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

Optimal. Leaf size=93 \[ \frac{b x \left (c+d x^4\right )^{q+1}}{d (4 q+5)}-\frac{x \left (c+d x^4\right )^q \left (\frac{d x^4}{c}+1\right )^{-q} (b c-a d (4 q+5)) \, _2F_1\left (\frac{1}{4},-q;\frac{5}{4};-\frac{d x^4}{c}\right )}{d (4 q+5)} \]

[Out]

(b*x*(c + d*x^4)^(1 + q))/(d*(5 + 4*q)) - ((b*c - a*d*(5 + 4*q))*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)

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Rubi [A]  time = 0.0407276, antiderivative size = 85, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {388, 246, 245} \[ x \left (c+d x^4\right )^q \left (\frac{d x^4}{c}+1\right )^{-q} \left (a-\frac{b c}{4 d q+5 d}\right ) \, _2F_1\left (\frac{1}{4},-q;\frac{5}{4};-\frac{d x^4}{c}\right )+\frac{b x \left (c+d x^4\right )^{q+1}}{d (4 q+5)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)*(c + d*x^4)^q,x]

[Out]

(b*x*(c + d*x^4)^(1 + q))/(d*(5 + 4*q)) + ((a - (b*c)/(5*d + 4*d*q))*x*(c + d*x^4)^q*Hypergeometric2F1[1/4, -q
, 5/4, -((d*x^4)/c)])/(1 + (d*x^4)/c)^q

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

Mathematica [A]  time = 0.0299453, size = 90, normalized size = 0.97 \[ \frac{x \left (c+d x^4\right )^q \left (\frac{d x^4}{c}+1\right )^{-q} \left ((a d (4 q+5)-b c) \, _2F_1\left (\frac{1}{4},-q;\frac{5}{4};-\frac{d x^4}{c}\right )+b \left (c+d x^4\right ) \left (\frac{d x^4}{c}+1\right )^q\right )}{d (4 q+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(c + d*x^4)^q*(b*(c + d*x^4)*(1 + (d*x^4)/c)^q + (-(b*c) + a*d*(5 + 4*q))*Hypergeometric2F1[1/4, -q, 5/4, -
((d*x^4)/c)]))/(d*(5 + 4*q)*(1 + (d*x^4)/c)^q)

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Maple [F]  time = 0.212, size = 0, normalized size = 0. \begin{align*} \int \left ( b{x}^{4}+a \right ) \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)*(d*x^4+c)^q,x)

[Out]

int((b*x^4+a)*(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 )}{\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)*(d*x^4+c)^q,x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [C]  time = 92.1663, size = 75, normalized size = 0.81 \begin{align*} \frac{a 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{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 )}}{4 \Gamma \left (\frac{9}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c**q*x*gamma(1/4)*hyper((1/4, -q), (5/4,), d*x**4*exp_polar(I*pi)/c)/(4*gamma(5/4)) + b*c**q*x**5*gamma(5/4)
*hyper((5/4, -q), (9/4,), d*x**4*exp_polar(I*pi)/c)/(4*gamma(9/4))

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

[Out]

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