3.433 \(\int \frac {\sin ^2(x) (3 \sin ^3(x)-\cos (x) \sin (4 x))}{\cos ^{\frac {7}{2}}(2 x)} \, dx\)

Optimal. Leaf size=87 \[ -\frac {11 \cos (x)}{20 \cos ^{\frac {3}{2}}(2 x)}-\frac {2 \cos ^3(x)}{3 \cos ^{\frac {3}{2}}(2 x)}+\frac {63 \cos (x)}{20 \sqrt {\cos (2 x)}}+\frac {3 \sin ^2(x) \cos (x)}{10 \cos ^{\frac {5}{2}}(2 x)}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \cos (x)}{\sqrt {\cos (2 x)}}\right )}{\sqrt {2}} \]

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Rubi [A]  time = 0.21, antiderivative size = 91, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {4377, 12, 452, 288, 217, 206, 4366, 378, 191} \[ -\frac {2 \cos ^3(x)}{3 \cos ^{\frac {3}{2}}(2 x)}+\frac {13 \cos (x)}{5 \sqrt {\cos (2 x)}}+\frac {3 \sin ^4(x) \cos (x)}{5 \cos ^{\frac {5}{2}}(2 x)}-\frac {4 \sin ^2(x) \cos (x)}{5 \cos ^{\frac {3}{2}}(2 x)}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \cos (x)}{\sqrt {\cos (2 x)}}\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[x]^2*(3*Sin[x]^3 - Cos[x]*Sin[4*x]))/Cos[2*x]^(7/2),x]

[Out]

-(ArcTanh[(Sqrt[2]*Cos[x])/Sqrt[Cos[2*x]]]/Sqrt[2]) - (2*Cos[x]^3)/(3*Cos[2*x]^(3/2)) + (13*Cos[x])/(5*Sqrt[Co
s[2*x]]) - (4*Cos[x]*Sin[x]^2)/(5*Cos[2*x]^(3/2)) + (3*Cos[x]*Sin[x]^4)/(5*Cos[2*x]^(5/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 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])

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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 378

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

Rule 452

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

Rule 4366

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_), x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dis
t[d/(b*c), Subst[Int[SubstFor[(1 - d^2*x^2)^((n - 1)/2), 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] && IntegerQ[(n - 1)/2] && NonsumQ[u] &&
(EqQ[F, Sin] || EqQ[F, sin])

Rule 4377

Int[(u_)*((v_) + (d_.)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_.)), x_Symbol] :> With[{e = FreeFactors[Cos[c*(a +
b*x)], x]}, Int[ActivateTrig[u*v], x] + Dist[d, Int[ActivateTrig[u]*Sin[c*(a + b*x)]^n, x], x] /; FunctionOfQ[
Cos[c*(a + b*x)]/e, u, x]] /; FreeQ[{a, b, c, d}, x] &&  !FreeQ[v, x] && IntegerQ[(n - 1)/2] && NonsumQ[u] &&
(EqQ[F, Sin] || EqQ[F, sin])

Rubi steps

\begin {align*} \int \frac {\sin ^2(x) \left (3 \sin ^3(x)-\cos (x) \sin (4 x)\right )}{\cos ^{\frac {7}{2}}(2 x)} \, dx &=3 \int \frac {\sin ^5(x)}{\cos ^{\frac {7}{2}}(2 x)} \, dx-\int \frac {\cos (x) \sin ^2(x) \sin (4 x)}{\cos ^{\frac {7}{2}}(2 x)} \, dx\\ &=-\left (3 \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2}{\left (-1+2 x^2\right )^{7/2}} \, dx,x,\cos (x)\right )\right )+\operatorname {Subst}\left (\int \frac {4 x^2 \left (1-x^2\right )}{\left (-1+2 x^2\right )^{5/2}} \, dx,x,\cos (x)\right )\\ &=\frac {3 \cos (x) \sin ^4(x)}{5 \cos ^{\frac {5}{2}}(2 x)}+\frac {12}{5} \operatorname {Subst}\left (\int \frac {1-x^2}{\left (-1+2 x^2\right )^{5/2}} \, dx,x,\cos (x)\right )+4 \operatorname {Subst}\left (\int \frac {x^2 \left (1-x^2\right )}{\left (-1+2 x^2\right )^{5/2}} \, dx,x,\cos (x)\right )\\ &=-\frac {2 \cos ^3(x)}{3 \cos ^{\frac {3}{2}}(2 x)}-\frac {4 \cos (x) \sin ^2(x)}{5 \cos ^{\frac {3}{2}}(2 x)}+\frac {3 \cos (x) \sin ^4(x)}{5 \cos ^{\frac {5}{2}}(2 x)}-\frac {8}{5} \operatorname {Subst}\left (\int \frac {1}{\left (-1+2 x^2\right )^{3/2}} \, dx,x,\cos (x)\right )-2 \operatorname {Subst}\left (\int \frac {x^2}{\left (-1+2 x^2\right )^{3/2}} \, dx,x,\cos (x)\right )\\ &=-\frac {2 \cos ^3(x)}{3 \cos ^{\frac {3}{2}}(2 x)}+\frac {13 \cos (x)}{5 \sqrt {\cos (2 x)}}-\frac {4 \cos (x) \sin ^2(x)}{5 \cos ^{\frac {3}{2}}(2 x)}+\frac {3 \cos (x) \sin ^4(x)}{5 \cos ^{\frac {5}{2}}(2 x)}-\operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+2 x^2}} \, dx,x,\cos (x)\right )\\ &=-\frac {2 \cos ^3(x)}{3 \cos ^{\frac {3}{2}}(2 x)}+\frac {13 \cos (x)}{5 \sqrt {\cos (2 x)}}-\frac {4 \cos (x) \sin ^2(x)}{5 \cos ^{\frac {3}{2}}(2 x)}+\frac {3 \cos (x) \sin ^4(x)}{5 \cos ^{\frac {5}{2}}(2 x)}-\operatorname {Subst}\left (\int \frac {1}{1-2 x^2} \, dx,x,\frac {\cos (x)}{\sqrt {\cos (2 x)}}\right )\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \cos (x)}{\sqrt {\cos (2 x)}}\right )}{\sqrt {2}}-\frac {2 \cos ^3(x)}{3 \cos ^{\frac {3}{2}}(2 x)}+\frac {13 \cos (x)}{5 \sqrt {\cos (2 x)}}-\frac {4 \cos (x) \sin ^2(x)}{5 \cos ^{\frac {3}{2}}(2 x)}+\frac {3 \cos (x) \sin ^4(x)}{5 \cos ^{\frac {5}{2}}(2 x)}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 62, normalized size = 0.71 \[ \frac {250 \cos (x)+45 \cos (3 x)+169 \cos (5 x)-120 \sqrt {2} \cos ^{\frac {5}{2}}(2 x) \log \left (\sqrt {2} \cos (x)+\sqrt {\cos (2 x)}\right )}{240 \cos ^{\frac {5}{2}}(2 x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[x]^2*(3*Sin[x]^3 - Cos[x]*Sin[4*x]))/Cos[2*x]^(7/2),x]

[Out]

(250*Cos[x] + 45*Cos[3*x] + 169*Cos[5*x] - 120*Sqrt[2]*Cos[2*x]^(5/2)*Log[Sqrt[2]*Cos[x] + Sqrt[Cos[2*x]]])/(2
40*Cos[2*x]^(5/2))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^2(x) \left (3 \sin ^3(x)-\cos (x) \sin (4 x)\right )}{\cos ^{\frac {7}{2}}(2 x)} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[(Sin[x]^2*(3*Sin[x]^3 - Cos[x]*Sin[4*x]))/Cos[2*x]^(7/2),x]

[Out]

Could not integrate

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fricas [B]  time = 1.37, size = 163, normalized size = 1.87 \[ \frac {15 \, {\left (8 \, \sqrt {2} \cos \relax (x)^{6} - 12 \, \sqrt {2} \cos \relax (x)^{4} + 6 \, \sqrt {2} \cos \relax (x)^{2} - \sqrt {2}\right )} \log \left (2048 \, \cos \relax (x)^{8} - 2048 \, \cos \relax (x)^{6} + 640 \, \cos \relax (x)^{4} - 64 \, \cos \relax (x)^{2} - 8 \, {\left (128 \, \sqrt {2} \cos \relax (x)^{7} - 96 \, \sqrt {2} \cos \relax (x)^{5} + 20 \, \sqrt {2} \cos \relax (x)^{3} - \sqrt {2} \cos \relax (x)\right )} \sqrt {2 \, \cos \relax (x)^{2} - 1} + 1\right ) + 16 \, {\left (169 \, \cos \relax (x)^{5} - 200 \, \cos \relax (x)^{3} + 60 \, \cos \relax (x)\right )} \sqrt {2 \, \cos \relax (x)^{2} - 1}}{240 \, {\left (8 \, \cos \relax (x)^{6} - 12 \, \cos \relax (x)^{4} + 6 \, \cos \relax (x)^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*sin(x)^3-cos(x)*sin(4*x))/cos(2*x)^(7/2)/csc(x)^2,x, algorithm="fricas")

[Out]

1/240*(15*(8*sqrt(2)*cos(x)^6 - 12*sqrt(2)*cos(x)^4 + 6*sqrt(2)*cos(x)^2 - sqrt(2))*log(2048*cos(x)^8 - 2048*c
os(x)^6 + 640*cos(x)^4 - 64*cos(x)^2 - 8*(128*sqrt(2)*cos(x)^7 - 96*sqrt(2)*cos(x)^5 + 20*sqrt(2)*cos(x)^3 - s
qrt(2)*cos(x))*sqrt(2*cos(x)^2 - 1) + 1) + 16*(169*cos(x)^5 - 200*cos(x)^3 + 60*cos(x))*sqrt(2*cos(x)^2 - 1))/
(8*cos(x)^6 - 12*cos(x)^4 + 6*cos(x)^2 - 1)

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giac [A]  time = 0.73, size = 55, normalized size = 0.63 \[ \frac {1}{2} \, \sqrt {2} \log \left ({\left | -\sqrt {2} \cos \relax (x) + \sqrt {2 \, \cos \relax (x)^{2} - 1} \right |}\right ) + \frac {{\left ({\left (169 \, \cos \relax (x)^{2} - 200\right )} \cos \relax (x)^{2} + 60\right )} \cos \relax (x)}{15 \, {\left (2 \, \cos \relax (x)^{2} - 1\right )}^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*sin(x)^3-cos(x)*sin(4*x))/cos(2*x)^(7/2)/csc(x)^2,x, algorithm="giac")

[Out]

1/2*sqrt(2)*log(abs(-sqrt(2)*cos(x) + sqrt(2*cos(x)^2 - 1))) + 1/15*((169*cos(x)^2 - 200)*cos(x)^2 + 60)*cos(x
)/(2*cos(x)^2 - 1)^(5/2)

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maple [B]  time = 0.41, size = 180, normalized size = 2.07

method	result	size
default	\(-\frac {120 \ln \left (\cos \relax (x ) \sqrt {2}+\sqrt {-2 \left (\sin ^{2}\relax (x )\right )+1}\right ) \sqrt {2}\, \left (\sin ^{6}\relax (x )\right )+338 \sqrt {-2 \left (\sin ^{2}\relax (x )\right )+1}\, \cos \relax (x ) \left (\sin ^{4}\relax (x )\right )-180 \ln \left (\cos \relax (x ) \sqrt {2}+\sqrt {-2 \left (\sin ^{2}\relax (x )\right )+1}\right ) \sqrt {2}\, \left (\sin ^{4}\relax (x )\right )-276 \sqrt {-2 \left (\sin ^{2}\relax (x )\right )+1}\, \left (\sin ^{2}\relax (x )\right ) \cos \relax (x )+90 \ln \left (\cos \relax (x ) \sqrt {2}+\sqrt {-2 \left (\sin ^{2}\relax (x )\right )+1}\right ) \sqrt {2}\, \left (\sin ^{2}\relax (x )\right )+58 \sqrt {-2 \left (\sin ^{2}\relax (x )\right )+1}\, \cos \relax (x )-15 \ln \left (\cos \relax (x ) \sqrt {2}+\sqrt {-2 \left (\sin ^{2}\relax (x )\right )+1}\right ) \sqrt {2}}{30 \left (8 \left (\sin ^{6}\relax (x )\right )-12 \left (\sin ^{4}\relax (x )\right )+6 \left (\sin ^{2}\relax (x )\right )-1\right )}\)	\(180\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*sin(x)^3-cos(x)*sin(4*x))/cos(2*x)^(7/2)/csc(x)^2,x,method=_RETURNVERBOSE)

[Out]

-1/30/(8*sin(x)^6-12*sin(x)^4+6*sin(x)^2-1)*(120*ln(cos(x)*2^(1/2)+(-2*sin(x)^2+1)^(1/2))*2^(1/2)*sin(x)^6+338
*(-2*sin(x)^2+1)^(1/2)*cos(x)*sin(x)^4-180*ln(cos(x)*2^(1/2)+(-2*sin(x)^2+1)^(1/2))*2^(1/2)*sin(x)^4-276*(-2*s
in(x)^2+1)^(1/2)*sin(x)^2*cos(x)+90*ln(cos(x)*2^(1/2)+(-2*sin(x)^2+1)^(1/2))*2^(1/2)*sin(x)^2+58*(-2*sin(x)^2+
1)^(1/2)*cos(x)-15*ln(cos(x)*2^(1/2)+(-2*sin(x)^2+1)^(1/2))*2^(1/2))

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maxima [B]  time = 1.61, size = 1359, normalized size = 15.62 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*sin(x)^3-cos(x)*sin(4*x))/cos(2*x)^(7/2)/csc(x)^2,x, algorithm="maxima")

[Out]

1/48*(4*(4*sqrt(2)*sin(4*x)*sin(5/2*arctan2(sin(4*x), cos(4*x))) + 4*(sqrt(2)*cos(4*x) + sqrt(2))*cos(5/2*arct
an2(sin(4*x), cos(4*x))) + 3*sqrt(2)*cos(8*x) + 7*sqrt(2)*cos(4*x) + 4*sqrt(2))*cos(5/2*arctan2(sin(4*x), cos(
4*x) + 1)) + 12*sqrt(2)*sqrt(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)*cos(3/2*arctan2(sin(4*x), cos(4*x) + 1)
) - 12*(sqrt(2)*cos(4*x)^2 + sqrt(2)*sin(4*x)^2 + 2*sqrt(2)*cos(4*x) + sqrt(2))*cos(1/2*arctan2(sin(4*x), cos(
4*x) + 1)) - 4*(4*sqrt(2)*cos(5/2*arctan2(sin(4*x), cos(4*x)))*sin(4*x) - 4*(sqrt(2)*cos(4*x) + sqrt(2))*sin(5
/2*arctan2(sin(4*x), cos(4*x))) - 3*sqrt(2)*sin(8*x) - 7*sqrt(2)*sin(4*x))*sin(5/2*arctan2(sin(4*x), cos(4*x)
+ 1)) - 3*(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1/4)*((sqrt(2)*cos(4*x)^2 + sqrt(2)*sin(4*x)^2 + 2*sqrt(
2)*cos(4*x) + sqrt(2))*log(sqrt(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)*cos(1/2*arctan2(sin(4*x), cos(4*x) +
1))^2 + sqrt(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)*sin(1/2*arctan2(sin(4*x), cos(4*x) + 1))^2 + 2*(cos(4*
x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1/4)*cos(1/2*arctan2(sin(4*x), cos(4*x) + 1)) + 1) - (sqrt(2)*cos(4*x)^2
+ sqrt(2)*sin(4*x)^2 + 2*sqrt(2)*cos(4*x) + sqrt(2))*log(sqrt(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)*cos(1/
2*arctan2(sin(4*x), cos(4*x) + 1))^2 + sqrt(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)*sin(1/2*arctan2(sin(4*x)
, cos(4*x) + 1))^2 - 2*(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1/4)*cos(1/2*arctan2(sin(4*x), cos(4*x) + 1
)) + 1) + (sqrt(2)*cos(4*x)^2 + sqrt(2)*sin(4*x)^2 + 2*sqrt(2)*cos(4*x) + sqrt(2))*log(((cos(1/2*arctan2(sin(4
*x), cos(4*x)))^2 + sin(1/2*arctan2(sin(4*x), cos(4*x)))^2)*cos(1/2*arctan2(sin(4*x), cos(4*x) + 1))^2 + (cos(
1/2*arctan2(sin(4*x), cos(4*x)))^2 + sin(1/2*arctan2(sin(4*x), cos(4*x)))^2)*sin(1/2*arctan2(sin(4*x), cos(4*x
) + 1))^2)*sqrt(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1) + 2*(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1/4)
*(cos(1/2*arctan2(sin(4*x), cos(4*x) + 1))*cos(1/2*arctan2(sin(4*x), cos(4*x))) + sin(1/2*arctan2(sin(4*x), co
s(4*x) + 1))*sin(1/2*arctan2(sin(4*x), cos(4*x)))) + 1) - (sqrt(2)*cos(4*x)^2 + sqrt(2)*sin(4*x)^2 + 2*sqrt(2)
*cos(4*x) + sqrt(2))*log(((cos(1/2*arctan2(sin(4*x), cos(4*x)))^2 + sin(1/2*arctan2(sin(4*x), cos(4*x)))^2)*co
s(1/2*arctan2(sin(4*x), cos(4*x) + 1))^2 + (cos(1/2*arctan2(sin(4*x), cos(4*x)))^2 + sin(1/2*arctan2(sin(4*x),
cos(4*x)))^2)*sin(1/2*arctan2(sin(4*x), cos(4*x) + 1))^2)*sqrt(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1) - 2*
(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1/4)*(cos(1/2*arctan2(sin(4*x), cos(4*x) + 1))*cos(1/2*arctan2(sin
(4*x), cos(4*x))) + sin(1/2*arctan2(sin(4*x), cos(4*x) + 1))*sin(1/2*arctan2(sin(4*x), cos(4*x)))) + 1)))/(cos
(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(5/4) + 1/20*(((15*cos(8*x) + 70*cos(4*x) + 43)*cos(5/2*arctan2(sin(4*x
), cos(4*x))) + 5*(3*sin(8*x) + 14*sin(4*x))*sin(5/2*arctan2(sin(4*x), cos(4*x))) - 12)*cos(5/2*arctan2(sin(4*
x), cos(4*x) + 1)) + 15*(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)*cos(1/2*arctan2(sin(4*x), cos(4*x) + 1)) -
(5*(3*sin(8*x) + 14*sin(4*x))*cos(5/2*arctan2(sin(4*x), cos(4*x))) - (15*cos(8*x) + 70*cos(4*x) + 43)*sin(5/2*
arctan2(sin(4*x), cos(4*x))))*sin(5/2*arctan2(sin(4*x), cos(4*x) + 1)) + 40*sqrt(cos(4*x)^2 + sin(4*x)^2 + 2*c
os(4*x) + 1)*cos(3/2*arctan2(sin(4*x), cos(4*x) + 1)))/((sqrt(2)*cos(4*x)^2 + sqrt(2)*sin(4*x)^2 + 2*sqrt(2)*c
os(4*x) + sqrt(2))*(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1/4))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\sin \relax (x)}^2\,\left (3\,{\sin \relax (x)}^3-\sin \left (4\,x\right )\,\cos \relax (x)\right )}{{\cos \left (2\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(x)^2*(3*sin(x)^3 - sin(4*x)*cos(x)))/cos(2*x)^(7/2),x)

[Out]

int((sin(x)^2*(3*sin(x)^3 - sin(4*x)*cos(x)))/cos(2*x)^(7/2), 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((3*sin(x)**3-cos(x)*sin(4*x))/cos(2*x)**(7/2)/csc(x)**2,x)

[Out]

Timed out

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