\(\int \frac {x^2 \sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx\) [884]

Optimal result

Integrand size = 22, antiderivative size = 268 \[ \int \frac {x^2 \sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx=\frac {\left (15 b^2 c^2+10 a b c d+7 a^2 d^2\right ) \sqrt [4]{a+b x} (c+d x)^{3/4}}{32 b^2 d^3}-\frac {(9 b c+7 a d) (a+b x)^{5/4} (c+d x)^{3/4}}{24 b^2 d^2}+\frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d}-\frac {(b c-a d) \left (15 b^2 c^2+10 a b c d+7 a^2 d^2\right ) \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{11/4} d^{13/4}}-\frac {(b c-a d) \left (15 b^2 c^2+10 a b c d+7 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{11/4} d^{13/4}} \]

[Out]

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^3-1/24*(7*a*d+9*b*c)*(b*x+a)^(5/4)*(d
*x+c)^(3/4)/b^2/d^2+1/3*x*(b*x+a)^(5/4)*(d*x+c)^(3/4)/b/d-1/64*(-a*d+b*c)*(7*a^2*d^2+10*a*b*c*d+15*b^2*c^2)*ar
ctan(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)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {92, 81, 52, 65, 246, 218, 214, 211} \[ \int \frac {x^2 \sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx=-\frac {(b c-a d) \left (7 a^2 d^2+10 a b c d+15 b^2 c^2\right ) \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{11/4} d^{13/4}}-\frac {(b c-a d) \left (7 a^2 d^2+10 a b c d+15 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{11/4} d^{13/4}}+\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} \left (7 a^2 d^2+10 a b c d+15 b^2 c^2\right )}{32 b^2 d^3}-\frac {(a+b x)^{5/4} (c+d x)^{3/4} (7 a d+9 b c)}{24 b^2 d^2}+\frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d} \]

[In]

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

[Out]

((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))/(32*b^2*d^3) - ((9*b*c + 7*a*d)*(a + b
*x)^(5/4)*(c + d*x)^(3/4))/(24*b^2*d^2) + (x*(a + b*x)^(5/4)*(c + d*x)^(3/4))/(3*b*d) - ((b*c - a*d)*(15*b^2*c
^2 + 10*a*b*c*d + 7*a^2*d^2)*ArcTan[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(64*b^(11/4)*d^(13/4
)) - ((b*c - a*d)*(15*b^2*c^2 + 10*a*b*c*d + 7*a^2*d^2)*ArcTanh[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(
1/4))])/(64*b^(11/4)*d^(13/4))

Rule 52

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] + Dist[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] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(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 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[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 211

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

Rule 214

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 218

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

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[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]

Rubi steps \begin{align*} \text {integral}& = \frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d}+\frac {\int \frac {\sqrt [4]{a+b x} \left (-a c-\frac {1}{4} (9 b c+7 a d) x\right )}{\sqrt [4]{c+d x}} \, dx}{3 b d} \\ & = -\frac {(9 b c+7 a d) (a+b x)^{5/4} (c+d x)^{3/4}}{24 b^2 d^2}+\frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d}+\frac {\left (15 b^2 c^2+10 a b c d+7 a^2 d^2\right ) \int \frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx}{32 b^2 d^2} \\ & = \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 {(9 b c+7 a d) (a+b x)^{5/4} (c+d x)^{3/4}}{24 b^2 d^2}+\frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d}-\frac {\left ((b c-a d) \left (15 b^2 c^2+10 a b c d+7 a^2 d^2\right )\right ) \int \frac {1}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx}{128 b^2 d^3} \\ & = \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 {(9 b c+7 a d) (a+b x)^{5/4} (c+d x)^{3/4}}{24 b^2 d^2}+\frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d}-\frac {\left ((b c-a d) \left (15 b^2 c^2+10 a b c d+7 a^2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{c-\frac {a d}{b}+\frac {d x^4}{b}}} \, dx,x,\sqrt [4]{a+b x}\right )}{32 b^3 d^3} \\ & = \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 {(9 b c+7 a d) (a+b x)^{5/4} (c+d x)^{3/4}}{24 b^2 d^2}+\frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d}-\frac {\left ((b c-a d) \left (15 b^2 c^2+10 a b c d+7 a^2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{1-\frac {d x^4}{b}} \, dx,x,\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{32 b^3 d^3} \\ & = \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 {(9 b c+7 a d) (a+b x)^{5/4} (c+d x)^{3/4}}{24 b^2 d^2}+\frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d}-\frac {\left ((b c-a d) \left (15 b^2 c^2+10 a b c d+7 a^2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b}-\sqrt {d} x^2} \, dx,x,\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{64 b^{5/2} d^3}-\frac {\left ((b c-a d) \left (15 b^2 c^2+10 a b c d+7 a^2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b}+\sqrt {d} x^2} \, dx,x,\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{64 b^{5/2} d^3} \\ & = \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 {(9 b c+7 a d) (a+b x)^{5/4} (c+d x)^{3/4}}{24 b^2 d^2}+\frac {x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d}-\frac {(b c-a d) \left (15 b^2 c^2+10 a b c d+7 a^2 d^2\right ) \tan ^{-1}\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 ) \tanh ^{-1}\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}} \\ \end{align*}

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 \[ \int \frac {x^2 \sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx=\frac {(a+b x)^{5/4} \left (-5 b (c+d x) (9 b c+7 a d-8 b d x)+3 \left (15 b^2 c^2+10 a b c d+7 a^2 d^2\right ) \sqrt [4]{\frac {b (c+d x)}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {5}{4},\frac {9}{4},\frac {d (a+b x)}{-b c+a d}\right )\right )}{120 b^3 d^2 \sqrt [4]{c+d x}} \]

[In]

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

[Out]

((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)*Hypergeometric2F1[1/4, 5/4, 9/4, (d*(a + b*x))/(-(b*c) + a*d)]))/(120*b^3*d^2*(c + d*
x)^(1/4))

Maple [F]

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

[In]

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

[Out]

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 = 0.28 (sec) , antiderivative size = 1814, normalized size of antiderivative = 6.77 \[ \int \frac {x^2 \sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-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 + 9
3775*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 + 9377
5*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 + 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*I*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) + (I*b^3*d^4*x + I*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*I*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 + 24
01*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) + (-I*b^3*d^4*x - I*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)) - 4*(32*b^2*d^2*x^2 + 45*b^2*c^2 - 6*a*b*c*d - 7*a^2*d^2 - 4*(9
*b^2*c*d - a*b*d^2)*x)*(b*x + a)^(1/4)*(d*x + c)^(3/4))/(b^2*d^3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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