\(\int \frac {(a+b x)^{5/4}}{\sqrt [4]{c+d x}} \, dx\) [588]

Optimal result

Integrand size = 19, antiderivative size = 167 \[ \int \frac {(a+b x)^{5/4}}{\sqrt [4]{c+d x}} \, dx=-\frac {5 (b c-a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 d^2}+\frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 d}+\frac {5 (b c-a d)^2 \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{3/4} d^{9/4}}+\frac {5 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{3/4} d^{9/4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b x)^{5/4}}{\sqrt [4]{c+d x}} \, dx=\frac {2 \sqrt [4]{d} \sqrt [4]{a+b x} (c+d x)^{3/4} (-5 b c+9 a d+4 b d x)-\frac {5 (b c-a d)^2 \arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{b^{3/4}}+\frac {5 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{b^{3/4}}}{16 d^{9/4}} \] Input:

Integrate[(a + b*x)^(5/4)/(c + d*x)^(1/4),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.18, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {60, 60, 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.

\(\displaystyle \int \frac {(a+b x)^{5/4}}{\sqrt [4]{c+d x}} \, dx\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 d}-\frac {5 (b c-a d) \int \frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}dx}{8 d}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 d}-\frac {5 (b c-a d) \left (\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{d}-\frac {(b c-a d) \int \frac {1}{(a+b x)^{3/4} \sqrt [4]{c+d x}}dx}{4 d}\right )}{8 d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 d}-\frac {5 (b c-a d) \left (\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}d\sqrt [4]{a+b x}}{b d}\right )}{8 d}\)

\(\Big \downarrow \) 770

\(\displaystyle \frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 d}-\frac {5 (b c-a d) \left (\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{d}-\frac {(b c-a d) \int \frac {1}{1-\frac {d (a+b x)}{b}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}}{b d}\right )}{8 d}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 d}-\frac {5 (b c-a d) \left (\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{d}-\frac {(b c-a d) \left (\frac {1}{2} \sqrt {b} \int \frac {1}{\sqrt {b}-\sqrt {d} \sqrt {a+b x}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}+\frac {1}{2} \sqrt {b} \int \frac {1}{\sqrt {b}+\sqrt {d} \sqrt {a+b x}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}\right )}{b d}\right )}{8 d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 d}-\frac {5 (b c-a d) \left (\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{d}-\frac {(b c-a d) \left (\frac {1}{2} \sqrt {b} \int \frac {1}{\sqrt {b}-\sqrt {d} \sqrt {a+b x}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c-\frac {a d}{b}+\frac {d (a+b x)}{b}}}+\frac {\sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{2 \sqrt [4]{d}}\right )}{b d}\right )}{8 d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 d}-\frac {5 (b c-a d) \left (\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{d}-\frac {(b c-a d) \left (\frac {\sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{2 \sqrt [4]{d}}+\frac {\sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{\frac {d (a+b x)}{b}-\frac {a d}{b}+c}}\right )}{2 \sqrt [4]{d}}\right )}{b d}\right )}{8 d}\)

Input:

Int[(a + b*x)^(5/4)/(c + d*x)^(1/4),x]
 

Output:

((a + b*x)^(5/4)*(c + d*x)^(3/4))/(2*d) - (5*(b*c - a*d)*(((a + b*x)^(1/4) 
*(c + d*x)^(3/4))/d - ((b*c - a*d)*((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))))/(b*d)))/(8*d)
 

Defintions of rubi rules used

rule 60

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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, 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 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]

\[\int \frac {\left (b x +a \right )^{\frac {5}{4}}}{\left (x d +c \right )^{\frac {1}{4}}}d x\]

Input:

int((b*x+a)^(5/4)/(d*x+c)^(1/4),x)
 

Output:

int((b*x+a)^(5/4)/(d*x+c)^(1/4),x)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 1219, normalized size of antiderivative = 7.30 \[ \int \frac {(a+b x)^{5/4}}{\sqrt [4]{c+d x}} \, dx=\text {Too large to display} \] Input:

integrate((b*x+a)^(5/4)/(d*x+c)^(1/4),x, algorithm="fricas")
 

Output:

1/32*(5*d^2*((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^ 
5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a 
^7*b*c*d^7 + a^8*d^8)/(b^3*d^9))^(1/4)*log(5*((b^2*c^2 - 2*a*b*c*d + a^2*d 
^2)*(b*x + a)^(1/4)*(d*x + c)^(3/4) + (b*d^3*x + b*c*d^2)*((b^8*c^8 - 8*a* 
b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 
 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b^3*d 
^9))^(1/4))/(d*x + c)) - 5*d^2*((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6* 
d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^ 
6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b^3*d^9))^(1/4)*log(5*((b^2*c^2 
- 2*a*b*c*d + a^2*d^2)*(b*x + a)^(1/4)*(d*x + c)^(3/4) - (b*d^3*x + b*c*d^ 
2)*((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 7 
0*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^ 
7 + a^8*d^8)/(b^3*d^9))^(1/4))/(d*x + c)) - 5*I*d^2*((b^8*c^8 - 8*a*b^7*c^ 
7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^ 
5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b^3*d^9))^( 
1/4)*log(5*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(b*x + a)^(1/4)*(d*x + c)^(3/4 
) - (I*b*d^3*x + I*b*c*d^2)*((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 
 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b 
^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b^3*d^9))^(1/4))/(d*x + c)) + 5*I*d 
^2*((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 ...
 

Sympy [F]

\[ \int \frac {(a+b x)^{5/4}}{\sqrt [4]{c+d x}} \, dx=\int \frac {\left (a + b x\right )^{\frac {5}{4}}}{\sqrt [4]{c + d x}}\, dx \] Input:

integrate((b*x+a)**(5/4)/(d*x+c)**(1/4),x)
 

Output:

Integral((a + b*x)**(5/4)/(c + d*x)**(1/4), x)
 

Maxima [F]

\[ \int \frac {(a+b x)^{5/4}}{\sqrt [4]{c+d x}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {5}{4}}}{{\left (d x + c\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate((b*x+a)^(5/4)/(d*x+c)^(1/4),x, algorithm="maxima")
 

Output:

integrate((b*x + a)^(5/4)/(d*x + c)^(1/4), x)
 

Giac [F]

\[ \int \frac {(a+b x)^{5/4}}{\sqrt [4]{c+d x}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {5}{4}}}{{\left (d x + c\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate((b*x+a)^(5/4)/(d*x+c)^(1/4),x, algorithm="giac")
 

Output:

integrate((b*x + a)^(5/4)/(d*x + c)^(1/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{5/4}}{\sqrt [4]{c+d x}} \, dx=\int \frac {{\left (a+b\,x\right )}^{5/4}}{{\left (c+d\,x\right )}^{1/4}} \,d x \] Input:

int((a + b*x)^(5/4)/(c + d*x)^(1/4),x)
 

Output:

int((a + b*x)^(5/4)/(c + d*x)^(1/4), x)
 

Reduce [F]

\[ \int \frac {(a+b x)^{5/4}}{\sqrt [4]{c+d x}} \, dx=\left (\int \frac {\left (b x +a \right )^{\frac {1}{4}}}{\left (d x +c \right )^{\frac {1}{4}}}d x \right ) a +\left (\int \frac {\left (b x +a \right )^{\frac {1}{4}} x}{\left (d x +c \right )^{\frac {1}{4}}}d x \right ) b \] Input:

int((b*x+a)^(5/4)/(d*x+c)^(1/4),x)
 

Output:

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