\(\int \frac {x^2}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx\) [901]

Optimal result

Integrand size = 22, antiderivative size = 201 \[ \int \frac {x^2}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=-\frac {(5 b c+7 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 b^2 d^2}+\frac {x \sqrt [4]{a+b x} (c+d x)^{3/4}}{2 b d}+\frac {\left (5 b^2 c^2+6 a b c d+21 a^2 d^2\right ) \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{11/4} d^{9/4}}+\frac {\left (5 b^2 c^2+6 a b c d+21 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{11/4} d^{9/4}} \]

[Out]

-1/8*(7*a*d+5*b*c)*(b*x+a)^(1/4)*(d*x+c)^(3/4)/b^2/d^2+1/2*x*(b*x+a)^(1/4)*(d*x+c)^(3/4)/b/d+1/16*(21*a^2*d^2+
6*a*b*c*d+5*b^2*c^2)*arctan(d^(1/4)*(b*x+a)^(1/4)/b^(1/4)/(d*x+c)^(1/4))/b^(11/4)/d^(9/4)+1/16*(21*a^2*d^2+6*a
*b*c*d+5*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^(9/4)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 201, 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 = {92, 81, 65, 246, 218, 214, 211} \[ \int \frac {x^2}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\frac {\left (21 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{11/4} d^{9/4}}+\frac {\left (21 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{11/4} d^{9/4}}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (7 a d+5 b c)}{8 b^2 d^2}+\frac {x \sqrt [4]{a+b x} (c+d x)^{3/4}}{2 b d} \]

[In]

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

[Out]

-1/8*((5*b*c + 7*a*d)*(a + b*x)^(1/4)*(c + d*x)^(3/4))/(b^2*d^2) + (x*(a + b*x)^(1/4)*(c + d*x)^(3/4))/(2*b*d)
+ ((5*b^2*c^2 + 6*a*b*c*d + 21*a^2*d^2)*ArcTan[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(16*b^(1
1/4)*d^(9/4)) + ((5*b^2*c^2 + 6*a*b*c*d + 21*a^2*d^2)*ArcTanh[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/
4))])/(16*b^(11/4)*d^(9/4))

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 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 \sqrt [4]{a+b x} (c+d x)^{3/4}}{2 b d}+\frac {\int \frac {-a c-\frac {1}{4} (5 b c+7 a d) x}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx}{2 b d} \\ & = -\frac {(5 b c+7 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 b^2 d^2}+\frac {x \sqrt [4]{a+b x} (c+d x)^{3/4}}{2 b d}+\frac {\left (5 b^2 c^2+6 a b c d+21 a^2 d^2\right ) \int \frac {1}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx}{32 b^2 d^2} \\ & = -\frac {(5 b c+7 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 b^2 d^2}+\frac {x \sqrt [4]{a+b x} (c+d x)^{3/4}}{2 b d}+\frac {\left (5 b^2 c^2+6 a b c d+21 a^2 d^2\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 )}{8 b^3 d^2} \\ & = -\frac {(5 b c+7 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 b^2 d^2}+\frac {x \sqrt [4]{a+b x} (c+d x)^{3/4}}{2 b d}+\frac {\left (5 b^2 c^2+6 a b c d+21 a^2 d^2\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 )}{8 b^3 d^2} \\ & = -\frac {(5 b c+7 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 b^2 d^2}+\frac {x \sqrt [4]{a+b x} (c+d x)^{3/4}}{2 b d}+\frac {\left (5 b^2 c^2+6 a b c d+21 a^2 d^2\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 )}{16 b^{5/2} d^2}+\frac {\left (5 b^2 c^2+6 a b c d+21 a^2 d^2\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 )}{16 b^{5/2} d^2} \\ & = -\frac {(5 b c+7 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 b^2 d^2}+\frac {x \sqrt [4]{a+b x} (c+d x)^{3/4}}{2 b d}+\frac {\left (5 b^2 c^2+6 a b c d+21 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 )}{16 b^{11/4} d^{9/4}}+\frac {\left (5 b^2 c^2+6 a b c d+21 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 )}{16 b^{11/4} d^{9/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.83 \[ \int \frac {x^2}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\frac {2 b^{3/4} \sqrt [4]{d} \sqrt [4]{a+b x} (c+d x)^{3/4} (-5 b c-7 a d+4 b d x)+\left (5 b^2 c^2+6 a b c d+21 a^2 d^2\right ) \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )+\left (5 b^2 c^2+6 a b c d+21 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{11/4} d^{9/4}} \]

[In]

Integrate[x^2/((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)*(-5*b*c - 7*a*d + 4*b*d*x) + (5*b^2*c^2 + 6*a*b*c*d + 21*a^
2*d^2)*ArcTan[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))] + (5*b^2*c^2 + 6*a*b*c*d + 21*a^2*d^2)*ArcT
anh[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(16*b^(11/4)*d^(9/4))

Maple [F]

\[\int \frac {x^{2}}{\left (b x +a \right )^{\frac {3}{4}} \left (d x +c \right )^{\frac {1}{4}}}d x\]

[In]

int(x^2/(b*x+a)^(3/4)/(d*x+c)^(1/4),x)

[Out]

int(x^2/(b*x+a)^(3/4)/(d*x+c)^(1/4),x)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 1269, normalized size of antiderivative = 6.31 \[ \int \frac {x^2}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\text {Too large to display} \]

[In]

integrate(x^2/(b*x+a)^(3/4)/(d*x+c)^(1/4),x, algorithm="fricas")

[Out]

1/32*(b^2*d^2*((625*b^8*c^8 + 3000*a*b^7*c^7*d + 15900*a^2*b^6*c^6*d^2 + 42120*a^3*b^5*c^5*d^3 + 112806*a^4*b^
4*c^4*d^4 + 176904*a^5*b^3*c^3*d^5 + 280476*a^6*b^2*c^2*d^6 + 222264*a^7*b*c*d^7 + 194481*a^8*d^8)/(b^11*d^9))
^(1/4)*log(((5*b^2*c^2 + 6*a*b*c*d + 21*a^2*d^2)*(b*x + a)^(1/4)*(d*x + c)^(3/4) + (b^3*d^3*x + b^3*c*d^2)*((6
25*b^8*c^8 + 3000*a*b^7*c^7*d + 15900*a^2*b^6*c^6*d^2 + 42120*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4 + 17690
4*a^5*b^3*c^3*d^5 + 280476*a^6*b^2*c^2*d^6 + 222264*a^7*b*c*d^7 + 194481*a^8*d^8)/(b^11*d^9))^(1/4))/(d*x + c)
) - b^2*d^2*((625*b^8*c^8 + 3000*a*b^7*c^7*d + 15900*a^2*b^6*c^6*d^2 + 42120*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*
c^4*d^4 + 176904*a^5*b^3*c^3*d^5 + 280476*a^6*b^2*c^2*d^6 + 222264*a^7*b*c*d^7 + 194481*a^8*d^8)/(b^11*d^9))^(
1/4)*log(((5*b^2*c^2 + 6*a*b*c*d + 21*a^2*d^2)*(b*x + a)^(1/4)*(d*x + c)^(3/4) - (b^3*d^3*x + b^3*c*d^2)*((625
*b^8*c^8 + 3000*a*b^7*c^7*d + 15900*a^2*b^6*c^6*d^2 + 42120*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4 + 176904*
a^5*b^3*c^3*d^5 + 280476*a^6*b^2*c^2*d^6 + 222264*a^7*b*c*d^7 + 194481*a^8*d^8)/(b^11*d^9))^(1/4))/(d*x + c))
- I*b^2*d^2*((625*b^8*c^8 + 3000*a*b^7*c^7*d + 15900*a^2*b^6*c^6*d^2 + 42120*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*
c^4*d^4 + 176904*a^5*b^3*c^3*d^5 + 280476*a^6*b^2*c^2*d^6 + 222264*a^7*b*c*d^7 + 194481*a^8*d^8)/(b^11*d^9))^(
1/4)*log(((5*b^2*c^2 + 6*a*b*c*d + 21*a^2*d^2)*(b*x + a)^(1/4)*(d*x + c)^(3/4) - (I*b^3*d^3*x + I*b^3*c*d^2)*(
(625*b^8*c^8 + 3000*a*b^7*c^7*d + 15900*a^2*b^6*c^6*d^2 + 42120*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4 + 176
904*a^5*b^3*c^3*d^5 + 280476*a^6*b^2*c^2*d^6 + 222264*a^7*b*c*d^7 + 194481*a^8*d^8)/(b^11*d^9))^(1/4))/(d*x +
c)) + I*b^2*d^2*((625*b^8*c^8 + 3000*a*b^7*c^7*d + 15900*a^2*b^6*c^6*d^2 + 42120*a^3*b^5*c^5*d^3 + 112806*a^4*
b^4*c^4*d^4 + 176904*a^5*b^3*c^3*d^5 + 280476*a^6*b^2*c^2*d^6 + 222264*a^7*b*c*d^7 + 194481*a^8*d^8)/(b^11*d^9
))^(1/4)*log(((5*b^2*c^2 + 6*a*b*c*d + 21*a^2*d^2)*(b*x + a)^(1/4)*(d*x + c)^(3/4) - (-I*b^3*d^3*x - I*b^3*c*d
^2)*((625*b^8*c^8 + 3000*a*b^7*c^7*d + 15900*a^2*b^6*c^6*d^2 + 42120*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4
+ 176904*a^5*b^3*c^3*d^5 + 280476*a^6*b^2*c^2*d^6 + 222264*a^7*b*c*d^7 + 194481*a^8*d^8)/(b^11*d^9))^(1/4))/(d
*x + c)) + 4*(4*b*d*x - 5*b*c - 7*a*d)*(b*x + a)^(1/4)*(d*x + c)^(3/4))/(b^2*d^2)

Sympy [F]

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

[In]

integrate(x**2/(b*x+a)**(3/4)/(d*x+c)**(1/4),x)

[Out]

Integral(x**2/((a + b*x)**(3/4)*(c + d*x)**(1/4)), x)

Maxima [F]

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

[In]

integrate(x^2/(b*x+a)^(3/4)/(d*x+c)^(1/4),x, algorithm="maxima")

[Out]

integrate(x^2/((b*x + a)^(3/4)*(d*x + c)^(1/4)), x)

Giac [F]

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

[In]

integrate(x^2/(b*x+a)^(3/4)/(d*x+c)^(1/4),x, algorithm="giac")

[Out]

integrate(x^2/((b*x + a)^(3/4)*(d*x + c)^(1/4)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\int \frac {x^2}{{\left (a+b\,x\right )}^{3/4}\,{\left (c+d\,x\right )}^{1/4}} \,d x \]

[In]

int(x^2/((a + b*x)^(3/4)*(c + d*x)^(1/4)),x)

[Out]

int(x^2/((a + b*x)^(3/4)*(c + d*x)^(1/4)), x)