3.18.17 \(\int \frac {(a+b x)^{5/4}}{(c+d x)^{5/4}} \, dx\) [1717]

3.18.17.1 Optimal result

Integrand size = 19, antiderivative size = 152 \[ \int \frac {(a+b x)^{5/4}}{(c+d x)^{5/4}} \, dx=-\frac {4 (a+b x)^{5/4}}{d \sqrt [4]{c+d x}}+\frac {5 b \sqrt [4]{a+b x} (c+d x)^{3/4}}{d^2}-\frac {5 \sqrt [4]{b} (b c-a d) \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 d^{9/4}}-\frac {5 \sqrt [4]{b} (b c-a d) \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 d^{9/4}} \]

output

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

3.18.17.2 Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.92 \[ \int \frac {(a+b x)^{5/4}}{(c+d x)^{5/4}} \, dx=\frac {\sqrt [4]{a+b x} (5 b c-4 a d+b d x)}{d^2 \sqrt [4]{c+d x}}+\frac {5 \sqrt [4]{b} (b c-a d) \arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{2 d^{9/4}}-\frac {5 \sqrt [4]{b} (b c-a d) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{2 d^{9/4}} \]

input

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

output

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

3.18.17.3 Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.22, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {57, 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.

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 770

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

input

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

output

(-4*(a + b*x)^(5/4))/(d*(c + d*x)^(1/4)) + (5*b*(((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)*ArcTan 
h[(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)))/d
 

3.18.17.3.1 Defintions of rubi rules used

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 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]
 

3.18.17.4 Maple [F]

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

input

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

output

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

3.18.17.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 754, normalized size of antiderivative = 4.96 \[ \int \frac {(a+b x)^{5/4}}{(c+d x)^{5/4}} \, dx=-\frac {5 \, {\left (d^{3} x + c d^{2}\right )} \left (\frac {b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}}{d^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {5 \, {\left ({\left (b c - a d\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}} + {\left (d^{3} x + c d^{2}\right )} \left (\frac {b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}}{d^{9}}\right )^{\frac {1}{4}}\right )}}{d x + c}\right ) - 5 \, {\left (d^{3} x + c d^{2}\right )} \left (\frac {b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}}{d^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {5 \, {\left ({\left (b c - a d\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}} - {\left (d^{3} x + c d^{2}\right )} \left (\frac {b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}}{d^{9}}\right )^{\frac {1}{4}}\right )}}{d x + c}\right ) - 5 \, {\left (-i \, d^{3} x - i \, c d^{2}\right )} \left (\frac {b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}}{d^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {5 \, {\left ({\left (b c - a d\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}} + {\left (i \, d^{3} x + i \, c d^{2}\right )} \left (\frac {b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}}{d^{9}}\right )^{\frac {1}{4}}\right )}}{d x + c}\right ) - 5 \, {\left (i \, d^{3} x + i \, c d^{2}\right )} \left (\frac {b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}}{d^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {5 \, {\left ({\left (b c - a d\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}} + {\left (-i \, d^{3} x - i \, c d^{2}\right )} \left (\frac {b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}}{d^{9}}\right )^{\frac {1}{4}}\right )}}{d x + c}\right ) - 4 \, {\left (b d x + 5 \, b c - 4 \, a d\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{4 \, {\left (d^{3} x + c d^{2}\right )}} \]

input

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

output

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

3.18.17.6 Sympy [F]

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

input

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

output

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

3.18.17.7 Maxima [F]

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

input

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

output

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

3.18.17.8 Giac [F]

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

input

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

output

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

3.18.17.9 Mupad [F(-1)]

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

input

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

output

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