3.18.12 \(\int \frac {(a+b x)^{5/4}}{(c+d x)^{3/4}} \, dx\) [1712]

3.18.12.1 Optimal result

Integrand size = 19, antiderivative size = 332 \[ \int \frac {(a+b x)^{5/4}}{(c+d x)^{3/4}} \, dx=-\frac {5 (b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{3 d^2}+\frac {2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{3 d}+\frac {5 (b c-a d)^{5/2} ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right ),\frac {1}{2}\right )}{6 \sqrt {2} \sqrt [4]{b} d^{9/4} (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}} \]

output

-5/3*(-a*d+b*c)*(b*x+a)^(1/4)*(d*x+c)^(1/4)/d^2+2/3*(b*x+a)^(5/4)*(d*x+c)^ 
(1/4)/d+5/12*(-a*d+b*c)^(5/2)*((b*x+a)*(d*x+c))^(3/4)*(cos(2*arctan(b^(1/4 
)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2)))^2)^(1/2)/cos( 
2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2)) 
)*EllipticF(sin(2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/( 
-a*d+b*c)^(1/2))),1/2*2^(1/2))*(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1/2 
)/(-a*d+b*c))*((2*b*d*x+a*d+b*c)^2)^(1/2)*((a*d+b*(2*d*x+c))^2/(-a*d+b*c)^ 
2/(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1/2)/(-a*d+b*c))^2)^(1/2)/b^(1/4 
)/d^(9/4)/(b*x+a)^(3/4)/(d*x+c)^(3/4)/(2*b*d*x+a*d+b*c)*2^(1/2)/((a*d+b*(2 
*d*x+c))^2)^(1/2)
 

3.18.12.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.22 \[ \int \frac {(a+b x)^{5/4}}{(c+d x)^{3/4}} \, dx=\frac {4 (a+b x)^{9/4} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {9}{4},\frac {13}{4},\frac {d (a+b x)}{-b c+a d}\right )}{9 b (c+d x)^{3/4}} \]

input

Integrate[(a + b*x)^(5/4)/(c + d*x)^(3/4),x]
 

output

(4*(a + b*x)^(9/4)*((b*(c + d*x))/(b*c - a*d))^(3/4)*Hypergeometric2F1[3/4 
, 9/4, 13/4, (d*(a + b*x))/(-(b*c) + a*d)])/(9*b*(c + d*x)^(3/4))
 

3.18.12.3 Rubi [A] (warning: unable to verify)

Time = 0.31 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.54, 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, 768, 858, 807, 229}

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.

\(\displaystyle \int \frac {(a+b x)^{5/4}}{(c+d x)^{3/4}} \, dx\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{3 d}-\frac {5 (b c-a d) \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{3/4}}dx}{6 d}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}}dx}{2 d}\right )}{6 d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}-\frac {2 (b c-a d) \int \frac {1}{\left (c-\frac {a d}{b}+\frac {d (a+b x)}{b}\right )^{3/4}}d\sqrt [4]{a+b x}}{b d}\right )}{6 d}\)

\(\Big \downarrow \) 768

\(\displaystyle \frac {2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}-\frac {2 (a+b x)^{3/4} (b c-a d) \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{(a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4}}d\sqrt [4]{a+b x}}{b d \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}\right )}{6 d}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {2 (a+b x)^{3/4} (b c-a d) \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{\sqrt [4]{a+b x} \left (\frac {(b c-a d) (a+b x)}{d}+1\right )^{3/4}}d\frac {1}{\sqrt [4]{a+b x}}}{b d \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}+\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}\right )}{6 d}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/4} (b c-a d) \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{\left (\frac {\sqrt {a+b x} (b c-a d)}{d}+1\right )^{3/4}}d\sqrt {a+b x}}{b d \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}+\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}\right )}{6 d}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {2 (a+b x)^{3/4} \sqrt {b c-a d} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d} \sqrt {a+b x}}{\sqrt {d}}\right ),2\right )}{b \sqrt {d} \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}+\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}\right )}{6 d}\)

input

Int[(a + b*x)^(5/4)/(c + d*x)^(3/4),x]
 

output

(2*(a + b*x)^(5/4)*(c + d*x)^(1/4))/(3*d) - (5*(b*c - a*d)*((2*(a + b*x)^( 
1/4)*(c + d*x)^(1/4))/d + (2*Sqrt[b*c - a*d]*(a + b*x)^(3/4)*(1 + (b*c - a 
*d)/(d*(a + b*x)))^(3/4)*EllipticF[ArcTan[(Sqrt[b*c - a*d]*Sqrt[a + b*x])/ 
Sqrt[d]]/2, 2])/(b*Sqrt[d]*(c - (a*d)/b + (d*(a + b*x))/b)^(3/4))))/(6*d)
 

3.18.12.3.1 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 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 768

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 
/4)/(a + b*x^4)^(3/4))   Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ 
[{a, b}, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

3.18.12.4 Maple [F]

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

input

int((b*x+a)^(5/4)/(d*x+c)^(3/4),x)
 

output

int((b*x+a)^(5/4)/(d*x+c)^(3/4),x)
 

3.18.12.5 Fricas [F]

\[ \int \frac {(a+b x)^{5/4}}{(c+d x)^{3/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {5}{4}}}{{\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]

input

integrate((b*x+a)^(5/4)/(d*x+c)^(3/4),x, algorithm="fricas")
 

output

integral((b*x + a)^(5/4)/(d*x + c)^(3/4), x)
 

3.18.12.6 Sympy [F]

\[ \int \frac {(a+b x)^{5/4}}{(c+d x)^{3/4}} \, dx=\int \frac {\left (a + b x\right )^{\frac {5}{4}}}{\left (c + d x\right )^{\frac {3}{4}}}\, dx \]

input

integrate((b*x+a)**(5/4)/(d*x+c)**(3/4),x)
 

output

Integral((a + b*x)**(5/4)/(c + d*x)**(3/4), x)
 

3.18.12.7 Maxima [F]

\[ \int \frac {(a+b x)^{5/4}}{(c+d x)^{3/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {5}{4}}}{{\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]

input

integrate((b*x+a)^(5/4)/(d*x+c)^(3/4),x, algorithm="maxima")
 

output

integrate((b*x + a)^(5/4)/(d*x + c)^(3/4), x)
 

3.18.12.8 Giac [F]

\[ \int \frac {(a+b x)^{5/4}}{(c+d x)^{3/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {5}{4}}}{{\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]

input

integrate((b*x+a)^(5/4)/(d*x+c)^(3/4),x, algorithm="giac")
 

output

integrate((b*x + a)^(5/4)/(d*x + c)^(3/4), x)
 

3.18.12.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{5/4}}{(c+d x)^{3/4}} \, dx=\int \frac {{\left (a+b\,x\right )}^{5/4}}{{\left (c+d\,x\right )}^{3/4}} \,d x \]

input

int((a + b*x)^(5/4)/(c + d*x)^(3/4),x)
 

output

int((a + b*x)^(5/4)/(c + d*x)^(3/4), x)
 

3.18.12.10 Reduce [F]

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

input

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

output

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