3.455 \(\int \frac {\sin ^6(x) \tan (x)}{\cos ^{\frac {3}{4}}(2 x)} \, dx\)

Optimal. Leaf size=102 \[ \frac {1}{36} \cos ^{\frac {9}{4}}(2 x)-\frac {1}{5} \cos ^{\frac {5}{4}}(2 x)+\frac {7}{4} \sqrt [4]{\cos (2 x)}+\frac {\tan ^{-1}\left (\frac {1-\sqrt {\cos (2 x)}}{\sqrt {2} \sqrt [4]{\cos (2 x)}}\right )}{\sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {\cos (2 x)}+1}{\sqrt {2} \sqrt [4]{\cos (2 x)}}\right )}{\sqrt {2}} \]

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Rubi [A]  time = 0.19, antiderivative size = 154, normalized size of antiderivative = 1.51, number of steps used = 14, number of rules used = 10, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {4361, 446, 88, 63, 211, 1165, 628, 1162, 617, 204} \[ \frac {1}{36} \cos ^{\frac {9}{4}}(2 x)-\frac {1}{5} \cos ^{\frac {5}{4}}(2 x)+\frac {7}{4} \sqrt [4]{\cos (2 x)}+\frac {\log \left (\sqrt {\cos (2 x)}-\sqrt {2} \sqrt [4]{\cos (2 x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {\cos (2 x)}+\sqrt {2} \sqrt [4]{\cos (2 x)}+1\right )}{2 \sqrt {2}}+\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{\cos (2 x)}\right )}{\sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {2} \sqrt [4]{\cos (2 x)}+1\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[x]^6*Tan[x])/Cos[2*x]^(3/4),x]

[Out]

ArcTan[1 - Sqrt[2]*Cos[2*x]^(1/4)]/Sqrt[2] - ArcTan[1 + Sqrt[2]*Cos[2*x]^(1/4)]/Sqrt[2] + (7*Cos[2*x]^(1/4))/4
- Cos[2*x]^(5/4)/5 + Cos[2*x]^(9/4)/36 + Log[1 - Sqrt[2]*Cos[2*x]^(1/4) + Sqrt[Cos[2*x]]]/(2*Sqrt[2]) - Log[1
+ Sqrt[2]*Cos[2*x]^(1/4) + Sqrt[Cos[2*x]]]/(2*Sqrt[2])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 204

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

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 446

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 4361

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dist[(b*
c)^(-1), Subst[Int[SubstFor[1/x, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[
c*(a + b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Tan] || EqQ[F, tan])

Rubi steps

\begin {align*} \int \frac {\sin ^6(x) \tan (x)}{\cos ^{\frac {3}{4}}(2 x)} \, dx &=-\operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^3}{x \left (-1+2 x^2\right )^{3/4}} \, dx,x,\cos (x)\right )\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {(1-x)^3}{x (-1+2 x)^{3/4}} \, dx,x,\cos ^2(x)\right )\right )\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {7}{4 (-1+2 x)^{3/4}}+\frac {1}{x (-1+2 x)^{3/4}}+\sqrt [4]{-1+2 x}-\frac {1}{4} (-1+2 x)^{5/4}\right ) \, dx,x,\cos ^2(x)\right )\right )\\ &=\frac {7}{4} \sqrt [4]{-1+2 \cos ^2(x)}-\frac {1}{5} \left (-1+2 \cos ^2(x)\right )^{5/4}+\frac {1}{36} \left (-1+2 \cos ^2(x)\right )^{9/4}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x (-1+2 x)^{3/4}} \, dx,x,\cos ^2(x)\right )\\ &=\frac {7}{4} \sqrt [4]{-1+2 \cos ^2(x)}-\frac {1}{5} \left (-1+2 \cos ^2(x)\right )^{5/4}+\frac {1}{36} \left (-1+2 \cos ^2(x)\right )^{9/4}-\operatorname {Subst}\left (\int \frac {1}{\frac {1}{2}+\frac {x^4}{2}} \, dx,x,\sqrt [4]{-1+2 \cos ^2(x)}\right )\\ &=\frac {7}{4} \sqrt [4]{-1+2 \cos ^2(x)}-\frac {1}{5} \left (-1+2 \cos ^2(x)\right )^{5/4}+\frac {1}{36} \left (-1+2 \cos ^2(x)\right )^{9/4}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1-x^2}{\frac {1}{2}+\frac {x^4}{2}} \, dx,x,\sqrt [4]{-1+2 \cos ^2(x)}\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1+x^2}{\frac {1}{2}+\frac {x^4}{2}} \, dx,x,\sqrt [4]{-1+2 \cos ^2(x)}\right )\\ &=\frac {7}{4} \sqrt [4]{-1+2 \cos ^2(x)}-\frac {1}{5} \left (-1+2 \cos ^2(x)\right )^{5/4}+\frac {1}{36} \left (-1+2 \cos ^2(x)\right )^{9/4}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+2 \cos ^2(x)}\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+2 \cos ^2(x)}\right )+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt [4]{-1+2 \cos ^2(x)}\right )}{2 \sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt [4]{-1+2 \cos ^2(x)}\right )}{2 \sqrt {2}}\\ &=\frac {7}{4} \sqrt [4]{-1+2 \cos ^2(x)}-\frac {1}{5} \left (-1+2 \cos ^2(x)\right )^{5/4}+\frac {1}{36} \left (-1+2 \cos ^2(x)\right )^{9/4}+\frac {\log \left (1-\sqrt {2} \sqrt [4]{-1+2 \cos ^2(x)}+\sqrt {-1+2 \cos ^2(x)}\right )}{2 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} \sqrt [4]{-1+2 \cos ^2(x)}+\sqrt {-1+2 \cos ^2(x)}\right )}{2 \sqrt {2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt [4]{-1+2 \cos ^2(x)}\right )}{\sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt [4]{-1+2 \cos ^2(x)}\right )}{\sqrt {2}}\\ &=\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{-1+2 \cos ^2(x)}\right )}{\sqrt {2}}-\frac {\tan ^{-1}\left (1+\sqrt {2} \sqrt [4]{-1+2 \cos ^2(x)}\right )}{\sqrt {2}}+\frac {7}{4} \sqrt [4]{-1+2 \cos ^2(x)}-\frac {1}{5} \left (-1+2 \cos ^2(x)\right )^{5/4}+\frac {1}{36} \left (-1+2 \cos ^2(x)\right )^{9/4}+\frac {\log \left (1-\sqrt {2} \sqrt [4]{-1+2 \cos ^2(x)}+\sqrt {-1+2 \cos ^2(x)}\right )}{2 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} \sqrt [4]{-1+2 \cos ^2(x)}+\sqrt {-1+2 \cos ^2(x)}\right )}{2 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 153, normalized size = 1.50 \[ \frac {1}{360} \left (-72 \cos ^{\frac {5}{4}}(2 x)+5 \cos (4 x) \sqrt [4]{\cos (2 x)}+635 \sqrt [4]{\cos (2 x)}+90 \sqrt {2} \log \left (\sqrt {\cos (2 x)}-\sqrt {2} \sqrt [4]{\cos (2 x)}+1\right )-90 \sqrt {2} \log \left (\sqrt {\cos (2 x)}+\sqrt {2} \sqrt [4]{\cos (2 x)}+1\right )+180 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{\cos (2 x)}\right )-180 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt [4]{\cos (2 x)}+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[x]^6*Tan[x])/Cos[2*x]^(3/4),x]

[Out]

(180*Sqrt[2]*ArcTan[1 - Sqrt[2]*Cos[2*x]^(1/4)] - 180*Sqrt[2]*ArcTan[1 + Sqrt[2]*Cos[2*x]^(1/4)] + 635*Cos[2*x
]^(1/4) - 72*Cos[2*x]^(5/4) + 5*Cos[2*x]^(1/4)*Cos[4*x] + 90*Sqrt[2]*Log[1 - Sqrt[2]*Cos[2*x]^(1/4) + Sqrt[Cos
[2*x]]] - 90*Sqrt[2]*Log[1 + Sqrt[2]*Cos[2*x]^(1/4) + Sqrt[Cos[2*x]]])/360

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^6(x) \tan (x)}{\cos ^{\frac {3}{4}}(2 x)} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[(Sin[x]^6*Tan[x])/Cos[2*x]^(3/4),x]

[Out]

Could not integrate

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^6*tan(x)/cos(2*x)^(3/4),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.68, size = 120, normalized size = 1.18 \[ \frac {1}{36} \, \cos \left (2 \, x\right )^{\frac {9}{4}} - \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \cos \left (2 \, x\right )^{\frac {1}{4}}\right )}\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \cos \left (2 \, x\right )^{\frac {1}{4}}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} \cos \left (2 \, x\right )^{\frac {1}{4}} + \sqrt {\cos \left (2 \, x\right )} + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} \cos \left (2 \, x\right )^{\frac {1}{4}} + \sqrt {\cos \left (2 \, x\right )} + 1\right ) - \frac {1}{5} \, \cos \left (2 \, x\right )^{\frac {5}{4}} + \frac {7}{4} \, \cos \left (2 \, x\right )^{\frac {1}{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^6*tan(x)/cos(2*x)^(3/4),x, algorithm="giac")

[Out]

1/36*cos(2*x)^(9/4) - 1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*cos(2*x)^(1/4))) - 1/2*sqrt(2)*arctan(-1/2*s
qrt(2)*(sqrt(2) - 2*cos(2*x)^(1/4))) - 1/4*sqrt(2)*log(sqrt(2)*cos(2*x)^(1/4) + sqrt(cos(2*x)) + 1) + 1/4*sqrt
(2)*log(-sqrt(2)*cos(2*x)^(1/4) + sqrt(cos(2*x)) + 1) - 1/5*cos(2*x)^(5/4) + 7/4*cos(2*x)^(1/4)

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maple [F]  time = 0.44, size = 0, normalized size = 0.00 \[\int \frac {\left (\sin ^{6}\relax (x )\right ) \tan \relax (x )}{\cos \left (2 x \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^6*tan(x)/cos(2*x)^(3/4),x)

[Out]

int(sin(x)^6*tan(x)/cos(2*x)^(3/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \relax (x)^{6} \tan \relax (x)}{\cos \left (2 \, x\right )^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^6*tan(x)/cos(2*x)^(3/4),x, algorithm="maxima")

[Out]

integrate(sin(x)^6*tan(x)/cos(2*x)^(3/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\sin \relax (x)}^6\,\mathrm {tan}\relax (x)}{{\cos \left (2\,x\right )}^{3/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(x)^6*tan(x))/cos(2*x)^(3/4),x)

[Out]

int((sin(x)^6*tan(x))/cos(2*x)^(3/4), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)**6*tan(x)/cos(2*x)**(3/4),x)

[Out]

Timed out

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