\(\int (e x)^{7/4} (c+d x^n)^q (a x^j+b x^{j+n})^{5/3} \, dx\) [281]

Optimal result

Integrand size = 34, antiderivative size = 129 \[ \int (e x)^{7/4} \left (c+d x^n\right )^q \left (a x^j+b x^{j+n}\right )^{5/3} \, dx=\frac {12 a e x^{2+j} (e x)^{3/4} \left (c+d x^n\right )^q \left (1+\frac {d x^n}{c}\right )^{-q} \left (a x^j+b x^{j+n}\right )^{2/3} \operatorname {AppellF1}\left (\frac {33+20 j}{12 n},-\frac {5}{3},-q,\frac {33+20 j+12 n}{12 n},-\frac {b x^n}{a},-\frac {d x^n}{c}\right )}{(33+20 j) \left (1+\frac {b x^n}{a}\right )^{2/3}} \] Output:

12*a*e*x^(2+j)*(e*x)^(3/4)*(c+d*x^n)^q*(a*x^j+b*x^(j+n))^(2/3)*AppellF1(1/ 
12*(33+20*j)/n,-5/3,-q,1/12*(33+20*j+12*n)/n,-b*x^n/a,-d*x^n/c)/(33+20*j)/ 
(1+b*x^n/a)^(2/3)/((1+d*x^n/c)^q)
 

Mathematica [A] (warning: unable to verify)

Time = 1.23 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.63 \[ \int (e x)^{7/4} \left (c+d x^n\right )^q \left (a x^j+b x^{j+n}\right )^{5/3} \, dx=\frac {12 x^{1+j} (e x)^{7/4} \left (x^j \left (a+b x^n\right )\right )^{2/3} \left (c+d x^n\right )^q \left (1+\frac {d x^n}{c}\right )^{-q} \left (a (33+20 j+12 n) \operatorname {AppellF1}\left (\frac {33+20 j}{12 n},-\frac {2}{3},-q,\frac {\frac {11}{4}+\frac {5 j}{3}+n}{n},-\frac {b x^n}{a},-\frac {d x^n}{c}\right )+b (33+20 j) x^n \operatorname {AppellF1}\left (\frac {33+20 j+12 n}{12 n},-\frac {2}{3},-q,\frac {33+20 j+24 n}{12 n},-\frac {b x^n}{a},-\frac {d x^n}{c}\right )\right )}{(33+20 j) (33+20 j+12 n) \left (1+\frac {b x^n}{a}\right )^{2/3}} \] Input:

Integrate[(e*x)^(7/4)*(c + d*x^n)^q*(a*x^j + b*x^(j + n))^(5/3),x]
 

Output:

(12*x^(1 + j)*(e*x)^(7/4)*(x^j*(a + b*x^n))^(2/3)*(c + d*x^n)^q*(a*(33 + 2 
0*j + 12*n)*AppellF1[(33 + 20*j)/(12*n), -2/3, -q, (11/4 + (5*j)/3 + n)/n, 
 -((b*x^n)/a), -((d*x^n)/c)] + b*(33 + 20*j)*x^n*AppellF1[(33 + 20*j + 12* 
n)/(12*n), -2/3, -q, (33 + 20*j + 24*n)/(12*n), -((b*x^n)/a), -((d*x^n)/c) 
]))/((33 + 20*j)*(33 + 20*j + 12*n)*(1 + (b*x^n)/a)^(2/3)*(1 + (d*x^n)/c)^ 
q)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1948, 1013, 1013, 1012}

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.

\(\displaystyle \int (e x)^{7/4} \left (a x^j+b x^{j+n}\right )^{5/3} \left (c+d x^n\right )^q \, dx\)

\(\Big \downarrow \) 1948

\(\displaystyle \frac {e (e x)^{3/4} x^{\frac {1}{12} (-8 j-9)} \left (a x^j+b x^{j+n}\right )^{2/3} \int x^{\frac {1}{12} (20 j+21)} \left (b x^n+a\right )^{5/3} \left (d x^n+c\right )^qdx}{\left (a+b x^n\right )^{2/3}}\)

\(\Big \downarrow \) 1013

\(\displaystyle \frac {a e (e x)^{3/4} x^{\frac {1}{12} (-8 j-9)} \left (a x^j+b x^{j+n}\right )^{2/3} \int x^{\frac {1}{12} (20 j+21)} \left (\frac {b x^n}{a}+1\right )^{5/3} \left (d x^n+c\right )^qdx}{\left (\frac {b x^n}{a}+1\right )^{2/3}}\)

\(\Big \downarrow \) 1013

\(\displaystyle \frac {a e (e x)^{3/4} x^{\frac {1}{12} (-8 j-9)} \left (a x^j+b x^{j+n}\right )^{2/3} \left (c+d x^n\right )^q \left (\frac {d x^n}{c}+1\right )^{-q} \int x^{\frac {1}{12} (20 j+21)} \left (\frac {b x^n}{a}+1\right )^{5/3} \left (\frac {d x^n}{c}+1\right )^qdx}{\left (\frac {b x^n}{a}+1\right )^{2/3}}\)

\(\Big \downarrow \) 1012

\(\displaystyle \frac {12 a e (e x)^{3/4} x^{\frac {1}{12} (-8 j-9)+\frac {1}{12} (20 j+33)} \left (a x^j+b x^{j+n}\right )^{2/3} \left (c+d x^n\right )^q \left (\frac {d x^n}{c}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {20 j+33}{12 n},-\frac {5}{3},-q,\frac {20 j+12 n+33}{12 n},-\frac {b x^n}{a},-\frac {d x^n}{c}\right )}{(20 j+33) \left (\frac {b x^n}{a}+1\right )^{2/3}}\)

Input:

Int[(e*x)^(7/4)*(c + d*x^n)^q*(a*x^j + b*x^(j + n))^(5/3),x]
 

Output:

(12*a*e*x^((-9 - 8*j)/12 + (33 + 20*j)/12)*(e*x)^(3/4)*(c + d*x^n)^q*(a*x^ 
j + b*x^(j + n))^(2/3)*AppellF1[(33 + 20*j)/(12*n), -5/3, -q, (33 + 20*j + 
 12*n)/(12*n), -((b*x^n)/a), -((d*x^n)/c)])/((33 + 20*j)*(1 + (b*x^n)/a)^( 
2/3)*(1 + (d*x^n)/c)^q)
 

Defintions of rubi rules used

rule 1012

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m 
+ 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, 
 b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n 
 - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1013

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^ 
n/a))^FracPart[p])   Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; 
 FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & 
& NeQ[m, n - 1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 1948

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Simp[e^IntPart[m]*(e*x)^FracPart[m]*( 
(a*x^j + b*x^(j + n))^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x 
^n)^FracPart[p]))   Int[x^(m + j*p)*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; 
FreeQ[{a, b, c, d, e, j, m, n, p, q}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] 
 && NeQ[b*c - a*d, 0] &&  !(EqQ[n, 1] && EqQ[j, 1])
 

Maple [F]

\[\int \left (e x \right )^{\frac {7}{4}} \left (c +d \,x^{n}\right )^{q} \left (a \,x^{j}+b \,x^{j +n}\right )^{\frac {5}{3}}d x\]

Input:

int((e*x)^(7/4)*(c+d*x^n)^q*(a*x^j+b*x^(j+n))^(5/3),x)
 

Output:

int((e*x)^(7/4)*(c+d*x^n)^q*(a*x^j+b*x^(j+n))^(5/3),x)
 

Fricas [F]

\[ \int (e x)^{7/4} \left (c+d x^n\right )^q \left (a x^j+b x^{j+n}\right )^{5/3} \, dx=\int { {\left (b x^{j + n} + a x^{j}\right )}^{\frac {5}{3}} \left (e x\right )^{\frac {7}{4}} {\left (d x^{n} + c\right )}^{q} \,d x } \] Input:

integrate((e*x)^(7/4)*(c+d*x^n)^q*(a*x^j+b*x^(j+n))^(5/3),x, algorithm="fr 
icas")
 

Output:

integral((b*e*x*x^(j + n) + a*e*x*x^j)*(b*x^(j + n) + a*x^j)^(2/3)*(e*x)^( 
3/4)*(d*x^n + c)^q, x)
 

Sympy [F(-1)]

Timed out. \[ \int (e x)^{7/4} \left (c+d x^n\right )^q \left (a x^j+b x^{j+n}\right )^{5/3} \, dx=\text {Timed out} \] Input:

integrate((e*x)**(7/4)*(c+d*x**n)**q*(a*x**j+b*x**(j+n))**(5/3),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (e x)^{7/4} \left (c+d x^n\right )^q \left (a x^j+b x^{j+n}\right )^{5/3} \, dx=\int { {\left (b x^{j + n} + a x^{j}\right )}^{\frac {5}{3}} \left (e x\right )^{\frac {7}{4}} {\left (d x^{n} + c\right )}^{q} \,d x } \] Input:

integrate((e*x)^(7/4)*(c+d*x^n)^q*(a*x^j+b*x^(j+n))^(5/3),x, algorithm="ma 
xima")
 

Output:

integrate((b*x^(j + n) + a*x^j)^(5/3)*(e*x)^(7/4)*(d*x^n + c)^q, x)
 

Giac [F]

\[ \int (e x)^{7/4} \left (c+d x^n\right )^q \left (a x^j+b x^{j+n}\right )^{5/3} \, dx=\int { {\left (b x^{j + n} + a x^{j}\right )}^{\frac {5}{3}} \left (e x\right )^{\frac {7}{4}} {\left (d x^{n} + c\right )}^{q} \,d x } \] Input:

integrate((e*x)^(7/4)*(c+d*x^n)^q*(a*x^j+b*x^(j+n))^(5/3),x, algorithm="gi 
ac")
 

Output:

integrate((b*x^(j + n) + a*x^j)^(5/3)*(e*x)^(7/4)*(d*x^n + c)^q, x)
 

Mupad [F(-1)]

Timed out. \[ \int (e x)^{7/4} \left (c+d x^n\right )^q \left (a x^j+b x^{j+n}\right )^{5/3} \, dx=\int {\left (a\,x^j+b\,x^{j+n}\right )}^{5/3}\,{\left (e\,x\right )}^{7/4}\,{\left (c+d\,x^n\right )}^q \,d x \] Input:

int((a*x^j + b*x^(j + n))^(5/3)*(e*x)^(7/4)*(c + d*x^n)^q,x)
 

Output:

int((a*x^j + b*x^(j + n))^(5/3)*(e*x)^(7/4)*(c + d*x^n)^q, x)
 

Reduce [F]

\[ \int (e x)^{7/4} \left (c+d x^n\right )^q \left (a x^j+b x^{j+n}\right )^{5/3} \, dx=\text {too large to display} \] Input:

int((e*x)^(7/4)*(c+d*x^n)^q*(a*x^j+b*x^(j+n))^(5/3),x)
 

Output:

(4*e**(3/4)*e*(60*x**((20*j + 12*n + 9)/12)*(x**n*d + c)**q*(x**n*b + a)** 
(2/3)*a*b*d*j*x**2 + 36*x**((20*j + 12*n + 9)/12)*(x**n*d + c)**q*(x**n*b 
+ a)**(2/3)*a*b*d*n*q*x**2 + 99*x**((20*j + 12*n + 9)/12)*(x**n*d + c)**q* 
(x**n*b + a)**(2/3)*a*b*d*x**2 + 60*x**((20*j + 12*n + 9)/12)*(x**n*d + c) 
**q*(x**n*b + a)**(2/3)*b**2*c*j*x**2 + 24*x**((20*j + 12*n + 9)/12)*(x**n 
*d + c)**q*(x**n*b + a)**(2/3)*b**2*c*n*x**2 + 99*x**((20*j + 12*n + 9)/12 
)*(x**n*d + c)**q*(x**n*b + a)**(2/3)*b**2*c*x**2 + 60*x**((20*j + 9)/12)* 
(x**n*d + c)**q*(x**n*b + a)**(2/3)*a**2*d*j*x**2 + 36*x**((20*j + 9)/12)* 
(x**n*d + c)**q*(x**n*b + a)**(2/3)*a**2*d*n*q*x**2 + 60*x**((20*j + 9)/12 
)*(x**n*d + c)**q*(x**n*b + a)**(2/3)*a**2*d*n*x**2 + 99*x**((20*j + 9)/12 
)*(x**n*d + c)**q*(x**n*b + a)**(2/3)*a**2*d*x**2 + 60*x**((20*j + 9)/12)* 
(x**n*d + c)**q*(x**n*b + a)**(2/3)*a*b*c*j*x**2 + 72*x**((20*j + 9)/12)*( 
x**n*d + c)**q*(x**n*b + a)**(2/3)*a*b*c*n*q*x**2 + 84*x**((20*j + 9)/12)* 
(x**n*d + c)**q*(x**n*b + a)**(2/3)*a*b*c*n*x**2 + 99*x**((20*j + 9)/12)*( 
x**n*d + c)**q*(x**n*b + a)**(2/3)*a*b*c*x**2 - 16000*int((x**((20*j + 24* 
n + 9)/12)*(x**n*d + c)**q*(x**n*b + a)**(2/3)*x)/(400*x**(2*n)*a*b*d**2*j 
**2 + 480*x**(2*n)*a*b*d**2*j*n*q + 400*x**(2*n)*a*b*d**2*j*n + 1320*x**(2 
*n)*a*b*d**2*j + 144*x**(2*n)*a*b*d**2*n**2*q**2 + 240*x**(2*n)*a*b*d**2*n 
**2*q + 792*x**(2*n)*a*b*d**2*n*q + 660*x**(2*n)*a*b*d**2*n + 1089*x**(2*n 
)*a*b*d**2 + 400*x**(2*n)*b**2*c*d*j**2 + 240*x**(2*n)*b**2*c*d*j*n*q +...