3.27.11 \(\int \frac {\sqrt {x+x^4} (b+a x^6)}{-d+c x^6} \, dx\) [2611]

Optimal. Leaf size=227 \[ \frac {a x \sqrt {x+x^4}}{3 c}+\frac {\sqrt {-\left (\left (\sqrt {c}+\sqrt {d}\right ) \sqrt {d}\right )} (b c+a d) \text {ArcTan}\left (\frac {\sqrt {-\sqrt {c} \sqrt {d}-d} x \sqrt {x+x^4}}{\sqrt {d} (1+x) \left (1-x+x^2\right )}\right )}{3 c^{3/2} d}-\frac {\sqrt {\left (\sqrt {c}-\sqrt {d}\right ) \sqrt {d}} (b c+a d) \text {ArcTan}\left (\frac {\sqrt {\sqrt {c} \sqrt {d}-d} x \sqrt {x+x^4}}{\sqrt {d} (1+x) \left (1-x+x^2\right )}\right )}{3 c^{3/2} d}+\frac {a \tanh ^{-1}\left (\frac {x^2}{\sqrt {x+x^4}}\right )}{3 c} \]

[Out]

1/3*a*x*(x^4+x)^(1/2)/c+1/3*(-(c^(1/2)+d^(1/2))*d^(1/2))^(1/2)*(a*d+b*c)*arctan((-c^(1/2)*d^(1/2)-d)^(1/2)*x*(
x^4+x)^(1/2)/d^(1/2)/(1+x)/(x^2-x+1))/c^(3/2)/d-1/3*((c^(1/2)-d^(1/2))*d^(1/2))^(1/2)*(a*d+b*c)*arctan((c^(1/2
)*d^(1/2)-d)^(1/2)*x*(x^4+x)^(1/2)/d^(1/2)/(1+x)/(x^2-x+1))/c^(3/2)/d+1/3*a*arctanh(x^2/(x^4+x)^(1/2))/c

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Rubi [A]
time = 0.50, antiderivative size = 250, normalized size of antiderivative = 1.10, number of steps used = 15, number of rules used = 10, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {2081, 6847, 1707, 201, 221, 1189, 399, 385, 214, 211} \begin {gather*} -\frac {\sqrt {x^4+x} \sqrt {\sqrt {c}-\sqrt {d}} (a d+b c) \text {ArcTan}\left (\frac {x^{3/2} \sqrt {\sqrt {c}-\sqrt {d}}}{\sqrt [4]{d} \sqrt {x^3+1}}\right )}{3 c^{3/2} d^{3/4} \sqrt {x^3+1} \sqrt {x}}-\frac {\sqrt {x^4+x} \sqrt {\sqrt {c}+\sqrt {d}} (a d+b c) \tanh ^{-1}\left (\frac {x^{3/2} \sqrt {\sqrt {c}+\sqrt {d}}}{\sqrt [4]{d} \sqrt {x^3+1}}\right )}{3 c^{3/2} d^{3/4} \sqrt {x^3+1} \sqrt {x}}+\frac {a \sqrt {x^4+x} x}{3 c}+\frac {a \sqrt {x^4+x} \sinh ^{-1}\left (x^{3/2}\right )}{3 c \sqrt {x^3+1} \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[x + x^4]*(b + a*x^6))/(-d + c*x^6),x]

[Out]

(a*x*Sqrt[x + x^4])/(3*c) + (a*Sqrt[x + x^4]*ArcSinh[x^(3/2)])/(3*c*Sqrt[x]*Sqrt[1 + x^3]) - (Sqrt[Sqrt[c] - S
qrt[d]]*(b*c + a*d)*Sqrt[x + x^4]*ArcTan[(Sqrt[Sqrt[c] - Sqrt[d]]*x^(3/2))/(d^(1/4)*Sqrt[1 + x^3])])/(3*c^(3/2
)*d^(3/4)*Sqrt[x]*Sqrt[1 + x^3]) - (Sqrt[Sqrt[c] + Sqrt[d]]*(b*c + a*d)*Sqrt[x + x^4]*ArcTanh[(Sqrt[Sqrt[c] +
Sqrt[d]]*x^(3/2))/(d^(1/4)*Sqrt[1 + x^3])])/(3*c^(3/2)*d^(3/4)*Sqrt[x]*Sqrt[1 + x^3])

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 211

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

Rule 214

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

Rule 221

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

Rule 385

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

Rule 399

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

Rule 1189

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{r = Rt[(-a)*c, 2]}, Dist[-c/(2*r), In
t[(d + e*x^2)^q/(r - c*x^2), x], x] - Dist[c/(2*r), Int[(d + e*x^2)^q/(r + c*x^2), x], x]] /; FreeQ[{a, c, d,
e, q}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[q]

Rule 1707

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(d + e*
x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e, q}, x] && PolyQ[Px, x^2] && NeQ[c*d^2 + a*e^2, 0] && Intege
rQ[p]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rule 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

\begin {align*} \int \frac {\sqrt {x+x^4} \left (b+a x^6\right )}{-d+c x^6} \, dx &=\frac {\sqrt {x+x^4} \int \frac {\sqrt {x} \sqrt {1+x^3} \left (b+a x^6\right )}{-d+c x^6} \, dx}{\sqrt {x} \sqrt {1+x^3}}\\ &=\frac {\left (2 \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {\sqrt {1+x^2} \left (b+a x^4\right )}{-d+c x^4} \, dx,x,x^{3/2}\right )}{3 \sqrt {x} \sqrt {1+x^3}}\\ &=\frac {\left (2 \sqrt {x+x^4}\right ) \text {Subst}\left (\int \left (\frac {a \sqrt {1+x^2}}{c}+\frac {(b c+a d) \sqrt {1+x^2}}{c \left (-d+c x^4\right )}\right ) \, dx,x,x^{3/2}\right )}{3 \sqrt {x} \sqrt {1+x^3}}\\ &=\frac {\left (2 a \sqrt {x+x^4}\right ) \text {Subst}\left (\int \sqrt {1+x^2} \, dx,x,x^{3/2}\right )}{3 c \sqrt {x} \sqrt {1+x^3}}+\frac {\left (2 (b c+a d) \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {\sqrt {1+x^2}}{-d+c x^4} \, dx,x,x^{3/2}\right )}{3 c \sqrt {x} \sqrt {1+x^3}}\\ &=\frac {a x \sqrt {x+x^4}}{3 c}+\frac {\left (a \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,x^{3/2}\right )}{3 c \sqrt {x} \sqrt {1+x^3}}-\frac {\left ((b c+a d) \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {\sqrt {1+x^2}}{\sqrt {c} \sqrt {d}-c x^2} \, dx,x,x^{3/2}\right )}{3 \sqrt {c} \sqrt {d} \sqrt {x} \sqrt {1+x^3}}-\frac {\left ((b c+a d) \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {\sqrt {1+x^2}}{\sqrt {c} \sqrt {d}+c x^2} \, dx,x,x^{3/2}\right )}{3 \sqrt {c} \sqrt {d} \sqrt {x} \sqrt {1+x^3}}\\ &=\frac {a x \sqrt {x+x^4}}{3 c}+\frac {a \sqrt {x+x^4} \sinh ^{-1}\left (x^{3/2}\right )}{3 c \sqrt {x} \sqrt {1+x^3}}+\frac {\left (\left (-1+\frac {\sqrt {d}}{\sqrt {c}}\right ) (b c+a d) \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2} \left (\sqrt {c} \sqrt {d}+c x^2\right )} \, dx,x,x^{3/2}\right )}{3 \sqrt {c} \sqrt {d} \sqrt {x} \sqrt {1+x^3}}-\frac {\left (\left (1+\frac {\sqrt {d}}{\sqrt {c}}\right ) (b c+a d) \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2} \left (\sqrt {c} \sqrt {d}-c x^2\right )} \, dx,x,x^{3/2}\right )}{3 \sqrt {c} \sqrt {d} \sqrt {x} \sqrt {1+x^3}}\\ &=\frac {a x \sqrt {x+x^4}}{3 c}+\frac {a \sqrt {x+x^4} \sinh ^{-1}\left (x^{3/2}\right )}{3 c \sqrt {x} \sqrt {1+x^3}}+\frac {\left (\left (-1+\frac {\sqrt {d}}{\sqrt {c}}\right ) (b c+a d) \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c} \sqrt {d}-\left (-c+\sqrt {c} \sqrt {d}\right ) x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {1+x^3}}\right )}{3 \sqrt {c} \sqrt {d} \sqrt {x} \sqrt {1+x^3}}-\frac {\left (\left (1+\frac {\sqrt {d}}{\sqrt {c}}\right ) (b c+a d) \sqrt {x+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c} \sqrt {d}-\left (c+\sqrt {c} \sqrt {d}\right ) x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {1+x^3}}\right )}{3 \sqrt {c} \sqrt {d} \sqrt {x} \sqrt {1+x^3}}\\ &=\frac {a x \sqrt {x+x^4}}{3 c}+\frac {a \sqrt {x+x^4} \sinh ^{-1}\left (x^{3/2}\right )}{3 c \sqrt {x} \sqrt {1+x^3}}-\frac {\sqrt {\sqrt {c}-\sqrt {d}} (b c+a d) \sqrt {x+x^4} \tan ^{-1}\left (\frac {\sqrt {\sqrt {c}-\sqrt {d}} x^{3/2}}{\sqrt [4]{d} \sqrt {1+x^3}}\right )}{3 c^{3/2} d^{3/4} \sqrt {x} \sqrt {1+x^3}}-\frac {\sqrt {\sqrt {c}+\sqrt {d}} (b c+a d) \sqrt {x+x^4} \tanh ^{-1}\left (\frac {\sqrt {\sqrt {c}+\sqrt {d}} x^{3/2}}{\sqrt [4]{d} \sqrt {1+x^3}}\right )}{3 c^{3/2} d^{3/4} \sqrt {x} \sqrt {1+x^3}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 0.35, size = 185, normalized size = 0.81 \begin {gather*} \frac {\sqrt {x+x^4} \left (a \left (x^{3/2} \sqrt {1+x^3}+\tanh ^{-1}\left (\frac {x^{3/2}}{\sqrt {1+x^3}}\right )\right )+(b c+a d) \text {RootSum}\left [16 c-16 d-32 c \text {$\#$1}+32 d \text {$\#$1}+24 c \text {$\#$1}^2-16 d \text {$\#$1}^2-8 c \text {$\#$1}^3+c \text {$\#$1}^4\&,\frac {\log \left (2+2 x^3+2 x^{3/2} \sqrt {1+x^3}-\text {$\#$1}\right ) \text {$\#$1}^2}{-8 c+8 d+12 c \text {$\#$1}-8 d \text {$\#$1}-6 c \text {$\#$1}^2+c \text {$\#$1}^3}\&\right ]\right )}{3 c \sqrt {x} \sqrt {1+x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[x + x^4]*(b + a*x^6))/(-d + c*x^6),x]

[Out]

(Sqrt[x + x^4]*(a*(x^(3/2)*Sqrt[1 + x^3] + ArcTanh[x^(3/2)/Sqrt[1 + x^3]]) + (b*c + a*d)*RootSum[16*c - 16*d -
32*c*#1 + 32*d*#1 + 24*c*#1^2 - 16*d*#1^2 - 8*c*#1^3 + c*#1^4 & , (Log[2 + 2*x^3 + 2*x^(3/2)*Sqrt[1 + x^3] -
#1]*#1^2)/(-8*c + 8*d + 12*c*#1 - 8*d*#1 - 6*c*#1^2 + c*#1^3) & ]))/(3*c*Sqrt[x]*Sqrt[1 + x^3])

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.44, size = 686, normalized size = 3.02 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4+x)^(1/2)*(a*x^6+b)/(c*x^6-d),x,method=_RETURNVERBOSE)

[Out]

a/c*(1/3*x*(x^4+x)^(1/2)-(-1/2-1/2*I*3^(1/2))*((3/2+1/2*I*3^(1/2))*x/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2)*(1+x)^2*
(-(x-1/2+1/2*I*3^(1/2))/(1/2-1/2*I*3^(1/2))/(1+x))^(1/2)*(-(x-1/2-1/2*I*3^(1/2))/(1/2+1/2*I*3^(1/2))/(1+x))^(1
/2)/(3/2+1/2*I*3^(1/2))/(x*(1+x)*(x-1/2+1/2*I*3^(1/2))*(x-1/2-1/2*I*3^(1/2)))^(1/2)*(-EllipticF(((3/2+1/2*I*3^
(1/2))*x/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2),((-3/2+1/2*I*3^(1/2))*(-1/2-1/2*I*3^(1/2))/(-1/2+1/2*I*3^(1/2))/(-3/
2-1/2*I*3^(1/2)))^(1/2))+EllipticPi(((3/2+1/2*I*3^(1/2))*x/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2),(1/2+1/2*I*3^(1/2)
)/(3/2+1/2*I*3^(1/2)),((-3/2+1/2*I*3^(1/2))*(-1/2-1/2*I*3^(1/2))/(-1/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1
/2))))+1/3*(a*d+b*c)/c*4^(1/2)*sum((-_alpha^3-1)/_alpha^4*(1+x)^2*(_alpha^5-_alpha^4+_alpha^3-_alpha^2+_alpha-
1)/(c-d)*(-1-I*3^(1/2))*(x/(1+x)*(3+I*3^(1/2))/(1+I*3^(1/2)))^(1/2)*(-1/(1+x)*(I*3^(1/2)+2*x-1)/(1-I*3^(1/2)))
^(1/2)*(-1/(1+x)*(-I*3^(1/2)+2*x-1)/(1+I*3^(1/2)))^(1/2)/(3+I*3^(1/2))/(x*(1+x)*(I*3^(1/2)+2*x-1)*(-I*3^(1/2)+
2*x-1))^(1/2)*(EllipticF(((3/2+1/2*I*3^(1/2))*x/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2),((-3/2+1/2*I*3^(1/2))*(-1/2-1
/2*I*3^(1/2))/(-1/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))+_alpha^5*c/d*EllipticPi(((3/2+1/2*I*3^(1/2))*x
/(1/2+1/2*I*3^(1/2))/(1+x))^(1/2),1/6*(I*_alpha^5*3^(1/2)*c+3*_alpha^5*c+I*3^(1/2)*d+3*d)/d,((-3/2+1/2*I*3^(1/
2))*(-1/2-1/2*I*3^(1/2))/(-1/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))),_alpha=RootOf(_Z^6*c-d))

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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((x^4+x)^(1/2)*(a*x^6+b)/(c*x^6-d),x, algorithm="maxima")

[Out]

integrate((a*x^6 + b)*sqrt(x^4 + x)/(c*x^6 - d), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+x)^(1/2)*(a*x^6+b)/(c*x^6-d),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x \left (x + 1\right ) \left (x^{2} - x + 1\right )} \left (a x^{6} + b\right )}{c x^{6} - d}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4+x)**(1/2)*(a*x**6+b)/(c*x**6-d),x)

[Out]

Integral(sqrt(x*(x + 1)*(x**2 - x + 1))*(a*x**6 + b)/(c*x**6 - d), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+x)^(1/2)*(a*x^6+b)/(c*x^6-d),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, need to choose a branch for the
root of a polynomial with parameters. This might be wrong.The choice was done assuming [sageVARc,sageVARd]=[3
5,45]Warning, need to

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int -\frac {\left (a\,x^6+b\right )\,\sqrt {x^4+x}}{d-c\,x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((b + a*x^6)*(x + x^4)^(1/2))/(d - c*x^6),x)

[Out]

int(-((b + a*x^6)*(x + x^4)^(1/2))/(d - c*x^6), x)

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