\(\int \frac {x^3}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx\) [510]

Optimal result

Integrand size = 22, antiderivative size = 278 \[ \int \frac {x^3}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\frac {3 \left (5 b^2 c^2+10 a b c d+17 a^2 d^2\right ) \sqrt [4]{a+b x} (c+d x)^{3/4}}{32 b^3 d^3}-\frac {3 (b c+3 a d) (a+b x)^{5/4} (c+d x)^{3/4}}{8 b^3 d^2}+\frac {(a+b x)^{9/4} (c+d x)^{3/4}}{3 b^3 d}-\frac {\left (15 b^3 c^3+15 a b^2 c^2 d+21 a^2 b c d^2+77 a^3 d^3\right ) \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{15/4} d^{13/4}}-\frac {\left (15 b^3 c^3+15 a b^2 c^2 d+21 a^2 b c d^2+77 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{15/4} d^{13/4}} \] Output:

3/32*(17*a^2*d^2+10*a*b*c*d+5*b^2*c^2)*(b*x+a)^(1/4)*(d*x+c)^(3/4)/b^3/d^3 
-3/8*(3*a*d+b*c)*(b*x+a)^(5/4)*(d*x+c)^(3/4)/b^3/d^2+1/3*(b*x+a)^(9/4)*(d* 
x+c)^(3/4)/b^3/d-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)*arctanh(d^(1/4)*( 
b*x+a)^(1/4)/b^(1/4)/(d*x+c)^(1/4))/b^(15/4)/d^(13/4)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.82 \[ \int \frac {x^3}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\frac {2 b^{3/4} \sqrt [4]{d} \sqrt [4]{a+b x} (c+d x)^{3/4} \left (77 a^2 d^2+2 a b d (27 c-22 d x)+b^2 \left (45 c^2-36 c d x+32 d^2 x^2\right )\right )-3 \left (15 b^3 c^3+15 a b^2 c^2 d+21 a^2 b c d^2+77 a^3 d^3\right ) \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )-3 \left (15 b^3 c^3+15 a b^2 c^2 d+21 a^2 b c d^2+77 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{192 b^{15/4} d^{13/4}} \] Input:

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

Output:

(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 + 1 
5*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))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 272, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {111, 27, 164, 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.

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

\(\Big \downarrow \) 111

\(\displaystyle \frac {\int -\frac {x (8 a c+(9 b c+11 a d) x)}{4 (a+b x)^{3/4} \sqrt [4]{c+d x}}dx}{3 b d}+\frac {x^2 \sqrt [4]{a+b x} (c+d x)^{3/4}}{3 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^2 \sqrt [4]{a+b x} (c+d x)^{3/4}}{3 b d}-\frac {\int \frac {x (8 a c+(9 b c+11 a d) x)}{(a+b x)^{3/4} \sqrt [4]{c+d x}}dx}{12 b d}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {x^2 \sqrt [4]{a+b x} (c+d x)^{3/4}}{3 b d}-\frac {\frac {3 \left (77 a^3 d^3+21 a^2 b c d^2+15 a b^2 c^2 d+15 b^3 c^3\right ) \int \frac {1}{(a+b x)^{3/4} \sqrt [4]{c+d x}}dx}{32 b^2 d^2}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} \left (77 a^2 d^2-4 b d x (11 a d+9 b c)+54 a b c d+45 b^2 c^2\right )}{8 b^2 d^2}}{12 b d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {x^2 \sqrt [4]{a+b x} (c+d x)^{3/4}}{3 b d}-\frac {\frac {3 \left (77 a^3 d^3+21 a^2 b c d^2+15 a b^2 c^2 d+15 b^3 c^3\right ) \int \frac {1}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}d\sqrt [4]{a+b x}}{8 b^3 d^2}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} \left (77 a^2 d^2-4 b d x (11 a d+9 b c)+54 a b c d+45 b^2 c^2\right )}{8 b^2 d^2}}{12 b d}\)

\(\Big \downarrow \) 770

\(\displaystyle \frac {x^2 \sqrt [4]{a+b x} (c+d x)^{3/4}}{3 b d}-\frac {\frac {3 \left (77 a^3 d^3+21 a^2 b c d^2+15 a b^2 c^2 d+15 b^3 c^3\right ) \int \frac {1}{1-\frac {d (a+b x)}{b}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}}{8 b^3 d^2}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} \left (77 a^2 d^2-4 b d x (11 a d+9 b c)+54 a b c d+45 b^2 c^2\right )}{8 b^2 d^2}}{12 b d}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {x^2 \sqrt [4]{a+b x} (c+d x)^{3/4}}{3 b d}-\frac {\frac {3 \left (77 a^3 d^3+21 a^2 b c d^2+15 a b^2 c^2 d+15 b^3 c^3\right ) \left (\frac {1}{2} \sqrt {b} \int \frac {1}{\sqrt {b}-\sqrt {d} \sqrt {a+b x}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}+\frac {1}{2} \sqrt {b} \int \frac {1}{\sqrt {b}+\sqrt {d} \sqrt {a+b x}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}\right )}{8 b^3 d^2}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} \left (77 a^2 d^2-4 b d x (11 a d+9 b c)+54 a b c d+45 b^2 c^2\right )}{8 b^2 d^2}}{12 b d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {x^2 \sqrt [4]{a+b x} (c+d x)^{3/4}}{3 b d}-\frac {\frac {3 \left (77 a^3 d^3+21 a^2 b c d^2+15 a b^2 c^2 d+15 b^3 c^3\right ) \left (\frac {1}{2} \sqrt {b} \int \frac {1}{\sqrt {b}-\sqrt {d} \sqrt {a+b x}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}+\frac {\sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{2 \sqrt [4]{d}}\right )}{8 b^3 d^2}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} \left (77 a^2 d^2-4 b d x (11 a d+9 b c)+54 a b c d+45 b^2 c^2\right )}{8 b^2 d^2}}{12 b d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {x^2 \sqrt [4]{a+b x} (c+d x)^{3/4}}{3 b d}-\frac {\frac {3 \left (77 a^3 d^3+21 a^2 b c d^2+15 a b^2 c^2 d+15 b^3 c^3\right ) \left (\frac {\sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{2 \sqrt [4]{d}}+\frac {\sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{2 \sqrt [4]{d}}\right )}{8 b^3 d^2}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} \left (77 a^2 d^2-4 b d x (11 a d+9 b c)+54 a b c d+45 b^2 c^2\right )}{8 b^2 d^2}}{12 b d}\)

Input:

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

Output:

(x^2*(a + b*x)^(1/4)*(c + d*x)^(3/4))/(3*b*d) - (-1/8*((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))/(b^2*d^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)*((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))))/( 
8*b^3*d^2))/(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 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 111

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1))   Int[(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 164

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ 
))*((g_.) + (h_.)*(x_)), x_] :> 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] + Simp[(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 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]

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

Input:

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

Output:

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 = 0.19 (sec) , antiderivative size = 1812, normalized size of antiderivative = 6.52 \[ \int \frac {x^3}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\text {Too large to display} \] Input:

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

Output:

-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 + 36 
722511*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 + 461497 
5*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 + 2 
7042120*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 + 20250 
0*a*b^11*c^11*d + 587250*a^2*b^10*c^10*d^2 + 2092500*a^3*b^9*c^9*d^3 + 461 
4975*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) - (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 + 2 
092500*a^3*b^9*c^9*d^3 + 4614975*a^4*b^8*c^8*d^4 + 8958600*a^5*b^7*c^7*...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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