\(\int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx\) [576]

Optimal result

Integrand size = 19, antiderivative size = 134 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx=-\frac {4 d \sqrt [4]{c+d x}}{b^2 \sqrt [4]{a+b x}}-\frac {4 (c+d x)^{5/4}}{5 b (a+b x)^{5/4}}-\frac {2 d^{5/4} \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{9/4}}+\frac {2 d^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{b^{9/4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.92 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx=-\frac {4 \sqrt [4]{c+d x} (b c+5 a d+6 b d x)}{5 b^2 (a+b x)^{5/4}}+\frac {2 d^{5/4} \arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{b^{9/4}}+\frac {2 d^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{b^{9/4}} \] Input:

Integrate[(c + d*x)^(5/4)/(a + b*x)^(9/4),x]
 

Output:

(-4*(c + d*x)^(1/4)*(b*c + 5*a*d + 6*b*d*x))/(5*b^2*(a + b*x)^(5/4)) + (2* 
d^(5/4)*ArcTan[(b^(1/4)*(c + d*x)^(1/4))/(d^(1/4)*(a + b*x)^(1/4))])/b^(9/ 
4) + (2*d^(5/4)*ArcTanh[(b^(1/4)*(c + d*x)^(1/4))/(d^(1/4)*(a + b*x)^(1/4) 
)])/b^(9/4)
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.33, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {57, 57, 73, 854, 27, 827, 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.

\(\displaystyle \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx\)

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 854

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

Int[(c + d*x)^(5/4)/(a + b*x)^(9/4),x]
 

Output:

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

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 57

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[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 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 854

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 
 1)/n)   Subst[Int[x^m/(1 - b*x^n)^(p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]
 

Maple [F]

\[\int \frac {\left (x d +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {9}{4}}}d x\]

Input:

int((d*x+c)^(5/4)/(b*x+a)^(9/4),x)
 

Output:

int((d*x+c)^(5/4)/(b*x+a)^(9/4),x)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 385, normalized size of antiderivative = 2.87 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx=\frac {5 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )} \left (\frac {d^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} d + {\left (b^{3} x + a b^{2}\right )} \left (\frac {d^{5}}{b^{9}}\right )^{\frac {1}{4}}}{b x + a}\right ) - 5 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )} \left (\frac {d^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} d - {\left (b^{3} x + a b^{2}\right )} \left (\frac {d^{5}}{b^{9}}\right )^{\frac {1}{4}}}{b x + a}\right ) - 5 \, {\left (-i \, b^{4} x^{2} - 2 i \, a b^{3} x - i \, a^{2} b^{2}\right )} \left (\frac {d^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} d + {\left (i \, b^{3} x + i \, a b^{2}\right )} \left (\frac {d^{5}}{b^{9}}\right )^{\frac {1}{4}}}{b x + a}\right ) - 5 \, {\left (i \, b^{4} x^{2} + 2 i \, a b^{3} x + i \, a^{2} b^{2}\right )} \left (\frac {d^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} d + {\left (-i \, b^{3} x - i \, a b^{2}\right )} \left (\frac {d^{5}}{b^{9}}\right )^{\frac {1}{4}}}{b x + a}\right ) - 4 \, {\left (6 \, b d x + b c + 5 \, a d\right )} {\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}}}{5 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \] Input:

integrate((d*x+c)^(5/4)/(b*x+a)^(9/4),x, algorithm="fricas")
 

Output:

1/5*(5*(b^4*x^2 + 2*a*b^3*x + a^2*b^2)*(d^5/b^9)^(1/4)*log(((b*x + a)^(3/4 
)*(d*x + c)^(1/4)*d + (b^3*x + a*b^2)*(d^5/b^9)^(1/4))/(b*x + a)) - 5*(b^4 
*x^2 + 2*a*b^3*x + a^2*b^2)*(d^5/b^9)^(1/4)*log(((b*x + a)^(3/4)*(d*x + c) 
^(1/4)*d - (b^3*x + a*b^2)*(d^5/b^9)^(1/4))/(b*x + a)) - 5*(-I*b^4*x^2 - 2 
*I*a*b^3*x - I*a^2*b^2)*(d^5/b^9)^(1/4)*log(((b*x + a)^(3/4)*(d*x + c)^(1/ 
4)*d + (I*b^3*x + I*a*b^2)*(d^5/b^9)^(1/4))/(b*x + a)) - 5*(I*b^4*x^2 + 2* 
I*a*b^3*x + I*a^2*b^2)*(d^5/b^9)^(1/4)*log(((b*x + a)^(3/4)*(d*x + c)^(1/4 
)*d + (-I*b^3*x - I*a*b^2)*(d^5/b^9)^(1/4))/(b*x + a)) - 4*(6*b*d*x + b*c 
+ 5*a*d)*(b*x + a)^(3/4)*(d*x + c)^(1/4))/(b^4*x^2 + 2*a*b^3*x + a^2*b^2)
 

Sympy [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{4}}}{\left (a + b x\right )^{\frac {9}{4}}}\, dx \] Input:

integrate((d*x+c)**(5/4)/(b*x+a)**(9/4),x)
 

Output:

Integral((c + d*x)**(5/4)/(a + b*x)**(9/4), x)
 

Maxima [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate((d*x+c)^(5/4)/(b*x+a)^(9/4),x, algorithm="maxima")
 

Output:

integrate((d*x + c)^(5/4)/(b*x + a)^(9/4), x)
 

Giac [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate((d*x+c)^(5/4)/(b*x+a)^(9/4),x, algorithm="giac")
 

Output:

integrate((d*x + c)^(5/4)/(b*x + a)^(9/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{9/4}} \,d x \] Input:

int((c + d*x)^(5/4)/(a + b*x)^(9/4),x)
 

Output:

int((c + d*x)^(5/4)/(a + b*x)^(9/4), x)
 

Reduce [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{9/4}} \, dx=\frac {5 \left (b x +a \right )^{\frac {1}{4}} \left (\int \frac {\left (d x +c \right )^{\frac {1}{4}} x}{\left (b x +a \right )^{\frac {1}{4}} a^{2}+2 \left (b x +a \right )^{\frac {1}{4}} a b x +\left (b x +a \right )^{\frac {1}{4}} b^{2} x^{2}}d x \right ) a^{2} d^{2}-5 \left (b x +a \right )^{\frac {1}{4}} \left (\int \frac {\left (d x +c \right )^{\frac {1}{4}} x}{\left (b x +a \right )^{\frac {1}{4}} a^{2}+2 \left (b x +a \right )^{\frac {1}{4}} a b x +\left (b x +a \right )^{\frac {1}{4}} b^{2} x^{2}}d x \right ) a b c d +5 \left (b x +a \right )^{\frac {1}{4}} \left (\int \frac {\left (d x +c \right )^{\frac {1}{4}} x}{\left (b x +a \right )^{\frac {1}{4}} a^{2}+2 \left (b x +a \right )^{\frac {1}{4}} a b x +\left (b x +a \right )^{\frac {1}{4}} b^{2} x^{2}}d x \right ) a b \,d^{2} x -5 \left (b x +a \right )^{\frac {1}{4}} \left (\int \frac {\left (d x +c \right )^{\frac {1}{4}} x}{\left (b x +a \right )^{\frac {1}{4}} a^{2}+2 \left (b x +a \right )^{\frac {1}{4}} a b x +\left (b x +a \right )^{\frac {1}{4}} b^{2} x^{2}}d x \right ) b^{2} c d x +4 \left (d x +c \right )^{\frac {1}{4}} c^{2}+4 \left (d x +c \right )^{\frac {1}{4}} c d x}{5 \left (b x +a \right )^{\frac {1}{4}} \left (a b d x -b^{2} c x +a^{2} d -a b c \right )} \] Input:

int((d*x+c)^(5/4)/(b*x+a)^(9/4),x)
 

Output:

(5*(a + b*x)**(1/4)*int(((c + d*x)**(1/4)*x)/((a + b*x)**(1/4)*a**2 + 2*(a 
 + b*x)**(1/4)*a*b*x + (a + b*x)**(1/4)*b**2*x**2),x)*a**2*d**2 - 5*(a + b 
*x)**(1/4)*int(((c + d*x)**(1/4)*x)/((a + b*x)**(1/4)*a**2 + 2*(a + b*x)** 
(1/4)*a*b*x + (a + b*x)**(1/4)*b**2*x**2),x)*a*b*c*d + 5*(a + b*x)**(1/4)* 
int(((c + d*x)**(1/4)*x)/((a + b*x)**(1/4)*a**2 + 2*(a + b*x)**(1/4)*a*b*x 
 + (a + b*x)**(1/4)*b**2*x**2),x)*a*b*d**2*x - 5*(a + b*x)**(1/4)*int(((c 
+ d*x)**(1/4)*x)/((a + b*x)**(1/4)*a**2 + 2*(a + b*x)**(1/4)*a*b*x + (a + 
b*x)**(1/4)*b**2*x**2),x)*b**2*c*d*x + 4*(c + d*x)**(1/4)*c**2 + 4*(c + d* 
x)**(1/4)*c*d*x)/(5*(a + b*x)**(1/4)*(a**2*d - a*b*c + a*b*d*x - b**2*c*x) 
)