\(\int (a+b x)^{5/4} (c+d x)^{5/4} \, dx\) [581]

Optimal result

Integrand size = 19, antiderivative size = 232 \[ \int (a+b x)^{5/4} (c+d x)^{5/4} \, dx=-\frac {5 (b c-a d)^3 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{84 b^2 d^2}+\frac {(b c-a d)^2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{42 b^2 d}+\frac {(b c-a d) (a+b x)^{9/4} \sqrt [4]{c+d x}}{7 b^2}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}-\frac {5 (b c-a d)^{7/2} (a+b x)^{3/4} \left (\frac {b (c+d x)}{d (a+b x)}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d}}{\sqrt {d} \sqrt {a+b x}}\right ),2\right )}{84 b^3 d^{3/2} (c+d x)^{3/4}} \] Output:

-5/84*(-a*d+b*c)^3*(b*x+a)^(1/4)*(d*x+c)^(1/4)/b^2/d^2+1/42*(-a*d+b*c)^2*( 
b*x+a)^(5/4)*(d*x+c)^(1/4)/b^2/d+1/7*(-a*d+b*c)*(b*x+a)^(9/4)*(d*x+c)^(1/4 
)/b^2+2/7*(b*x+a)^(9/4)*(d*x+c)^(5/4)/b-5/84*(-a*d+b*c)^(7/2)*(b*x+a)^(3/4 
)*(b*(d*x+c)/d/(b*x+a))^(3/4)*InverseJacobiAM(1/2*arctan(1/d^(1/2)/(b*x+a) 
^(1/2)*(-a*d+b*c)^(1/2)),2^(1/2))/b^3/d^(3/2)/(d*x+c)^(3/4)
 

Mathematica [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.31 \[ \int (a+b x)^{5/4} (c+d x)^{5/4} \, dx=\frac {4 (a+b x)^{9/4} (c+d x)^{5/4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {9}{4},\frac {13}{4},\frac {d (a+b x)}{-b c+a d}\right )}{9 b \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \] Input:

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

Output:

(4*(a + b*x)^(9/4)*(c + d*x)^(5/4)*Hypergeometric2F1[-5/4, 9/4, 13/4, (d*( 
a + b*x))/(-(b*c) + a*d)])/(9*b*((b*(c + d*x))/(b*c - a*d))^(5/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.37 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {60, 60, 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.

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

\(\Big \downarrow \) 60

\(\displaystyle \frac {5 (b c-a d) \int (a+b x)^{5/4} \sqrt [4]{c+d x}dx}{14 b}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {5 (b c-a d) \left (\frac {(b c-a d) \int \frac {(a+b x)^{5/4}}{(c+d x)^{3/4}}dx}{10 b}+\frac {2 (a+b x)^{9/4} \sqrt [4]{c+d x}}{5 b}\right )}{14 b}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {5 (b c-a d) \left (\frac {(b c-a d) \left (\frac {2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{3 d}-\frac {5 (b c-a d) \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{3/4}}dx}{6 d}\right )}{10 b}+\frac {2 (a+b x)^{9/4} \sqrt [4]{c+d x}}{5 b}\right )}{14 b}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {5 (b c-a d) \left (\frac {(b c-a d) \left (\frac {2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}}dx}{2 d}\right )}{6 d}\right )}{10 b}+\frac {2 (a+b x)^{9/4} \sqrt [4]{c+d x}}{5 b}\right )}{14 b}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {5 (b c-a d) \left (\frac {(b c-a d) \left (\frac {2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}-\frac {2 (b c-a d) \int \frac {1}{\left (c-\frac {a d}{b}+\frac {d (a+b x)}{b}\right )^{3/4}}d\sqrt [4]{a+b x}}{b d}\right )}{6 d}\right )}{10 b}+\frac {2 (a+b x)^{9/4} \sqrt [4]{c+d x}}{5 b}\right )}{14 b}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}\)

\(\Big \downarrow \) 768

\(\displaystyle \frac {5 (b c-a d) \left (\frac {(b c-a d) \left (\frac {2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}-\frac {2 (a+b x)^{3/4} (b c-a d) \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{(a+b x)^{3/4} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4}}d\sqrt [4]{a+b x}}{b d \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}\right )}{6 d}\right )}{10 b}+\frac {2 (a+b x)^{9/4} \sqrt [4]{c+d x}}{5 b}\right )}{14 b}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {5 (b c-a d) \left (\frac {(b c-a d) \left (\frac {2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {2 (a+b x)^{3/4} (b c-a d) \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{\sqrt [4]{a+b x} \left (\frac {(b c-a d) (a+b x)}{d}+1\right )^{3/4}}d\frac {1}{\sqrt [4]{a+b x}}}{b d \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}+\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}\right )}{6 d}\right )}{10 b}+\frac {2 (a+b x)^{9/4} \sqrt [4]{c+d x}}{5 b}\right )}{14 b}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {5 (b c-a d) \left (\frac {(b c-a d) \left (\frac {2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/4} (b c-a d) \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \int \frac {1}{\left (\frac {\sqrt {a+b x} (b c-a d)}{d}+1\right )^{3/4}}d\sqrt {a+b x}}{b d \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}+\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}\right )}{6 d}\right )}{10 b}+\frac {2 (a+b x)^{9/4} \sqrt [4]{c+d x}}{5 b}\right )}{14 b}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {5 (b c-a d) \left (\frac {(b c-a d) \left (\frac {2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {2 (a+b x)^{3/4} \sqrt {b c-a d} \left (\frac {b c-a d}{d (a+b x)}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d} \sqrt {a+b x}}{\sqrt {d}}\right ),2\right )}{b \sqrt {d} \left (\frac {d (a+b x)}{b}-\frac {a d}{b}+c\right )^{3/4}}+\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}\right )}{6 d}\right )}{10 b}+\frac {2 (a+b x)^{9/4} \sqrt [4]{c+d x}}{5 b}\right )}{14 b}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}\)

Input:

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

Output:

(2*(a + b*x)^(9/4)*(c + d*x)^(5/4))/(7*b) + (5*(b*c - a*d)*((2*(a + b*x)^( 
9/4)*(c + d*x)^(1/4))/(5*b) + ((b*c - a*d)*((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*S 
qrt[b*c - a*d]*(a + b*x)^(3/4)*(1 + (b*c - a*d)/(d*(a + b*x)))^(3/4)*Ellip 
ticF[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)))/(10*b)))/(14*b)
 

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]
 

Maple [F]

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

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)
 

Fricas [F]

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

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

Output:

integral((b*d*x^2 + a*c + (b*c + a*d)*x)*(b*x + a)^(1/4)*(d*x + c)^(1/4), 
x)
 

Sympy [F]

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

Maxima [F]

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

Giac [F]

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

Mupad [F(-1)]

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

Reduce [F]

\[ \int (a+b x)^{5/4} (c+d x)^{5/4} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 16*(c + d*x)**(1/4)*(a + b*x)**(1/4)*a**3*c*d**2 + 4*(c + d*x)**(1/4)* 
(a + b*x)**(1/4)*a**3*d**3*x + 128*(c + d*x)**(1/4)*(a + b*x)**(1/4)*a**2* 
b*c**2*d + 140*(c + d*x)**(1/4)*(a + b*x)**(1/4)*a**2*b*c*d**2*x + 72*(c + 
 d*x)**(1/4)*(a + b*x)**(1/4)*a**2*b*d**3*x**2 - 16*(c + d*x)**(1/4)*(a + 
b*x)**(1/4)*a*b**2*c**3 + 140*(c + d*x)**(1/4)*(a + b*x)**(1/4)*a*b**2*c** 
2*d*x + 144*(c + d*x)**(1/4)*(a + b*x)**(1/4)*a*b**2*c*d**2*x**2 + 48*(c + 
 d*x)**(1/4)*(a + b*x)**(1/4)*a*b**2*d**3*x**3 + 4*(c + d*x)**(1/4)*(a + b 
*x)**(1/4)*b**3*c**3*x + 72*(c + d*x)**(1/4)*(a + b*x)**(1/4)*b**3*c**2*d* 
x**2 + 48*(c + d*x)**(1/4)*(a + b*x)**(1/4)*b**3*c*d**2*x**3 - 5*int(((c + 
 d*x)**(1/4)*(a + b*x)**(1/4)*x)/(a**2*c*d + a**2*d**2*x + a*b*c**2 + 2*a* 
b*c*d*x + a*b*d**2*x**2 + b**2*c**2*x + b**2*c*d*x**2),x)*a**5*d**5 + 15*i 
nt(((c + d*x)**(1/4)*(a + b*x)**(1/4)*x)/(a**2*c*d + a**2*d**2*x + a*b*c** 
2 + 2*a*b*c*d*x + a*b*d**2*x**2 + b**2*c**2*x + b**2*c*d*x**2),x)*a**4*b*c 
*d**4 - 10*int(((c + d*x)**(1/4)*(a + b*x)**(1/4)*x)/(a**2*c*d + a**2*d**2 
*x + a*b*c**2 + 2*a*b*c*d*x + a*b*d**2*x**2 + b**2*c**2*x + b**2*c*d*x**2) 
,x)*a**3*b**2*c**2*d**3 - 10*int(((c + d*x)**(1/4)*(a + b*x)**(1/4)*x)/(a* 
*2*c*d + a**2*d**2*x + a*b*c**2 + 2*a*b*c*d*x + a*b*d**2*x**2 + b**2*c**2* 
x + b**2*c*d*x**2),x)*a**2*b**3*c**3*d**2 + 15*int(((c + d*x)**(1/4)*(a + 
b*x)**(1/4)*x)/(a**2*c*d + a**2*d**2*x + a*b*c**2 + 2*a*b*c*d*x + a*b*d**2 
*x**2 + b**2*c**2*x + b**2*c*d*x**2),x)*a*b**4*c**4*d - 5*int(((c + d*x...