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

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

[Out]

1/2*arctanh((d*x+c)^2*b^(1/2)/(a+b*(d*x+c)^4)^(1/2))/d^2/b^(1/2)-1/2*c*(cos(2*arctan(b^(1/4)*(d*x+c)/a^(1/4)))
^2)^(1/2)/cos(2*arctan(b^(1/4)*(d*x+c)/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*(d*x+c)/a^(1/4))),1/2*2^(1/2))
*(a^(1/2)+(d*x+c)^2*b^(1/2))*((a+b*(d*x+c)^4)/(a^(1/2)+(d*x+c)^2*b^(1/2))^2)^(1/2)/a^(1/4)/b^(1/4)/d^2/(a+b*(d
*x+c)^4)^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 154, normalized size of antiderivative = 1.00, 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 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])

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 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 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)]

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*}

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Mathematica [C]  time = 1.01, size = 330, normalized size = 2.14 \[ \frac {\sqrt [4]{-1} \sqrt {2} \sqrt {-\frac {i \left (\sqrt [4]{-1} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}} \left (\sqrt {b} (c+d x)^2+i \sqrt {a}\right ) \left (\left (\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} c\right ) F\left (\left .\sin ^{-1}\left (\sqrt {-\frac {i \left (\sqrt [4]{b} (c+d x)+\sqrt [4]{-1} \sqrt [4]{a}\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}}\right )\right |-1\right )-2 \sqrt [4]{-1} \sqrt [4]{a} \Pi \left (-i;\left .\sin ^{-1}\left (\sqrt {-\frac {i \left (\sqrt [4]{b} (c+d x)+\sqrt [4]{-1} \sqrt [4]{a}\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}}\right )\right |-1\right )\right )}{\sqrt [4]{a} \sqrt {b} d^2 \sqrt {\frac {\sqrt {b} (c+d x)^2+i \sqrt {a}}{\left (\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)\right )^2}} \sqrt {a+b (c+d x)^4}} \]

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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fricas [F]  time = 1.22, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x}{\sqrt {b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {{\left (d x + c\right )}^{4} b + a}}\,{d x} \]

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)

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maple [C]  time = 0.03, size = 1528, normalized size = 9.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {{\left (d x + c\right )}^{4} b + a}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{\sqrt {a+b\,{\left (c+d\,x\right )}^4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*(c + d*x)^4)^(1/2),x)

[Out]

int(x/(a + b*(c + d*x)^4)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {a + b c^{4} + 4 b c^{3} d x + 6 b c^{2} d^{2} x^{2} + 4 b c d^{3} x^{3} + b d^{4} x^{4}}}\, dx \]

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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