3.3.19 \(\int (a+b x^4)^2 (c+d x^4)^q \, dx\) [219]

Optimal. Leaf size=176 \[ -\frac {b (5 b c-a d (13+4 q)) x \left (c+d x^4\right )^{1+q}}{d^2 (5+4 q) (9+4 q)}+\frac {b x \left (a+b x^4\right ) \left (c+d x^4\right )^{1+q}}{d (9+4 q)}+\frac {\left (5 b^2 c^2-2 a b c d (9+4 q)+a^2 d^2 \left (45+56 q+16 q^2\right )\right ) x \left (c+d x^4\right )^q \left (1+\frac {d x^4}{c}\right )^{-q} \, _2F_1\left (\frac {1}{4},-q;\frac {5}{4};-\frac {d x^4}{c}\right )}{d^2 (5+4 q) (9+4 q)} \]

[Out]

-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)

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Rubi [A]
time = 0.09, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {427, 396, 252, 251} \begin {gather*} \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 ) \, _2F_1\left (\frac {1}{4},-q;\frac {5}{4};-\frac {d x^4}{c}\right )}{d^2 (4 q+5) (4 q+9)}-\frac {b x \left (c+d x^4\right )^{q+1} (5 b c-a d (4 q+13))}{d^2 (4 q+5) (4 q+9)}+\frac {b x \left (a+b x^4\right ) \left (c+d x^4\right )^{q+1}}{d (4 q+9)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^2*(c + d*x^4)^q,x]

[Out]

-((b*(5*b*c - a*d*(13 + 4*q))*x*(c + d*x^4)^(1 + q))/(d^2*(5 + 4*q)*(9 + 4*q))) + (b*x*(a + b*x^4)*(c + d*x^4)
^(1 + q))/(d*(9 + 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*Hy
pergeometric2F1[1/4, -q, 5/4, -((d*x^4)/c)])/(d^2*(5 + 4*q)*(9 + 4*q)*(1 + (d*x^4)/c)^q)

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-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])

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(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}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 427

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] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[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] /; F
reeQ[{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] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rubi steps

\begin {align*} \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^q \, dx &=\frac {b x \left (a+b x^4\right ) \left (c+d x^4\right )^{1+q}}{d (9+4 q)}+\frac {\int \left (c+d x^4\right )^q \left (-a (b c-a d (9+4 q))-b (5 b c-a d (13+4 q)) x^4\right ) \, dx}{d (9+4 q)}\\ &=-\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 ) \int \left (c+d x^4\right )^q \, dx}{d^2 (5+4 q) (9+4 q)}\\ &=-\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 (\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 ) \left (c+d x^4\right )^q \left (1+\frac {d x^4}{c}\right )^{-q}\right ) \int \left (1+\frac {d x^4}{c}\right )^q \, dx}{d^2 (5+4 q) (9+4 q)}\\ &=-\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} \, _2F_1\left (\frac {1}{4},-q;\frac {5}{4};-\frac {d x^4}{c}\right )}{d^2 (5+4 q) (9+4 q)}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 106, normalized size = 0.60 \begin {gather*} \frac {1}{45} x \left (c+d x^4\right )^q \left (1+\frac {d x^4}{c}\right )^{-q} \left (45 a^2 \, _2F_1\left (\frac {1}{4},-q;\frac {5}{4};-\frac {d x^4}{c}\right )+b x^4 \left (18 a \, _2F_1\left (\frac {5}{4},-q;\frac {9}{4};-\frac {d x^4}{c}\right )+5 b x^4 \, _2F_1\left (\frac {9}{4},-q;\frac {13}{4};-\frac {d x^4}{c}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^2*(c + d*x^4)^q,x]

[Out]

(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*Hypergeometric2F1[9/4, -q, 13/4, -((d*x^4)/c)])))/(45*(1 + (d*x^4)/c)^q)

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \left (b \,x^{4}+a \right )^{2} \left (d \,x^{4}+c \right )^{q}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)^2*(d*x^4+c)^q,x)

[Out]

int((b*x^4+a)^2*(d*x^4+c)^q,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^2*(d*x^4+c)^q,x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^2*(d*x^4 + c)^q, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^2*(d*x^4+c)^q,x, algorithm="fricas")

[Out]

integral((b^2*x^8 + 2*a*b*x^4 + a^2)*(d*x^4 + c)^q, x)

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Sympy [C] Result contains complex when optimal does not.
time = 83.88, size = 119, normalized size = 0.68 \begin {gather*} \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)**2*(d*x**4+c)**q,x)

[Out]

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))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^2*(d*x^4+c)^q,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^2*(d*x^4 + c)^q, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (b\,x^4+a\right )}^2\,{\left (d\,x^4+c\right )}^q \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^2*(c + d*x^4)^q,x)

[Out]

int((a + b*x^4)^2*(c + d*x^4)^q, x)

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