 a^2 d^2+10 a b c d+15 b^2 c^2\right ) \int \frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}dx}{8 b d}}{12 b d}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d}-\frac {\frac {(a+b x)^{5/4} (c+d x)^{3/4} (7 a d+9 b c)}{2 b d}-\frac {3 \left (7 a^2 d^2+10 a b c d+15 b^2 c^2\right ) \left (\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{d}-\frac {(b c-a d) \int \frac {1}{(a+b x)^{3/4} \sqrt [4]{c+d x}}dx}{4 d}\right )}{8 b d}}{12 b d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d}-\frac {\frac {(a+b x)^{5/4} (c+d x)^{3/4} (7 a d+9 b c)}{2 b d}-\frac {3 \left (7 a^2 d^2+10 a b c d+15 b^2 c^2\right ) \left (\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}d\sqrt [4]{a+b x}}{b d}\right )}{8 b d}}{12 b d}\)

\(\Big \downarrow \) 770

\(\displaystyle \frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d}-\frac {\frac {(a+b x)^{5/4} (c+d x)^{3/4} (7 a d+9 b c)}{2 b d}-\frac {3 \left (7 a^2 d^2+10 a b c d+15 b^2 c^2\right ) \left (\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{d}-\frac {(b c-a d) \int \frac {1}{1-\frac {d (a+b x)}{b}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}}{b d}\right )}{8 b d}}{12 b d}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d}-\frac {\frac {(a+b x)^{5/4} (c+d x)^{3/4} (7 a d+9 b c)}{2 b d}-\frac {3 \left (7 a^2 d^2+10 a b c d+15 b^2 c^2\right ) \left (\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{d}-\frac {(b c-a d) \left (\frac {1}{2} \sqrt {b} \int \frac {1}{\sqrt {b}-\sqrt {d} \sqrt {a+b x}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}+\frac {1}{2} \sqrt {b} \int \frac {1}{\sqrt {b}+\sqrt {d} \sqrt {a+b x}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}\right )}{b d}\right )}{8 b d}}{12 b d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d}-\frac {\frac {(a+b x)^{5/4} (c+d x)^{3/4} (7 a d+9 b c)}{2 b d}-\frac {3 \left (7 a^2 d^2+10 a b c d+15 b^2 c^2\right ) \left (\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{d}-\frac {(b c-a d) \left (\frac {1}{2} \sqrt {b} \int \frac {1}{\sqrt {b}-\sqrt {d} \sqrt {a+b x}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}+\frac {\sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{2 \sqrt [4]{d}}\right )}{b d}\right )}{8 b d}}{12 b d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d}-\frac {\frac {(a+b x)^{5/4} (c+d x)^{3/4} (7 a d+9 b c)}{2 b d}-\frac {3 \left (7 a^2 d^2+10 a b c d+15 b^2 c^2\right ) \left (\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{d}-\frac {(b c-a d) \left (\frac {\sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{2 \sqrt [4]{d}}+\frac {\sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{2 \sqrt [4]{d}}\right )}{b d}\right )}{8 b d}}{12 b d}\)

Input:

Int[(x^2*(a + b*x)^(1/4))/(c + d*x)^(1/4),x]
 

Output:

(x*(a + b*x)^(5/4)*(c + d*x)^(3/4))/(3*b*d) - (((9*b*c + 7*a*d)*(a + b*x)^ 
(5/4)*(c + d*x)^(3/4))/(2*b*d) - (3*(15*b^2*c^2 + 10*a*b*c*d + 7*a^2*d^2)* 
(((a + b*x)^(1/4)*(c + d*x)^(3/4))/d - ((b*c - a*d)*((b^(1/4)*ArcTan[(d^(1 
/4)*(a + b*x)^(1/4))/(b^(1/4)*(c - (a*d)/b + (d*(a + b*x))/b)^(1/4))])/(2* 
d^(1/4)) + (b^(1/4)*ArcTanh[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c - (a*d)/ 
b + (d*(a + b*x))/b)^(1/4))])/(2*d^(1/4))))/(b*d)))/(8*b*d))/(12*b*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 770

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n)   Subst[In 
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, 
 b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 
/n]
 

Maple [F]

\[\int \frac {x^{2} \left (b x +a \right )^{\frac {1}{4}}}{\left (x d +c \right )^{\frac {1}{4}}}d x\]

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 1814, normalized size of antiderivative = 6.79 \[ \int \frac {x^2 \sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/384*(3*b^2*d^3*((50625*b^12*c^12 - 67500*a*b^11*c^11*d - 6750*a^2*b^10* 
c^10*d^2 - 61500*a^3*b^9*c^9*d^3 + 93775*a^4*b^8*c^8*d^4 + 18600*a^5*b^7*c 
^7*d^5 + 31580*a^6*b^6*c^6*d^6 - 48600*a^7*b^5*c^5*d^7 - 15249*a^8*b^4*c^4 
*d^8 - 11004*a^9*b^3*c^3*d^9 + 9506*a^10*b^2*c^2*d^10 + 4116*a^11*b*c*d^11 
 + 2401*a^12*d^12)/(b^11*d^13))^(1/4)*log(-((15*b^3*c^3 - 5*a*b^2*c^2*d - 
3*a^2*b*c*d^2 - 7*a^3*d^3)*(b*x + a)^(1/4)*(d*x + c)^(3/4) + (b^3*d^4*x + 
b^3*c*d^3)*((50625*b^12*c^12 - 67500*a*b^11*c^11*d - 6750*a^2*b^10*c^10*d^ 
2 - 61500*a^3*b^9*c^9*d^3 + 93775*a^4*b^8*c^8*d^4 + 18600*a^5*b^7*c^7*d^5 
+ 31580*a^6*b^6*c^6*d^6 - 48600*a^7*b^5*c^5*d^7 - 15249*a^8*b^4*c^4*d^8 - 
11004*a^9*b^3*c^3*d^9 + 9506*a^10*b^2*c^2*d^10 + 4116*a^11*b*c*d^11 + 2401 
*a^12*d^12)/(b^11*d^13))^(1/4))/(d*x + c)) - 3*b^2*d^3*((50625*b^12*c^12 - 
 67500*a*b^11*c^11*d - 6750*a^2*b^10*c^10*d^2 - 61500*a^3*b^9*c^9*d^3 + 93 
775*a^4*b^8*c^8*d^4 + 18600*a^5*b^7*c^7*d^5 + 31580*a^6*b^6*c^6*d^6 - 4860 
0*a^7*b^5*c^5*d^7 - 15249*a^8*b^4*c^4*d^8 - 11004*a^9*b^3*c^3*d^9 + 9506*a 
^10*b^2*c^2*d^10 + 4116*a^11*b*c*d^11 + 2401*a^12*d^12)/(b^11*d^13))^(1/4) 
*log(-((15*b^3*c^3 - 5*a*b^2*c^2*d - 3*a^2*b*c*d^2 - 7*a^3*d^3)*(b*x + a)^ 
(1/4)*(d*x + c)^(3/4) - (b^3*d^4*x + b^3*c*d^3)*((50625*b^12*c^12 - 67500* 
a*b^11*c^11*d - 6750*a^2*b^10*c^10*d^2 - 61500*a^3*b^9*c^9*d^3 + 93775*a^4 
*b^8*c^8*d^4 + 18600*a^5*b^7*c^7*d^5 + 31580*a^6*b^6*c^6*d^6 - 48600*a^7*b 
^5*c^5*d^7 - 15249*a^8*b^4*c^4*d^8 - 11004*a^9*b^3*c^3*d^9 + 9506*a^10*...
 

Sympy [F]

\[ \int \frac {x^2 \sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx=\int \frac {x^{2} \sqrt [4]{a + b x}}{\sqrt [4]{c + d x}}\, dx \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx=\int \frac {x^2\,{\left (a+b\,x\right )}^{1/4}}{{\left (c+d\,x\right )}^{1/4}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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