\(\int \frac {x^3}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx\) [900]

Optimal result

Integrand size = 22, antiderivative size = 259 \[ \int \frac {x^3}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\frac {x^2 \sqrt [4]{a+b x} (c+d x)^{3/4}}{3 b d}+\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} \left (45 b^2 c^2+54 a b c d+77 a^2 d^2-4 b d (9 b c+11 a d) x\right )}{96 b^3 d^3}-\frac {\left (15 b^3 c^3+15 a b^2 c^2 d+21 a^2 b c d^2+77 a^3 d^3\right ) \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{15/4} d^{13/4}}-\frac {\left (15 b^3 c^3+15 a b^2 c^2 d+21 a^2 b c d^2+77 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{15/4} d^{13/4}} \]

[Out]

1/3*x^2*(b*x+a)^(1/4)*(d*x+c)^(3/4)/b/d+1/96*(b*x+a)^(1/4)*(d*x+c)^(3/4)*(45*b^2*c^2+54*a*b*c*d+77*a^2*d^2-4*b
*d*(11*a*d+9*b*c)*x)/b^3/d^3-1/64*(77*a^3*d^3+21*a^2*b*c*d^2+15*a*b^2*c^2*d+15*b^3*c^3)*arctan(d^(1/4)*(b*x+a)
^(1/4)/b^(1/4)/(d*x+c)^(1/4))/b^(15/4)/d^(13/4)-1/64*(77*a^3*d^3+21*a^2*b*c*d^2+15*a*b^2*c^2*d+15*b^3*c^3)*arc
tanh(d^(1/4)*(b*x+a)^(1/4)/b^(1/4)/(d*x+c)^(1/4))/b^(15/4)/d^(13/4)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {102, 152, 65, 246, 218, 214, 211} \[ \int \frac {x^3}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} \left (77 a^2 d^2-4 b d x (11 a d+9 b c)+54 a b c d+45 b^2 c^2\right )}{96 b^3 d^3}-\frac {\left (77 a^3 d^3+21 a^2 b c d^2+15 a b^2 c^2 d+15 b^3 c^3\right ) \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{15/4} d^{13/4}}-\frac {\left (77 a^3 d^3+21 a^2 b c d^2+15 a b^2 c^2 d+15 b^3 c^3\right ) \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{15/4} d^{13/4}}+\frac {x^2 \sqrt [4]{a+b x} (c+d x)^{3/4}}{3 b d} \]

[In]

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

[Out]

(x^2*(a + b*x)^(1/4)*(c + d*x)^(3/4))/(3*b*d) + ((a + b*x)^(1/4)*(c + d*x)^(3/4)*(45*b^2*c^2 + 54*a*b*c*d + 77
*a^2*d^2 - 4*b*d*(9*b*c + 11*a*d)*x))/(96*b^3*d^3) - ((15*b^3*c^3 + 15*a*b^2*c^2*d + 21*a^2*b*c*d^2 + 77*a^3*d
^3)*ArcTan[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(64*b^(15/4)*d^(13/4)) - ((15*b^3*c^3 + 15*a*
b^2*c^2*d + 21*a^2*b*c*d^2 + 77*a^3*d^3)*ArcTanh[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(64*b^(
15/4)*d^(13/4))

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 102

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

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)
^(m + 1)*((c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d
*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1
)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)
^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 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^2 \sqrt [4]{a+b x} (c+d x)^{3/4}}{3 b d}+\frac {\int \frac {x \left (-2 a c+\frac {1}{4} (-9 b c-11 a d) x\right )}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx}{3 b d} \\ & = \frac {x^2 \sqrt [4]{a+b x} (c+d x)^{3/4}}{3 b d}+\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} \left (45 b^2 c^2+54 a b c d+77 a^2 d^2-4 b d (9 b c+11 a d) x\right )}{96 b^3 d^3}-\frac {\left (15 b^3 c^3+15 a b^2 c^2 d+21 a^2 b c d^2+77 a^3 d^3\right ) \int \frac {1}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx}{128 b^3 d^3} \\ & = \frac {x^2 \sqrt [4]{a+b x} (c+d x)^{3/4}}{3 b d}+\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} \left (45 b^2 c^2+54 a b c d+77 a^2 d^2-4 b d (9 b c+11 a d) x\right )}{96 b^3 d^3}-\frac {\left (15 b^3 c^3+15 a b^2 c^2 d+21 a^2 b c d^2+77 a^3 d^3\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^4 d^3} \\ & = \frac {x^2 \sqrt [4]{a+b x} (c+d x)^{3/4}}{3 b d}+\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} \left (45 b^2 c^2+54 a b c d+77 a^2 d^2-4 b d (9 b c+11 a d) x\right )}{96 b^3 d^3}-\frac {\left (15 b^3 c^3+15 a b^2 c^2 d+21 a^2 b c d^2+77 a^3 d^3\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^4 d^3} \\ & = \frac {x^2 \sqrt [4]{a+b x} (c+d x)^{3/4}}{3 b d}+\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} \left (45 b^2 c^2+54 a b c d+77 a^2 d^2-4 b d (9 b c+11 a d) x\right )}{96 b^3 d^3}-\frac {\left (15 b^3 c^3+15 a b^2 c^2 d+21 a^2 b c d^2+77 a^3 d^3\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^{7/2} d^3}-\frac {\left (15 b^3 c^3+15 a b^2 c^2 d+21 a^2 b c d^2+77 a^3 d^3\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^{7/2} d^3} \\ & = \frac {x^2 \sqrt [4]{a+b x} (c+d x)^{3/4}}{3 b d}+\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} \left (45 b^2 c^2+54 a b c d+77 a^2 d^2-4 b d (9 b c+11 a d) x\right )}{96 b^3 d^3}-\frac {\left (15 b^3 c^3+15 a b^2 c^2 d+21 a^2 b c d^2+77 a^3 d^3\right ) \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{15/4} d^{13/4}}-\frac {\left (15 b^3 c^3+15 a b^2 c^2 d+21 a^2 b c d^2+77 a^3 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{15/4} d^{13/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\frac {2 b^{3/4} \sqrt [4]{d} \sqrt [4]{a+b x} (c+d x)^{3/4} \left (77 a^2 d^2+2 a b d (27 c-22 d x)+b^2 \left (45 c^2-36 c d x+32 d^2 x^2\right )\right )-3 \left (15 b^3 c^3+15 a b^2 c^2 d+21 a^2 b c d^2+77 a^3 d^3\right ) \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )-3 \left (15 b^3 c^3+15 a b^2 c^2 d+21 a^2 b c d^2+77 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{192 b^{15/4} d^{13/4}} \]

[In]

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

[Out]

(2*b^(3/4)*d^(1/4)*(a + b*x)^(1/4)*(c + d*x)^(3/4)*(77*a^2*d^2 + 2*a*b*d*(27*c - 22*d*x) + b^2*(45*c^2 - 36*c*
d*x + 32*d^2*x^2)) - 3*(15*b^3*c^3 + 15*a*b^2*c^2*d + 21*a^2*b*c*d^2 + 77*a^3*d^3)*ArcTan[(d^(1/4)*(a + b*x)^(
1/4))/(b^(1/4)*(c + d*x)^(1/4))] - 3*(15*b^3*c^3 + 15*a*b^2*c^2*d + 21*a^2*b*c*d^2 + 77*a^3*d^3)*ArcTanh[(d^(1
/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(192*b^(15/4)*d^(13/4))

Maple [F]

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

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 1812, normalized size of antiderivative = 7.00 \[ \int \frac {x^3}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/384*(3*b^3*d^3*((50625*b^12*c^12 + 202500*a*b^11*c^11*d + 587250*a^2*b^10*c^10*d^2 + 2092500*a^3*b^9*c^9*d^
3 + 4614975*a^4*b^8*c^8*d^4 + 8958600*a^5*b^7*c^7*d^5 + 18926460*a^6*b^6*c^6*d^6 + 27042120*a^7*b^5*c^5*d^7 +
36722511*a^8*b^4*c^4*d^8 + 52655988*a^9*b^3*c^3*d^9 + 43080114*a^10*b^2*c^2*d^10 + 38348772*a^11*b*c*d^11 + 35
153041*a^12*d^12)/(b^15*d^13))^(1/4)*log(((15*b^3*c^3 + 15*a*b^2*c^2*d + 21*a^2*b*c*d^2 + 77*a^3*d^3)*(b*x + a
)^(1/4)*(d*x + c)^(3/4) + (b^4*d^4*x + b^4*c*d^3)*((50625*b^12*c^12 + 202500*a*b^11*c^11*d + 587250*a^2*b^10*c
^10*d^2 + 2092500*a^3*b^9*c^9*d^3 + 4614975*a^4*b^8*c^8*d^4 + 8958600*a^5*b^7*c^7*d^5 + 18926460*a^6*b^6*c^6*d
^6 + 27042120*a^7*b^5*c^5*d^7 + 36722511*a^8*b^4*c^4*d^8 + 52655988*a^9*b^3*c^3*d^9 + 43080114*a^10*b^2*c^2*d^
10 + 38348772*a^11*b*c*d^11 + 35153041*a^12*d^12)/(b^15*d^13))^(1/4))/(d*x + c)) - 3*b^3*d^3*((50625*b^12*c^12
+ 202500*a*b^11*c^11*d + 587250*a^2*b^10*c^10*d^2 + 2092500*a^3*b^9*c^9*d^3 + 4614975*a^4*b^8*c^8*d^4 + 89586
00*a^5*b^7*c^7*d^5 + 18926460*a^6*b^6*c^6*d^6 + 27042120*a^7*b^5*c^5*d^7 + 36722511*a^8*b^4*c^4*d^8 + 52655988
*a^9*b^3*c^3*d^9 + 43080114*a^10*b^2*c^2*d^10 + 38348772*a^11*b*c*d^11 + 35153041*a^12*d^12)/(b^15*d^13))^(1/4
)*log(((15*b^3*c^3 + 15*a*b^2*c^2*d + 21*a^2*b*c*d^2 + 77*a^3*d^3)*(b*x + a)^(1/4)*(d*x + c)^(3/4) - (b^4*d^4*
x + b^4*c*d^3)*((50625*b^12*c^12 + 202500*a*b^11*c^11*d + 587250*a^2*b^10*c^10*d^2 + 2092500*a^3*b^9*c^9*d^3 +
4614975*a^4*b^8*c^8*d^4 + 8958600*a^5*b^7*c^7*d^5 + 18926460*a^6*b^6*c^6*d^6 + 27042120*a^7*b^5*c^5*d^7 + 367
22511*a^8*b^4*c^4*d^8 + 52655988*a^9*b^3*c^3*d^9 + 43080114*a^10*b^2*c^2*d^10 + 38348772*a^11*b*c*d^11 + 35153
041*a^12*d^12)/(b^15*d^13))^(1/4))/(d*x + c)) - 3*I*b^3*d^3*((50625*b^12*c^12 + 202500*a*b^11*c^11*d + 587250*
a^2*b^10*c^10*d^2 + 2092500*a^3*b^9*c^9*d^3 + 4614975*a^4*b^8*c^8*d^4 + 8958600*a^5*b^7*c^7*d^5 + 18926460*a^6
*b^6*c^6*d^6 + 27042120*a^7*b^5*c^5*d^7 + 36722511*a^8*b^4*c^4*d^8 + 52655988*a^9*b^3*c^3*d^9 + 43080114*a^10*
b^2*c^2*d^10 + 38348772*a^11*b*c*d^11 + 35153041*a^12*d^12)/(b^15*d^13))^(1/4)*log(((15*b^3*c^3 + 15*a*b^2*c^2
*d + 21*a^2*b*c*d^2 + 77*a^3*d^3)*(b*x + a)^(1/4)*(d*x + c)^(3/4) - (I*b^4*d^4*x + I*b^4*c*d^3)*((50625*b^12*c
^12 + 202500*a*b^11*c^11*d + 587250*a^2*b^10*c^10*d^2 + 2092500*a^3*b^9*c^9*d^3 + 4614975*a^4*b^8*c^8*d^4 + 89
58600*a^5*b^7*c^7*d^5 + 18926460*a^6*b^6*c^6*d^6 + 27042120*a^7*b^5*c^5*d^7 + 36722511*a^8*b^4*c^4*d^8 + 52655
988*a^9*b^3*c^3*d^9 + 43080114*a^10*b^2*c^2*d^10 + 38348772*a^11*b*c*d^11 + 35153041*a^12*d^12)/(b^15*d^13))^(
1/4))/(d*x + c)) + 3*I*b^3*d^3*((50625*b^12*c^12 + 202500*a*b^11*c^11*d + 587250*a^2*b^10*c^10*d^2 + 2092500*a
^3*b^9*c^9*d^3 + 4614975*a^4*b^8*c^8*d^4 + 8958600*a^5*b^7*c^7*d^5 + 18926460*a^6*b^6*c^6*d^6 + 27042120*a^7*b
^5*c^5*d^7 + 36722511*a^8*b^4*c^4*d^8 + 52655988*a^9*b^3*c^3*d^9 + 43080114*a^10*b^2*c^2*d^10 + 38348772*a^11*
b*c*d^11 + 35153041*a^12*d^12)/(b^15*d^13))^(1/4)*log(((15*b^3*c^3 + 15*a*b^2*c^2*d + 21*a^2*b*c*d^2 + 77*a^3*
d^3)*(b*x + a)^(1/4)*(d*x + c)^(3/4) - (-I*b^4*d^4*x - I*b^4*c*d^3)*((50625*b^12*c^12 + 202500*a*b^11*c^11*d +
587250*a^2*b^10*c^10*d^2 + 2092500*a^3*b^9*c^9*d^3 + 4614975*a^4*b^8*c^8*d^4 + 8958600*a^5*b^7*c^7*d^5 + 1892
6460*a^6*b^6*c^6*d^6 + 27042120*a^7*b^5*c^5*d^7 + 36722511*a^8*b^4*c^4*d^8 + 52655988*a^9*b^3*c^3*d^9 + 430801
14*a^10*b^2*c^2*d^10 + 38348772*a^11*b*c*d^11 + 35153041*a^12*d^12)/(b^15*d^13))^(1/4))/(d*x + c)) - 4*(32*b^2
*d^2*x^2 + 45*b^2*c^2 + 54*a*b*c*d + 77*a^2*d^2 - 4*(9*b^2*c*d + 11*a*b*d^2)*x)*(b*x + a)^(1/4)*(d*x + c)^(3/4
))/(b^3*d^3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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