3.219 \(\int (a+b x^4)^2 (c+d x^4)^q \, dx\)

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

[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)

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Rubi [A]  time = 0.133064, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {416, 388, 246, 245} \[ \frac{x \left (c+d x^4\right )^q \left (\frac{d x^4}{c}+1\right )^{-q} \left (a^2 d^2 \left (16 q^2+56 q+45\right )-2 a b c d (4 q+9)+5 b^2 c^2\right ) \, _2F_1\left (\frac{1}{4},-q;\frac{5}{4};-\frac{d x^4}{c}\right )}{d^2 (4 q+5) (4 q+9)}-\frac{b x \left (c+d x^4\right )^{q+1} (5 b c-a d (4 q+13))}{d^2 (4 q+5) (4 q+9)}+\frac{b x \left (a+b x^4\right ) \left (c+d x^4\right )^{q+1}}{d (4 q+9)} \]

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 416

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]

Rule 388

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 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[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 245

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])

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*}

Mathematica [A]  time = 0.052535, size = 106, normalized size = 0.6 \[ \frac{1}{45} x \left (c+d x^4\right )^q \left (\frac{d x^4}{c}+1\right )^{-q} \left (45 a^2 \, _2F_1\left (\frac{1}{4},-q;\frac{5}{4};-\frac{d x^4}{c}\right )+b x^4 \left (18 a \, _2F_1\left (\frac{5}{4},-q;\frac{9}{4};-\frac{d x^4}{c}\right )+5 b x^4 \, _2F_1\left (\frac{9}{4},-q;\frac{13}{4};-\frac{d x^4}{c}\right )\right )\right ) \]

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 = 0.361, size = 0, normalized size = 0. \begin{align*} \int \left ( b{x}^{4}+a \right ) ^{2} \left ( d{x}^{4}+c \right ) ^{q}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{4} + a\right )}^{2}{\left (d x^{4} + c\right )}^{q}\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{2} x^{8} + 2 \, a b x^{4} + a^{2}\right )}{\left (d x^{4} + c\right )}^{q}, x\right ) \end{align*}

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 [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{4} + a\right )}^{2}{\left (d x^{4} + c\right )}^{q}\,{d x} \end{align*}

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)