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

Optimal. Leaf size=268 \[ \frac{\sqrt [4]{a+b x} (c+d x)^{3/4} \left (7 a^2 d^2+10 a b c d+15 b^2 c^2\right )}{32 b^2 d^3}-\frac{(b c-a d) \left (7 a^2 d^2+10 a b c d+15 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{11/4} d^{13/4}}-\frac{(b c-a d) \left (7 a^2 d^2+10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{11/4} d^{13/4}}-\frac{(a+b x)^{5/4} (c+d x)^{3/4} (7 a d+9 b c)}{24 b^2 d^2}+\frac{x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d} \]

[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))

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Rubi [A]  time = 0.249621, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {90, 80, 50, 63, 240, 212, 208, 205} \[ \frac{\sqrt [4]{a+b x} (c+d x)^{3/4} \left (7 a^2 d^2+10 a b c d+15 b^2 c^2\right )}{32 b^2 d^3}-\frac{(b c-a d) \left (7 a^2 d^2+10 a b c d+15 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{11/4} d^{13/4}}-\frac{(b c-a d) \left (7 a^2 d^2+10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{11/4} d^{13/4}}-\frac{(a+b x)^{5/4} (c+d x)^{3/4} (7 a d+9 b c)}{24 b^2 d^2}+\frac{x (a+b x)^{5/4} (c+d x)^{3/4}}{3 b d} \]

Antiderivative was successfully verified.

[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 90

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 80

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 50

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 63

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 240

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]

Rule 212

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] &&
!GtQ[a/b, 0]

Rule 208

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

Rule 205

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

Rubi steps

\begin{align*} \int \frac{x^2 \sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx &=\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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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]  time = 0.139518, size = 124, normalized size = 0.46 \[ \frac{(a+b x)^{5/4} \left (3 \left (7 a^2 d^2+10 a b c d+15 b^2 c^2\right ) \sqrt [4]{\frac{b (c+d x)}{b c-a d}} \, _2F_1\left (\frac{1}{4},\frac{5}{4};\frac{9}{4};\frac{d (a+b x)}{a d-b c}\right )-5 b (c+d x) (7 a d+9 b c-8 b d x)\right )}{120 b^3 d^2 \sqrt [4]{c+d x}} \]

Antiderivative was successfully verified.

[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))

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Maple [F]  time = 0.021, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2}\sqrt [4]{bx+a}{\frac{1}{\sqrt [4]{dx+c}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{\frac{1}{4}} x^{2}}{{\left (d x + c\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [B]  time = 3.37122, size = 5284, normalized size = 19.72 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*(12*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)*arctan(((15*b^11*c^3*d^10 - 5*a*b^10*c^2*d^11 - 3*a^2*b^9*c*d^12 - 7*a^3*b^8*d^13)*(b*x + a)^(1/4)*(d*x +
c)^(3/4)*((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))^(3/4) + (b
^8*d^11*x + b^8*c*d^10)*sqrt(((225*b^6*c^6 - 150*a*b^5*c^5*d - 65*a^2*b^4*c^4*d^2 - 180*a^3*b^3*c^3*d^3 + 79*a
^4*b^2*c^2*d^4 + 42*a^5*b*c*d^5 + 49*a^6*d^6)*sqrt(b*x + a)*sqrt(d*x + c) + (b^6*d^7*x + b^6*c*d^6)*sqrt((5062
5*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 + 1
8600*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)))/(d*x + c))*((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))^(3/4))/(50625*b^12*c^13 - 67500*a*
b^11*c^12*d - 6750*a^2*b^10*c^11*d^2 - 61500*a^3*b^9*c^10*d^3 + 93775*a^4*b^8*c^9*d^4 + 18600*a^5*b^7*c^8*d^5
+ 31580*a^6*b^6*c^7*d^6 - 48600*a^7*b^5*c^6*d^7 - 15249*a^8*b^4*c^5*d^8 - 11004*a^9*b^3*c^4*d^9 + 9506*a^10*b^
2*c^3*d^10 + 4116*a^11*b*c^2*d^11 + 2401*a^12*c*d^12 + (50625*b^12*c^12*d - 67500*a*b^11*c^11*d^2 - 6750*a^2*b
^10*c^10*d^3 - 61500*a^3*b^9*c^9*d^4 + 93775*a^4*b^8*c^8*d^5 + 18600*a^5*b^7*c^7*d^6 + 31580*a^6*b^6*c^6*d^7 -
48600*a^7*b^5*c^5*d^8 - 15249*a^8*b^4*c^4*d^9 - 11004*a^9*b^3*c^3*d^10 + 9506*a^10*b^2*c^2*d^11 + 4116*a^11*b
*c*d^12 + 2401*a^12*d^13)*x)) + 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 - 6
1500*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 - 6150
0*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 + 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 + 240
1*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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \sqrt [4]{a + b x}}{\sqrt [4]{c + d x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out