Integrand size = 19, antiderivative size = 176 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^q \, dx=-\frac {b (5 b c-a d (13+4 q)) x \left (c+d x^4\right )^{1+q}}{d^2 (5+4 q) (9+4 q)}+\frac {b x \left (a+b x^4\right ) \left (c+d x^4\right )^{1+q}}{d (9+4 q)}+\frac {\left (5 b^2 c^2-2 a b c d (9+4 q)+a^2 d^2 \left (45+56 q+16 q^2\right )\right ) x \left (c+d x^4\right )^q \left (1+\frac {d x^4}{c}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-q,\frac {5}{4},-\frac {d x^4}{c}\right )}{d^2 (5+4 q) (9+4 q)} \]
-b*(5*b*c-a*d*(13+4*q))*x*(d*x^4+c)^(1+q)/d^2/(16*q^2+56*q+45)+b*x*(b*x^4+ a)*(d*x^4+c)^(1+q)/d/(9+4*q)+(5*b^2*c^2-2*a*b*c*d*(9+4*q)+a^2*d^2*(16*q^2+ 56*q+45))*x*(d*x^4+c)^q*hypergeom([1/4, -q],[5/4],-d*x^4/c)/d^2/(16*q^2+56 *q+45)/((1+d*x^4/c)^q)
Time = 0.28 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.60 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^q \, dx=\frac {1}{45} x \left (c+d x^4\right )^q \left (1+\frac {d x^4}{c}\right )^{-q} \left (45 a^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-q,\frac {5}{4},-\frac {d x^4}{c}\right )+b x^4 \left (18 a \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-q,\frac {9}{4},-\frac {d x^4}{c}\right )+5 b x^4 \operatorname {Hypergeometric2F1}\left (\frac {9}{4},-q,\frac {13}{4},-\frac {d x^4}{c}\right )\right )\right ) \]
(x*(c + d*x^4)^q*(45*a^2*Hypergeometric2F1[1/4, -q, 5/4, -((d*x^4)/c)] + b *x^4*(18*a*Hypergeometric2F1[5/4, -q, 9/4, -((d*x^4)/c)] + 5*b*x^4*Hyperge ometric2F1[9/4, -q, 13/4, -((d*x^4)/c)])))/(45*(1 + (d*x^4)/c)^q)
Time = 0.34 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {933, 25, 913, 779, 778}
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.
\(\displaystyle \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^q \, dx\) |
\(\Big \downarrow \) 933 |
\(\displaystyle \frac {\int -\left (d x^4+c\right )^q \left (b (5 b c-a d (4 q+13)) x^4+a (b c-a d (4 q+9))\right )dx}{d (4 q+9)}+\frac {b x \left (a+b x^4\right ) \left (c+d x^4\right )^{q+1}}{d (4 q+9)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b x \left (a+b x^4\right ) \left (c+d x^4\right )^{q+1}}{d (4 q+9)}-\frac {\int \left (d x^4+c\right )^q \left (b (5 b c-a d (4 q+13)) x^4+a (b c-a d (4 q+9))\right )dx}{d (4 q+9)}\) |
\(\Big \downarrow \) 913 |
\(\displaystyle \frac {b x \left (a+b x^4\right ) \left (c+d x^4\right )^{q+1}}{d (4 q+9)}-\frac {\frac {b x \left (c+d x^4\right )^{q+1} (5 b c-a d (4 q+13))}{d (4 q+5)}-\frac {\left (a^2 d^2 \left (16 q^2+56 q+45\right )-2 a b c d (4 q+9)+5 b^2 c^2\right ) \int \left (d x^4+c\right )^qdx}{d (4 q+5)}}{d (4 q+9)}\) |
\(\Big \downarrow \) 779 |
\(\displaystyle \frac {b x \left (a+b x^4\right ) \left (c+d x^4\right )^{q+1}}{d (4 q+9)}-\frac {\frac {b x \left (c+d x^4\right )^{q+1} (5 b c-a d (4 q+13))}{d (4 q+5)}-\frac {\left (c+d x^4\right )^q \left (\frac {d x^4}{c}+1\right )^{-q} \left (a^2 d^2 \left (16 q^2+56 q+45\right )-2 a b c d (4 q+9)+5 b^2 c^2\right ) \int \left (\frac {d x^4}{c}+1\right )^qdx}{d (4 q+5)}}{d (4 q+9)}\) |
\(\Big \downarrow \) 778 |
\(\displaystyle \frac {b x \left (a+b x^4\right ) \left (c+d x^4\right )^{q+1}}{d (4 q+9)}-\frac {\frac {b x \left (c+d x^4\right )^{q+1} (5 b c-a d (4 q+13))}{d (4 q+5)}-\frac {x \left (c+d x^4\right )^q \left (\frac {d x^4}{c}+1\right )^{-q} \left (a^2 d^2 \left (16 q^2+56 q+45\right )-2 a b c d (4 q+9)+5 b^2 c^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-q,\frac {5}{4},-\frac {d x^4}{c}\right )}{d (4 q+5)}}{d (4 q+9)}\) |
(b*x*(a + b*x^4)*(c + d*x^4)^(1 + q))/(d*(9 + 4*q)) - ((b*(5*b*c - a*d*(13 + 4*q))*x*(c + d*x^4)^(1 + q))/(d*(5 + 4*q)) - ((5*b^2*c^2 - 2*a*b*c*d*(9 + 4*q) + a^2*d^2*(45 + 56*q + 16*q^2))*x*(c + d*x^4)^q*Hypergeometric2F1[ 1/4, -q, 5/4, -((d*x^4)/c)])/(d*(5 + 4*q)*(1 + (d*x^4)/c)^q))/(d*(9 + 4*q) )
3.3.19.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] && !IGtQ[p , 0] && !IntegerQ[1/n] && !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || GtQ[a, 0])
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x ^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) Int[(1 + b*(x^n/a))^p, x], x ] /; FreeQ[{a, b, n, p}, x] && !IGtQ[p, 0] && !IntegerQ[1/n] && !ILtQ[Si mplify[1/n + p], 0] && !(IntegerQ[p] || GtQ[a, 0])
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si mp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(p + 1) + 1))), x] - Simp[(a*d - b*c*(n*( p + 1) + 1))/(b*(n*(p + 1) + 1)) Int[(a + b*x^n)^p, x], x] /; FreeQ[{a, b , c, d, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), x] + Simp[1/(b*(n*(p + q) + 1)) Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Sim p[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d , 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] && !IGtQ[p, 1] && IntBinomialQ[ a, b, c, d, n, p, q, x]
\[\int \left (b \,x^{4}+a \right )^{2} \left (d \,x^{4}+c \right )^{q}d x\]
\[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^q \, dx=\int { {\left (b x^{4} + a\right )}^{2} {\left (d x^{4} + c\right )}^{q} \,d x } \]
Result contains complex when optimal does not.
Time = 77.69 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.68 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^q \, dx=\frac {a^{2} c^{q} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - q \\ \frac {5}{4} \end {matrix}\middle | {\frac {d x^{4} e^{i \pi }}{c}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {a b c^{q} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, - q \\ \frac {9}{4} \end {matrix}\middle | {\frac {d x^{4} e^{i \pi }}{c}} \right )}}{2 \Gamma \left (\frac {9}{4}\right )} + \frac {b^{2} c^{q} x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, - q \\ \frac {13}{4} \end {matrix}\middle | {\frac {d x^{4} e^{i \pi }}{c}} \right )}}{4 \Gamma \left (\frac {13}{4}\right )} \]
a**2*c**q*x*gamma(1/4)*hyper((1/4, -q), (5/4,), d*x**4*exp_polar(I*pi)/c)/ (4*gamma(5/4)) + a*b*c**q*x**5*gamma(5/4)*hyper((5/4, -q), (9/4,), d*x**4* exp_polar(I*pi)/c)/(2*gamma(9/4)) + b**2*c**q*x**9*gamma(9/4)*hyper((9/4, -q), (13/4,), d*x**4*exp_polar(I*pi)/c)/(4*gamma(13/4))
\[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^q \, dx=\int { {\left (b x^{4} + a\right )}^{2} {\left (d x^{4} + c\right )}^{q} \,d x } \]
\[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^q \, dx=\int { {\left (b x^{4} + a\right )}^{2} {\left (d x^{4} + c\right )}^{q} \,d x } \]
Timed out. \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^q \, dx=\int {\left (b\,x^4+a\right )}^2\,{\left (d\,x^4+c\right )}^q \,d x \]