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 \[ -\frac{2 \cos ^3(x)}{3 \cos ^{\frac{3}{2}}(2 x)}-\frac{11 \cos (x)}{20 \cos ^{\frac{3}{2}}(2 x)}+\frac{63 \cos (x)}{20 \sqrt{\cos (2 x)}}+\frac{3 \sin ^2(x) \cos (x)}{10 \cos ^{\frac{5}{2}}(2 x)}-\frac{\tanh ^{-1}\left (\frac{\sqrt{2} \cos (x)}{\sqrt{\cos (2 x)}}\right )}{\sqrt{2}} \]
[Out]
-(ArcTanh[(Sqrt[2]*Cos[x])/Sqrt[Cos[2*x]]]/Sqrt[2]) - (11*Cos[x])/(20*Cos[2*x]^(3/2)) - (2*Cos[x]^3)/(3*Cos[2*
x]^(3/2)) + (63*Cos[x])/(20*Sqrt[Cos[2*x]]) + (3*Cos[x]*Sin[x]^2)/(10*Cos[2*x]^(5/2))
________________________________________________________________________________________
Rubi [A] time = 0.207827, 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 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])
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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 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 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])
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 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 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]
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*}
Mathematica [A] time = 0.200984, 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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Maple [B] time = 0.168, size = 180, normalized size = 2.1 \begin{align*} -{\frac{1}{240\, \left ( \sin \left ( x \right ) \right ) ^{6}-360\, \left ( \sin \left ( x \right ) \right ) ^{4}+180\, \left ( \sin \left ( x \right ) \right ) ^{2}-30} \left ( 120\,\ln \left ( \cos \left ( x \right ) \sqrt{2}+\sqrt{-2\, \left ( \sin \left ( x \right ) \right ) ^{2}+1} \right ) \sqrt{2} \left ( \sin \left ( x \right ) \right ) ^{6}+338\,\sqrt{-2\, \left ( \sin \left ( x \right ) \right ) ^{2}+1}\cos \left ( x \right ) \left ( \sin \left ( x \right ) \right ) ^{4}-180\,\ln \left ( \cos \left ( x \right ) \sqrt{2}+\sqrt{-2\, \left ( \sin \left ( x \right ) \right ) ^{2}+1} \right ) \sqrt{2} \left ( \sin \left ( x \right ) \right ) ^{4}-276\,\sqrt{-2\, \left ( \sin \left ( x \right ) \right ) ^{2}+1} \left ( \sin \left ( x \right ) \right ) ^{2}\cos \left ( x \right ) +90\,\ln \left ( \cos \left ( x \right ) \sqrt{2}+\sqrt{-2\, \left ( \sin \left ( x \right ) \right ) ^{2}+1} \right ) \sqrt{2} \left ( \sin \left ( x \right ) \right ) ^{2}+58\,\sqrt{-2\, \left ( \sin \left ( x \right ) \right ) ^{2}+1}\cos \left ( x \right ) -15\,\ln \left ( \cos \left ( x \right ) \sqrt{2}+\sqrt{-2\, \left ( \sin \left ( x \right ) \right ) ^{2}+1} \right ) \sqrt{2} \right ) } \end{align*}
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)
[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 = 2.03273, size = 1835, normalized size = 21.09 \begin{align*} \text{result too large to display} \end{align*}
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))
________________________________________________________________________________________
Fricas [B] time = 3.79861, size = 506, normalized size = 5.82 \begin{align*} \frac{15 \,{\left (8 \, \sqrt{2} \cos \left (x\right )^{6} - 12 \, \sqrt{2} \cos \left (x\right )^{4} + 6 \, \sqrt{2} \cos \left (x\right )^{2} - \sqrt{2}\right )} \log \left (2048 \, \cos \left (x\right )^{8} - 2048 \, \cos \left (x\right )^{6} + 640 \, \cos \left (x\right )^{4} - 64 \, \cos \left (x\right )^{2} - 8 \,{\left (128 \, \sqrt{2} \cos \left (x\right )^{7} - 96 \, \sqrt{2} \cos \left (x\right )^{5} + 20 \, \sqrt{2} \cos \left (x\right )^{3} - \sqrt{2} \cos \left (x\right )\right )} \sqrt{2 \, \cos \left (x\right )^{2} - 1} + 1\right ) + 16 \,{\left (169 \, \cos \left (x\right )^{5} - 200 \, \cos \left (x\right )^{3} + 60 \, \cos \left (x\right )\right )} \sqrt{2 \, \cos \left (x\right )^{2} - 1}}{240 \,{\left (8 \, \cos \left (x\right )^{6} - 12 \, \cos \left (x\right )^{4} + 6 \, \cos \left (x\right )^{2} - 1\right )}} \end{align*}
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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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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
________________________________________________________________________________________
Giac [A] time = 1.13867, size = 74, normalized size = 0.85 \begin{align*} \frac{1}{2} \, \sqrt{2} \log \left ({\left | -\sqrt{2} \cos \left (x\right ) + \sqrt{2 \, \cos \left (x\right )^{2} - 1} \right |}\right ) + \frac{{\left ({\left (169 \, \cos \left (x\right )^{2} - 200\right )} \cos \left (x\right )^{2} + 60\right )} \cos \left (x\right )}{15 \,{\left (2 \, \cos \left (x\right )^{2} - 1\right )}^{\frac{5}{2}}} \end{align*}
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)