3.1250 \(\int \cot ^6(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=175 \[ \frac {5 \left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{16 d}+\frac {\left (13 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{24 d}-\frac {\left (11 a^2-18 b^2\right ) \cot (c+d x) \csc (c+d x)}{16 d}-\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{6 d}-\frac {2 a b \cot ^5(c+d x)}{5 d}+\frac {2 a b \cot ^3(c+d x)}{3 d}-\frac {2 a b \cot (c+d x)}{d}-2 a b x+\frac {b^2 \cos (c+d x)}{d} \]

[Out]

-2*a*b*x+5/16*(a^2-6*b^2)*arctanh(cos(d*x+c))/d+b^2*cos(d*x+c)/d-2*a*b*cot(d*x+c)/d+2/3*a*b*cot(d*x+c)^3/d-2/5
*a*b*cot(d*x+c)^5/d-1/16*(11*a^2-18*b^2)*cot(d*x+c)*csc(d*x+c)/d+1/24*(13*a^2-6*b^2)*cot(d*x+c)*csc(d*x+c)^3/d
-1/6*a^2*cot(d*x+c)*csc(d*x+c)^5/d

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Rubi [A]  time = 0.26, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2911, 3473, 8, 4366, 455, 1814, 1157, 388, 206} \[ \frac {5 \left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{16 d}+\frac {\left (13 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{24 d}-\frac {\left (11 a^2-18 b^2\right ) \cot (c+d x) \csc (c+d x)}{16 d}-\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{6 d}-\frac {2 a b \cot ^5(c+d x)}{5 d}+\frac {2 a b \cot ^3(c+d x)}{3 d}-\frac {2 a b \cot (c+d x)}{d}-2 a b x+\frac {b^2 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6*Csc[c + d*x]*(a + b*Sin[c + d*x])^2,x]

[Out]

-2*a*b*x + (5*(a^2 - 6*b^2)*ArcTanh[Cos[c + d*x]])/(16*d) + (b^2*Cos[c + d*x])/d - (2*a*b*Cot[c + d*x])/d + (2
*a*b*Cot[c + d*x]^3)/(3*d) - (2*a*b*Cot[c + d*x]^5)/(5*d) - ((11*a^2 - 18*b^2)*Cot[c + d*x]*Csc[c + d*x])/(16*
d) + ((13*a^2 - 6*b^2)*Cot[c + d*x]*Csc[c + d*x]^3)/(24*d) - (a^2*Cot[c + d*x]*Csc[c + d*x]^5)/(6*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 388

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

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 2911

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^2, x_Symbol] :> Dist[(2*a*b)/d, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e
+ f*x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 -
b^2, 0]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 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])

Rubi steps

\begin {align*} \int \cot ^6(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx &=(2 a b) \int \cot ^6(c+d x) \, dx+\int \cot ^6(c+d x) \csc (c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right ) \, dx\\ &=-\frac {2 a b \cot ^5(c+d x)}{5 d}-(2 a b) \int \cot ^4(c+d x) \, dx-\frac {\operatorname {Subst}\left (\int \frac {x^6 \left (a^2+b^2-b^2 x^2\right )}{\left (1-x^2\right )^4} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {2 a b \cot ^3(c+d x)}{3 d}-\frac {2 a b \cot ^5(c+d x)}{5 d}-\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{6 d}+(2 a b) \int \cot ^2(c+d x) \, dx+\frac {\operatorname {Subst}\left (\int \frac {a^2+6 a^2 x^2+6 a^2 x^4-6 b^2 x^6}{\left (1-x^2\right )^3} \, dx,x,\cos (c+d x)\right )}{6 d}\\ &=-\frac {2 a b \cot (c+d x)}{d}+\frac {2 a b \cot ^3(c+d x)}{3 d}-\frac {2 a b \cot ^5(c+d x)}{5 d}+\frac {\left (13 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{24 d}-\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{6 d}-(2 a b) \int 1 \, dx-\frac {\operatorname {Subst}\left (\int \frac {3 \left (3 a^2-2 b^2\right )+24 \left (a^2-b^2\right ) x^2-24 b^2 x^4}{\left (1-x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{24 d}\\ &=-2 a b x-\frac {2 a b \cot (c+d x)}{d}+\frac {2 a b \cot ^3(c+d x)}{3 d}-\frac {2 a b \cot ^5(c+d x)}{5 d}-\frac {\left (11 a^2-18 b^2\right ) \cot (c+d x) \csc (c+d x)}{16 d}+\frac {\left (13 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{24 d}-\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{6 d}+\frac {\operatorname {Subst}\left (\int \frac {3 \left (5 a^2-14 b^2\right )-48 b^2 x^2}{1-x^2} \, dx,x,\cos (c+d x)\right )}{48 d}\\ &=-2 a b x+\frac {b^2 \cos (c+d x)}{d}-\frac {2 a b \cot (c+d x)}{d}+\frac {2 a b \cot ^3(c+d x)}{3 d}-\frac {2 a b \cot ^5(c+d x)}{5 d}-\frac {\left (11 a^2-18 b^2\right ) \cot (c+d x) \csc (c+d x)}{16 d}+\frac {\left (13 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{24 d}-\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{6 d}+\frac {\left (5 \left (a^2-6 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{16 d}\\ &=-2 a b x+\frac {5 \left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{16 d}+\frac {b^2 \cos (c+d x)}{d}-\frac {2 a b \cot (c+d x)}{d}+\frac {2 a b \cot ^3(c+d x)}{3 d}-\frac {2 a b \cot ^5(c+d x)}{5 d}-\frac {\left (11 a^2-18 b^2\right ) \cot (c+d x) \csc (c+d x)}{16 d}+\frac {\left (13 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{24 d}-\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{6 d}\\ \end {align*}

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Mathematica [B]  time = 1.05, size = 384, normalized size = 2.19 \[ \frac {-5 a^2 \csc ^6\left (\frac {1}{2} (c+d x)\right )+60 a^2 \csc ^4\left (\frac {1}{2} (c+d x)\right )-330 a^2 \csc ^2\left (\frac {1}{2} (c+d x)\right )+5 a^2 \sec ^6\left (\frac {1}{2} (c+d x)\right )-60 a^2 \sec ^4\left (\frac {1}{2} (c+d x)\right )+330 a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )-600 a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+600 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2944 a b \tan \left (\frac {1}{2} (c+d x)\right )-2944 a b \cot \left (\frac {1}{2} (c+d x)\right )-12 a b \sin (c+d x) \csc ^6\left (\frac {1}{2} (c+d x)\right )+768 a b \sin ^6\left (\frac {1}{2} (c+d x)\right ) \csc ^5(c+d x)+164 a b \sin (c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right )-2624 a b \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)-3840 a b c-3840 a b d x+1920 b^2 \cos (c+d x)-30 b^2 \csc ^4\left (\frac {1}{2} (c+d x)\right )+540 b^2 \csc ^2\left (\frac {1}{2} (c+d x)\right )+30 b^2 \sec ^4\left (\frac {1}{2} (c+d x)\right )-540 b^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )+3600 b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-3600 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{1920 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6*Csc[c + d*x]*(a + b*Sin[c + d*x])^2,x]

[Out]

(-3840*a*b*c - 3840*a*b*d*x + 1920*b^2*Cos[c + d*x] - 2944*a*b*Cot[(c + d*x)/2] - 330*a^2*Csc[(c + d*x)/2]^2 +
540*b^2*Csc[(c + d*x)/2]^2 + 60*a^2*Csc[(c + d*x)/2]^4 - 30*b^2*Csc[(c + d*x)/2]^4 - 5*a^2*Csc[(c + d*x)/2]^6
+ 600*a^2*Log[Cos[(c + d*x)/2]] - 3600*b^2*Log[Cos[(c + d*x)/2]] - 600*a^2*Log[Sin[(c + d*x)/2]] + 3600*b^2*L
og[Sin[(c + d*x)/2]] + 330*a^2*Sec[(c + d*x)/2]^2 - 540*b^2*Sec[(c + d*x)/2]^2 - 60*a^2*Sec[(c + d*x)/2]^4 + 3
0*b^2*Sec[(c + d*x)/2]^4 + 5*a^2*Sec[(c + d*x)/2]^6 - 2624*a*b*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + 768*a*b*Csc
[c + d*x]^5*Sin[(c + d*x)/2]^6 + 164*a*b*Csc[(c + d*x)/2]^4*Sin[c + d*x] - 12*a*b*Csc[(c + d*x)/2]^6*Sin[c + d
*x] + 2944*a*b*Tan[(c + d*x)/2])/(1920*d)

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fricas [B]  time = 0.57, size = 360, normalized size = 2.06 \[ -\frac {960 \, a b d x \cos \left (d x + c\right )^{6} - 480 \, b^{2} \cos \left (d x + c\right )^{7} - 2880 \, a b d x \cos \left (d x + c\right )^{4} + 2880 \, a b d x \cos \left (d x + c\right )^{2} - 330 \, {\left (a^{2} - 6 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 960 \, a b d x + 400 \, {\left (a^{2} - 6 \, b^{2}\right )} \cos \left (d x + c\right )^{3} - 150 \, {\left (a^{2} - 6 \, b^{2}\right )} \cos \left (d x + c\right ) - 75 \, {\left ({\left (a^{2} - 6 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (a^{2} - 6 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (a^{2} - 6 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + 6 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 75 \, {\left ({\left (a^{2} - 6 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (a^{2} - 6 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (a^{2} - 6 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + 6 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 64 \, {\left (23 \, a b \cos \left (d x + c\right )^{5} - 35 \, a b \cos \left (d x + c\right )^{3} + 15 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{480 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^7*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/480*(960*a*b*d*x*cos(d*x + c)^6 - 480*b^2*cos(d*x + c)^7 - 2880*a*b*d*x*cos(d*x + c)^4 + 2880*a*b*d*x*cos(d
*x + c)^2 - 330*(a^2 - 6*b^2)*cos(d*x + c)^5 - 960*a*b*d*x + 400*(a^2 - 6*b^2)*cos(d*x + c)^3 - 150*(a^2 - 6*b
^2)*cos(d*x + c) - 75*((a^2 - 6*b^2)*cos(d*x + c)^6 - 3*(a^2 - 6*b^2)*cos(d*x + c)^4 + 3*(a^2 - 6*b^2)*cos(d*x
+ c)^2 - a^2 + 6*b^2)*log(1/2*cos(d*x + c) + 1/2) + 75*((a^2 - 6*b^2)*cos(d*x + c)^6 - 3*(a^2 - 6*b^2)*cos(d*
x + c)^4 + 3*(a^2 - 6*b^2)*cos(d*x + c)^2 - a^2 + 6*b^2)*log(-1/2*cos(d*x + c) + 1/2) - 64*(23*a*b*cos(d*x + c
)^5 - 35*a*b*cos(d*x + c)^3 + 15*a*b*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^6 - 3*d*cos(d*x + c)^4 + 3*d*
cos(d*x + c)^2 - d)

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giac [B]  time = 0.35, size = 337, normalized size = 1.93 \[ \frac {5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 24 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 45 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 30 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 280 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 225 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 480 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3840 \, {\left (d x + c\right )} a b + 2640 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 600 \, {\left (a^{2} - 6 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {3840 \, b^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + \frac {1470 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 8820 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 2640 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 225 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 480 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 280 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 45 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 30 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6}}}{1920 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^7*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/1920*(5*a^2*tan(1/2*d*x + 1/2*c)^6 + 24*a*b*tan(1/2*d*x + 1/2*c)^5 - 45*a^2*tan(1/2*d*x + 1/2*c)^4 + 30*b^2*
tan(1/2*d*x + 1/2*c)^4 - 280*a*b*tan(1/2*d*x + 1/2*c)^3 + 225*a^2*tan(1/2*d*x + 1/2*c)^2 - 480*b^2*tan(1/2*d*x
+ 1/2*c)^2 - 3840*(d*x + c)*a*b + 2640*a*b*tan(1/2*d*x + 1/2*c) - 600*(a^2 - 6*b^2)*log(abs(tan(1/2*d*x + 1/2
*c))) + 3840*b^2/(tan(1/2*d*x + 1/2*c)^2 + 1) + (1470*a^2*tan(1/2*d*x + 1/2*c)^6 - 8820*b^2*tan(1/2*d*x + 1/2*
c)^6 - 2640*a*b*tan(1/2*d*x + 1/2*c)^5 - 225*a^2*tan(1/2*d*x + 1/2*c)^4 + 480*b^2*tan(1/2*d*x + 1/2*c)^4 + 280
*a*b*tan(1/2*d*x + 1/2*c)^3 + 45*a^2*tan(1/2*d*x + 1/2*c)^2 - 30*b^2*tan(1/2*d*x + 1/2*c)^2 - 24*a*b*tan(1/2*d
*x + 1/2*c) - 5*a^2)/tan(1/2*d*x + 1/2*c)^6)/d

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maple [A]  time = 0.46, size = 318, normalized size = 1.82 \[ -\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{6 d \sin \left (d x +c \right )^{6}}+\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{24 d \sin \left (d x +c \right )^{4}}-\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{16 d \sin \left (d x +c \right )^{2}}-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{16 d}-\frac {5 a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{48 d}-\frac {5 a^{2} \cos \left (d x +c \right )}{16 d}-\frac {5 a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{16 d}-\frac {2 a b \left (\cot ^{5}\left (d x +c \right )\right )}{5 d}+\frac {2 a b \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}-\frac {2 a b \cot \left (d x +c \right )}{d}-2 a b x -\frac {2 a b c}{d}-\frac {b^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{4}}+\frac {3 b^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{2}}+\frac {3 b^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{8 d}+\frac {5 b^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{8 d}+\frac {15 b^{2} \cos \left (d x +c \right )}{8 d}+\frac {15 b^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^7*(a+b*sin(d*x+c))^2,x)

[Out]

-1/6/d*a^2/sin(d*x+c)^6*cos(d*x+c)^7+1/24/d*a^2/sin(d*x+c)^4*cos(d*x+c)^7-1/16/d*a^2/sin(d*x+c)^2*cos(d*x+c)^7
-1/16*a^2*cos(d*x+c)^5/d-5/48*a^2*cos(d*x+c)^3/d-5/16*a^2*cos(d*x+c)/d-5/16/d*a^2*ln(csc(d*x+c)-cot(d*x+c))-2/
5*a*b*cot(d*x+c)^5/d+2/3*a*b*cot(d*x+c)^3/d-2*a*b*cot(d*x+c)/d-2*a*b*x-2/d*a*b*c-1/4/d*b^2/sin(d*x+c)^4*cos(d*
x+c)^7+3/8/d*b^2/sin(d*x+c)^2*cos(d*x+c)^7+3/8*b^2*cos(d*x+c)^5/d+5/8*b^2*cos(d*x+c)^3/d+15/8*b^2*cos(d*x+c)/d
+15/8/d*b^2*ln(csc(d*x+c)-cot(d*x+c))

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maxima [A]  time = 0.42, size = 219, normalized size = 1.25 \[ -\frac {64 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{\tan \left (d x + c\right )^{5}}\right )} a b - 5 \, a^{2} {\left (\frac {2 \, {\left (33 \, \cos \left (d x + c\right )^{5} - 40 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1} + 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 30 \, b^{2} {\left (\frac {2 \, {\left (9 \, \cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 16 \, \cos \left (d x + c\right ) + 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^7*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/480*(64*(15*d*x + 15*c + (15*tan(d*x + c)^4 - 5*tan(d*x + c)^2 + 3)/tan(d*x + c)^5)*a*b - 5*a^2*(2*(33*cos(
d*x + c)^5 - 40*cos(d*x + c)^3 + 15*cos(d*x + c))/(cos(d*x + c)^6 - 3*cos(d*x + c)^4 + 3*cos(d*x + c)^2 - 1) +
15*log(cos(d*x + c) + 1) - 15*log(cos(d*x + c) - 1)) + 30*b^2*(2*(9*cos(d*x + c)^3 - 7*cos(d*x + c))/(cos(d*x
+ c)^4 - 2*cos(d*x + c)^2 + 1) - 16*cos(d*x + c) + 15*log(cos(d*x + c) + 1) - 15*log(cos(d*x + c) - 1)))/d

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mupad [B]  time = 15.06, size = 985, normalized size = 5.63 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^6*(a + b*sin(c + d*x))^2)/sin(c + d*x)^7,x)

[Out]

(5*a^2*sin(c/2 + (d*x)/2)^14 - 5*a^2*cos(c/2 + (d*x)/2)^14 - 40*a^2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^12
+ 180*a^2*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^10 + 225*a^2*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^8 - 22
5*a^2*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^6 - 180*a^2*cos(c/2 + (d*x)/2)^10*sin(c/2 + (d*x)/2)^4 + 40*a^2*
cos(c/2 + (d*x)/2)^12*sin(c/2 + (d*x)/2)^2 + 30*b^2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^12 - 450*b^2*cos(c
/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^10 - 480*b^2*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^8 + 4320*b^2*cos(c/2 +
(d*x)/2)^8*sin(c/2 + (d*x)/2)^6 + 450*b^2*cos(c/2 + (d*x)/2)^10*sin(c/2 + (d*x)/2)^4 - 30*b^2*cos(c/2 + (d*x)
/2)^12*sin(c/2 + (d*x)/2)^2 + 24*a*b*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^13 - 24*a*b*cos(c/2 + (d*x)/2)^13*s
in(c/2 + (d*x)/2) - 600*a^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)
^8 - 600*a^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^6 + 3600*b^2*l
og(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^8 + 3600*b^2*log(sin(c/2 + (
d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^6 - 256*a*b*cos(c/2 + (d*x)/2)^3*sin(c/2 +
(d*x)/2)^11 + 2360*a*b*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^9 - 2360*a*b*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d
*x)/2)^5 + 256*a*b*cos(c/2 + (d*x)/2)^11*sin(c/2 + (d*x)/2)^3 + 7680*a*b*atan((5*a^2*sin(c/2 + (d*x)/2) - 30*b
^2*sin(c/2 + (d*x)/2) + 32*a*b*cos(c/2 + (d*x)/2))/(30*b^2*cos(c/2 + (d*x)/2) - 5*a^2*cos(c/2 + (d*x)/2) + 32*
a*b*sin(c/2 + (d*x)/2)))*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^8 + 7680*a*b*atan((5*a^2*sin(c/2 + (d*x)/2) -
30*b^2*sin(c/2 + (d*x)/2) + 32*a*b*cos(c/2 + (d*x)/2))/(30*b^2*cos(c/2 + (d*x)/2) - 5*a^2*cos(c/2 + (d*x)/2)
+ 32*a*b*sin(c/2 + (d*x)/2)))*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^6)/(1920*d*cos(c/2 + (d*x)/2)^6*sin(c/2
+ (d*x)/2)^6*(cos(c/2 + (d*x)/2)^2 + sin(c/2 + (d*x)/2)^2))

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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(cos(d*x+c)**6*csc(d*x+c)**7*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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