3.2918 \(\int \frac{x}{\sqrt{a+b (c+d x)^4}} \, dx\)

Optimal. Leaf size=154 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} (c+d x)^2}{\sqrt{a+b (c+d x)^4}}\right )}{2 \sqrt{b} d^2}-\frac{c \left (\sqrt{a}+\sqrt{b} (c+d x)^2\right ) \sqrt{\frac{a+b (c+d x)^4}{\left (\sqrt{a}+\sqrt{b} (c+d x)^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d^2 \sqrt{a+b (c+d x)^4}} \]

[Out]

ArcTanh[(Sqrt[b]*(c + d*x)^2)/Sqrt[a + b*(c + d*x)^4]]/(2*Sqrt[b]*d^2) - (c*(Sqrt[a] + Sqrt[b]*(c + d*x)^2)*Sq
rt[(a + b*(c + d*x)^4)/(Sqrt[a] + Sqrt[b]*(c + d*x)^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*(c + d*x))/a^(1/4)], 1/2
])/(2*a^(1/4)*b^(1/4)*d^2*Sqrt[a + b*(c + d*x)^4])

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Rubi [A]  time = 0.16883, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {371, 1885, 220, 275, 217, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} (c+d x)^2}{\sqrt{a+b (c+d x)^4}}\right )}{2 \sqrt{b} d^2}-\frac{c \left (\sqrt{a}+\sqrt{b} (c+d x)^2\right ) \sqrt{\frac{a+b (c+d x)^4}{\left (\sqrt{a}+\sqrt{b} (c+d x)^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d^2 \sqrt{a+b (c+d x)^4}} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a + b*(c + d*x)^4],x]

[Out]

ArcTanh[(Sqrt[b]*(c + d*x)^2)/Sqrt[a + b*(c + d*x)^4]]/(2*Sqrt[b]*d^2) - (c*(Sqrt[a] + Sqrt[b]*(c + d*x)^2)*Sq
rt[(a + b*(c + d*x)^4)/(Sqrt[a] + Sqrt[b]*(c + d*x)^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*(c + d*x))/a^(1/4)], 1/2
])/(2*a^(1/4)*b^(1/4)*d^2*Sqrt[a + b*(c + d*x)^4])

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 1885

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[Sum[x^j*Sum[Coeff[P
q, x, j + (k*n)/2]*x^((k*n)/2), {k, 0, (2*(q - j))/n + 1}]*(a + b*x^n)^p, {j, 0, n/2 - 1}], x]] /; FreeQ[{a, b
, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x}{\sqrt{a+b (c+d x)^4}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{-c+x}{\sqrt{a+b x^4}} \, dx,x,c+d x\right )}{d^2}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{c}{\sqrt{a+b x^4}}+\frac{x}{\sqrt{a+b x^4}}\right ) \, dx,x,c+d x\right )}{d^2}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x}{\sqrt{a+b x^4}} \, dx,x,c+d x\right )}{d^2}-\frac{c \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^4}} \, dx,x,c+d x\right )}{d^2}\\ &=-\frac{c \left (\sqrt{a}+\sqrt{b} (c+d x)^2\right ) \sqrt{\frac{a+b (c+d x)^4}{\left (\sqrt{a}+\sqrt{b} (c+d x)^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d^2 \sqrt{a+b (c+d x)^4}}+\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,(c+d x)^2\right )}{2 d^2}\\ &=-\frac{c \left (\sqrt{a}+\sqrt{b} (c+d x)^2\right ) \sqrt{\frac{a+b (c+d x)^4}{\left (\sqrt{a}+\sqrt{b} (c+d x)^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d^2 \sqrt{a+b (c+d x)^4}}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{(c+d x)^2}{\sqrt{a+b (c+d x)^4}}\right )}{2 d^2}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{b} (c+d x)^2}{\sqrt{a+b (c+d x)^4}}\right )}{2 \sqrt{b} d^2}-\frac{c \left (\sqrt{a}+\sqrt{b} (c+d x)^2\right ) \sqrt{\frac{a+b (c+d x)^4}{\left (\sqrt{a}+\sqrt{b} (c+d x)^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d^2 \sqrt{a+b (c+d x)^4}}\\ \end{align*}

Mathematica [C]  time = 0.562349, size = 330, normalized size = 2.14 \[ \frac{\sqrt [4]{-1} \sqrt{2} \sqrt{-\frac{i \left (\sqrt [4]{-1} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}} \left (\sqrt{b} (c+d x)^2+i \sqrt{a}\right ) \left (\left (\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} c\right ) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{-\frac{i \left (\sqrt [4]{-1} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}}\right ),-1\right )-2 \sqrt [4]{-1} \sqrt [4]{a} \Pi \left (-i;\left .\sin ^{-1}\left (\sqrt{-\frac{i \left (\sqrt [4]{b} (c+d x)+\sqrt [4]{-1} \sqrt [4]{a}\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}}\right )\right |-1\right )\right )}{\sqrt [4]{a} \sqrt{b} d^2 \sqrt{\frac{\sqrt{b} (c+d x)^2+i \sqrt{a}}{\left (\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)\right )^2}} \sqrt{a+b (c+d x)^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1)^(1/4)*Sqrt[2]*Sqrt[((-I)*((-1)^(1/4)*a^(1/4) + b^(1/4)*(c + d*x)))/((-1)^(1/4)*a^(1/4) - b^(1/4)*(c + d*
x))]*(I*Sqrt[a] + Sqrt[b]*(c + d*x)^2)*(((-1)^(1/4)*a^(1/4) - b^(1/4)*c)*EllipticF[ArcSin[Sqrt[((-I)*((-1)^(1/
4)*a^(1/4) + b^(1/4)*(c + d*x)))/((-1)^(1/4)*a^(1/4) - b^(1/4)*(c + d*x))]], -1] - 2*(-1)^(1/4)*a^(1/4)*Ellipt
icPi[-I, ArcSin[Sqrt[((-I)*((-1)^(1/4)*a^(1/4) + b^(1/4)*(c + d*x)))/((-1)^(1/4)*a^(1/4) - b^(1/4)*(c + d*x))]
], -1]))/(a^(1/4)*Sqrt[b]*d^2*Sqrt[(I*Sqrt[a] + Sqrt[b]*(c + d*x)^2)/((-1)^(1/4)*a^(1/4) - b^(1/4)*(c + d*x))^
2]*Sqrt[a + b*(c + d*x)^4])

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Maple [C]  time = 0.021, size = 1528, normalized size = 9.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a+b*(d*x+c)^4)^(1/2),x)

[Out]

2*((1/b*(-a*b^3)^(1/4)-c)/d-(-I/b*(-a*b^3)^(1/4)-c)/d)*(((-I/b*(-a*b^3)^(1/4)-c)/d-(I/b*(-a*b^3)^(1/4)-c)/d)*(
x-(1/b*(-a*b^3)^(1/4)-c)/d)/((-I/b*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)/(x-(I/b*(-a*b^3)^(1/4)-c)/d))
^(1/2)*(x-(I/b*(-a*b^3)^(1/4)-c)/d)^2*(((I/b*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)*(x-(-1/b*(-a*b^3)^(
1/4)-c)/d)/((-1/b*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)/(x-(I/b*(-a*b^3)^(1/4)-c)/d))^(1/2)*(((I/b*(-a
*b^3)^(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)*(x-(-I/b*(-a*b^3)^(1/4)-c)/d)/((-I/b*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b
^3)^(1/4)-c)/d)/(x-(I/b*(-a*b^3)^(1/4)-c)/d))^(1/2)/((-I/b*(-a*b^3)^(1/4)-c)/d-(I/b*(-a*b^3)^(1/4)-c)/d)/((I/b
*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)/(b*d^4*(x-(1/b*(-a*b^3)^(1/4)-c)/d)*(x-(I/b*(-a*b^3)^(1/4)-c)/d
)*(x-(-1/b*(-a*b^3)^(1/4)-c)/d)*(x-(-I/b*(-a*b^3)^(1/4)-c)/d))^(1/2)*((I/b*(-a*b^3)^(1/4)-c)/d*EllipticF((((-I
/b*(-a*b^3)^(1/4)-c)/d-(I/b*(-a*b^3)^(1/4)-c)/d)*(x-(1/b*(-a*b^3)^(1/4)-c)/d)/((-I/b*(-a*b^3)^(1/4)-c)/d-(1/b*
(-a*b^3)^(1/4)-c)/d)/(x-(I/b*(-a*b^3)^(1/4)-c)/d))^(1/2),(((I/b*(-a*b^3)^(1/4)-c)/d-(-1/b*(-a*b^3)^(1/4)-c)/d)
*((1/b*(-a*b^3)^(1/4)-c)/d-(-I/b*(-a*b^3)^(1/4)-c)/d)/((1/b*(-a*b^3)^(1/4)-c)/d-(-1/b*(-a*b^3)^(1/4)-c)/d)/((I
/b*(-a*b^3)^(1/4)-c)/d-(-I/b*(-a*b^3)^(1/4)-c)/d))^(1/2))+((1/b*(-a*b^3)^(1/4)-c)/d-(I/b*(-a*b^3)^(1/4)-c)/d)*
EllipticPi((((-I/b*(-a*b^3)^(1/4)-c)/d-(I/b*(-a*b^3)^(1/4)-c)/d)*(x-(1/b*(-a*b^3)^(1/4)-c)/d)/((-I/b*(-a*b^3)^
(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)/(x-(I/b*(-a*b^3)^(1/4)-c)/d))^(1/2),((-I/b*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b
^3)^(1/4)-c)/d)/((-I/b*(-a*b^3)^(1/4)-c)/d-(I/b*(-a*b^3)^(1/4)-c)/d),(((I/b*(-a*b^3)^(1/4)-c)/d-(-1/b*(-a*b^3)
^(1/4)-c)/d)*((1/b*(-a*b^3)^(1/4)-c)/d-(-I/b*(-a*b^3)^(1/4)-c)/d)/((1/b*(-a*b^3)^(1/4)-c)/d-(-1/b*(-a*b^3)^(1/
4)-c)/d)/((I/b*(-a*b^3)^(1/4)-c)/d-(-I/b*(-a*b^3)^(1/4)-c)/d))^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*(d*x+c)^4)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x}{\sqrt{b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*(d*x+c)^4)^(1/2),x, algorithm="fricas")

[Out]

integral(x/sqrt(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4 + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{a + b c^{4} + 4 b c^{3} d x + 6 b c^{2} d^{2} x^{2} + 4 b c d^{3} x^{3} + b d^{4} x^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*(d*x+c)**4)**(1/2),x)

[Out]

Integral(x/sqrt(a + b*c**4 + 4*b*c**3*d*x + 6*b*c**2*d**2*x**2 + 4*b*c*d**3*x**3 + b*d**4*x**4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*(d*x+c)^4)^(1/2),x, algorithm="giac")

[Out]

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