3.1246 \(\int \cos ^3(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=180 \[ -\frac {\left (a^2-b^2\right ) \cos ^3(c+d x)}{3 d}-\frac {\left (2 a^2-b^2\right ) \cos (c+d x)}{d}+\frac {\left (5 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {15 a b \cot (c+d x)}{4 d}+\frac {a b \cos ^4(c+d x) \cot (c+d x)}{2 d}+\frac {5 a b \cos ^2(c+d x) \cot (c+d x)}{4 d}-\frac {15 a b x}{4}+\frac {b^2 \cos ^5(c+d x)}{5 d} \]

[Out]

-15/4*a*b*x+1/2*(5*a^2-2*b^2)*arctanh(cos(d*x+c))/d-(2*a^2-b^2)*cos(d*x+c)/d-1/3*(a^2-b^2)*cos(d*x+c)^3/d+1/5*
b^2*cos(d*x+c)^5/d-15/4*a*b*cot(d*x+c)/d+5/4*a*b*cos(d*x+c)^2*cot(d*x+c)/d+1/2*a*b*cos(d*x+c)^4*cot(d*x+c)/d-1
/2*a^2*cot(d*x+c)*csc(d*x+c)/d

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Rubi [A]  time = 0.30, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2911, 2591, 288, 321, 203, 455, 1810, 206} \[ -\frac {\left (a^2-b^2\right ) \cos ^3(c+d x)}{3 d}-\frac {\left (2 a^2-b^2\right ) \cos (c+d x)}{d}+\frac {\left (5 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {15 a b \cot (c+d x)}{4 d}+\frac {a b \cos ^4(c+d x) \cot (c+d x)}{2 d}+\frac {5 a b \cos ^2(c+d x) \cot (c+d x)}{4 d}-\frac {15 a b x}{4}+\frac {b^2 \cos ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*Cot[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]

[Out]

(-15*a*b*x)/4 + ((5*a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]])/(2*d) - ((2*a^2 - b^2)*Cos[c + d*x])/d - ((a^2 - b^2)*
Cos[c + d*x]^3)/(3*d) + (b^2*Cos[c + d*x]^5)/(5*d) - (15*a*b*Cot[c + d*x])/(4*d) + (5*a*b*Cos[c + d*x]^2*Cot[c
+ d*x])/(4*d) + (a*b*Cos[c + d*x]^4*Cot[c + d*x])/(2*d) - (a^2*Cot[c + d*x]*Csc[c + d*x])/(2*d)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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 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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

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]

Rubi steps

\begin {align*} \int \cos ^3(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^2 \, dx &=(2 a b) \int \cos ^4(c+d x) \cot ^2(c+d x) \, dx+\int \cos ^3(c+d x) \cot ^3(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right ) \, dx\\ &=-\frac {\operatorname {Subst}\left (\int \frac {x^6 \left (a^2+b^2-b^2 x^2\right )}{\left (1-x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{d}-\frac {(2 a b) \operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^2\right )^3} \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac {a b \cos ^4(c+d x) \cot (c+d x)}{2 d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {\operatorname {Subst}\left (\int \frac {a^2+2 a^2 x^2+2 a^2 x^4-2 b^2 x^6}{1-x^2} \, dx,x,\cos (c+d x)\right )}{2 d}-\frac {(5 a b) \operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=\frac {5 a b \cos ^2(c+d x) \cot (c+d x)}{4 d}+\frac {a b \cos ^4(c+d x) \cot (c+d x)}{2 d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {\operatorname {Subst}\left (\int \left (-2 \left (2 a^2-b^2\right )-2 \left (a^2-b^2\right ) x^2+2 b^2 x^4+\frac {5 a^2-2 b^2}{1-x^2}\right ) \, dx,x,\cos (c+d x)\right )}{2 d}-\frac {(15 a b) \operatorname {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,\cot (c+d x)\right )}{4 d}\\ &=-\frac {\left (2 a^2-b^2\right ) \cos (c+d x)}{d}-\frac {\left (a^2-b^2\right ) \cos ^3(c+d x)}{3 d}+\frac {b^2 \cos ^5(c+d x)}{5 d}-\frac {15 a b \cot (c+d x)}{4 d}+\frac {5 a b \cos ^2(c+d x) \cot (c+d x)}{4 d}+\frac {a b \cos ^4(c+d x) \cot (c+d x)}{2 d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {(15 a b) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{4 d}+\frac {\left (5 a^2-2 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{2 d}\\ &=-\frac {15}{4} a b x+\frac {\left (5 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {\left (2 a^2-b^2\right ) \cos (c+d x)}{d}-\frac {\left (a^2-b^2\right ) \cos ^3(c+d x)}{3 d}+\frac {b^2 \cos ^5(c+d x)}{5 d}-\frac {15 a b \cot (c+d x)}{4 d}+\frac {5 a b \cos ^2(c+d x) \cot (c+d x)}{4 d}+\frac {a b \cos ^4(c+d x) \cot (c+d x)}{2 d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 6.19, size = 250, normalized size = 1.39 \[ -\frac {\left (18 a^2-11 b^2\right ) \cos (c+d x)}{8 d}-\frac {\left (4 a^2-7 b^2\right ) \cos (3 (c+d x))}{48 d}+\frac {\left (2 b^2-5 a^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {\left (5 a^2-2 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}-\frac {a^2 \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}-\frac {15 a b (c+d x)}{4 d}-\frac {a b \sin (2 (c+d x))}{d}-\frac {a b \sin (4 (c+d x))}{16 d}+\frac {a b \tan \left (\frac {1}{2} (c+d x)\right )}{d}-\frac {a b \cot \left (\frac {1}{2} (c+d x)\right )}{d}+\frac {b^2 \cos (5 (c+d x))}{80 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*Cot[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]

[Out]

(-15*a*b*(c + d*x))/(4*d) - ((18*a^2 - 11*b^2)*Cos[c + d*x])/(8*d) - ((4*a^2 - 7*b^2)*Cos[3*(c + d*x)])/(48*d)
+ (b^2*Cos[5*(c + d*x)])/(80*d) - (a*b*Cot[(c + d*x)/2])/d - (a^2*Csc[(c + d*x)/2]^2)/(8*d) + ((5*a^2 - 2*b^2
)*Log[Cos[(c + d*x)/2]])/(2*d) + ((-5*a^2 + 2*b^2)*Log[Sin[(c + d*x)/2]])/(2*d) + (a^2*Sec[(c + d*x)/2]^2)/(8*
d) - (a*b*Sin[2*(c + d*x)])/d - (a*b*Sin[4*(c + d*x)])/(16*d) + (a*b*Tan[(c + d*x)/2])/d

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fricas [A]  time = 1.24, size = 244, normalized size = 1.36 \[ \frac {12 \, b^{2} \cos \left (d x + c\right )^{7} - 225 \, a b d x \cos \left (d x + c\right )^{2} - 4 \, {\left (5 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{5} + 225 \, a b d x - 20 \, {\left (5 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 30 \, {\left (5 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right ) + 15 \, {\left ({\left (5 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 5 \, a^{2} + 2 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 15 \, {\left ({\left (5 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 5 \, a^{2} + 2 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 15 \, {\left (2 \, a b \cos \left (d x + c\right )^{5} + 5 \, a b \cos \left (d x + c\right )^{3} - 15 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/60*(12*b^2*cos(d*x + c)^7 - 225*a*b*d*x*cos(d*x + c)^2 - 4*(5*a^2 - 2*b^2)*cos(d*x + c)^5 + 225*a*b*d*x - 20
*(5*a^2 - 2*b^2)*cos(d*x + c)^3 + 30*(5*a^2 - 2*b^2)*cos(d*x + c) + 15*((5*a^2 - 2*b^2)*cos(d*x + c)^2 - 5*a^2
+ 2*b^2)*log(1/2*cos(d*x + c) + 1/2) - 15*((5*a^2 - 2*b^2)*cos(d*x + c)^2 - 5*a^2 + 2*b^2)*log(-1/2*cos(d*x +
c) + 1/2) - 15*(2*a*b*cos(d*x + c)^5 + 5*a*b*cos(d*x + c)^3 - 15*a*b*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x +
c)^2 - d)

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giac [B]  time = 0.33, size = 346, normalized size = 1.92 \[ \frac {15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 450 \, {\left (d x + c\right )} a b + 120 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 60 \, {\left (5 \, a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {15 \, {\left (30 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + \frac {4 \, {\left (135 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 180 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 180 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 150 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 600 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 360 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 800 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 560 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 150 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 520 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 280 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 135 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 140 \, a^{2} + 92 \, b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/120*(15*a^2*tan(1/2*d*x + 1/2*c)^2 - 450*(d*x + c)*a*b + 120*a*b*tan(1/2*d*x + 1/2*c) - 60*(5*a^2 - 2*b^2)*l
og(abs(tan(1/2*d*x + 1/2*c))) + 15*(30*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(1/2*d*x + 1/2*c)^2 - 8*a*b*tan(
1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^2 + 4*(135*a*b*tan(1/2*d*x + 1/2*c)^9 - 180*a^2*tan(1/2*d*x + 1/2
*c)^8 + 180*b^2*tan(1/2*d*x + 1/2*c)^8 + 150*a*b*tan(1/2*d*x + 1/2*c)^7 - 600*a^2*tan(1/2*d*x + 1/2*c)^6 + 360
*b^2*tan(1/2*d*x + 1/2*c)^6 - 800*a^2*tan(1/2*d*x + 1/2*c)^4 + 560*b^2*tan(1/2*d*x + 1/2*c)^4 - 150*a*b*tan(1/
2*d*x + 1/2*c)^3 - 520*a^2*tan(1/2*d*x + 1/2*c)^2 + 280*b^2*tan(1/2*d*x + 1/2*c)^2 - 135*a*b*tan(1/2*d*x + 1/2
*c) - 140*a^2 + 92*b^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d

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maple [A]  time = 0.55, size = 261, normalized size = 1.45 \[ -\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{2 d}-\frac {5 a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{6 d}-\frac {5 a^{2} \cos \left (d x +c \right )}{2 d}-\frac {5 a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}-\frac {2 a b \left (\cos ^{7}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {2 a b \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}-\frac {5 a b \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{2 d}-\frac {15 a b \cos \left (d x +c \right ) \sin \left (d x +c \right )}{4 d}-\frac {15 a b x}{4}-\frac {15 a b c}{4 d}+\frac {b^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5 d}+\frac {b^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}+\frac {b^{2} \cos \left (d x +c \right )}{d}+\frac {b^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^3*(a+b*sin(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.59, size = 190, normalized size = 1.06 \[ -\frac {5 \, {\left (4 \, \cos \left (d x + c\right )^{3} - \frac {6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} + 15 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a b - 2 \, {\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} b^{2}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/60*(5*(4*cos(d*x + c)^3 - 6*cos(d*x + c)/(cos(d*x + c)^2 - 1) + 24*cos(d*x + c) - 15*log(cos(d*x + c) + 1)
+ 15*log(cos(d*x + c) - 1))*a^2 + 15*(15*d*x + 15*c + (15*tan(d*x + c)^4 + 25*tan(d*x + c)^2 + 8)/(tan(d*x + c
)^5 + 2*tan(d*x + c)^3 + tan(d*x + c)))*a*b - 2*(6*cos(d*x + c)^5 + 10*cos(d*x + c)^3 + 30*cos(d*x + c) - 15*l
og(cos(d*x + c) + 1) + 15*log(cos(d*x + c) - 1))*b^2)/d

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mupad [B]  time = 11.76, size = 484, normalized size = 2.69 \[ \frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {5\,a^2}{2}-b^2\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {49\,a^2}{2}-24\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {165\,a^2}{2}-48\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {127\,a^2}{6}-\frac {184\,b^2}{15}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {223\,a^2}{3}-\frac {112\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {335\,a^2}{3}-\frac {224\,b^2}{3}\right )+\frac {a^2}{2}+38\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+60\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+40\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-14\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {15\,a\,b\,\mathrm {atan}\left (\frac {225\,a^2\,b^2}{4\,\left (-\frac {75\,a^3\,b}{2}+\frac {225\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^2}{4}+15\,a\,b^3\right )}-\frac {15\,a\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{-\frac {75\,a^3\,b}{2}+\frac {225\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^2}{4}+15\,a\,b^3}+\frac {75\,a^3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (-\frac {75\,a^3\,b}{2}+\frac {225\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^2}{4}+15\,a\,b^3\right )}\right )}{2\,d}+\frac {a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d} \]

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)^3,x)

[Out]

(a^2*tan(c/2 + (d*x)/2)^2)/(8*d) - (log(tan(c/2 + (d*x)/2))*((5*a^2)/2 - b^2))/d - (tan(c/2 + (d*x)/2)^10*((49
*a^2)/2 - 24*b^2) + tan(c/2 + (d*x)/2)^8*((165*a^2)/2 - 48*b^2) + tan(c/2 + (d*x)/2)^2*((127*a^2)/6 - (184*b^2
)/15) + tan(c/2 + (d*x)/2)^4*((223*a^2)/3 - (112*b^2)/3) + tan(c/2 + (d*x)/2)^6*((335*a^2)/3 - (224*b^2)/3) +
a^2/2 + 38*a*b*tan(c/2 + (d*x)/2)^3 + 60*a*b*tan(c/2 + (d*x)/2)^5 + 40*a*b*tan(c/2 + (d*x)/2)^7 - 14*a*b*tan(c
/2 + (d*x)/2)^11 + 4*a*b*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 + 20*tan(c/2 + (d*x)/2)^4 + 40*tan(c/2
+ (d*x)/2)^6 + 40*tan(c/2 + (d*x)/2)^8 + 20*tan(c/2 + (d*x)/2)^10 + 4*tan(c/2 + (d*x)/2)^12)) + (15*a*b*atan(
(225*a^2*b^2)/(4*(15*a*b^3 - (75*a^3*b)/2 + (225*a^2*b^2*tan(c/2 + (d*x)/2))/4)) - (15*a*b^3*tan(c/2 + (d*x)/2
))/(15*a*b^3 - (75*a^3*b)/2 + (225*a^2*b^2*tan(c/2 + (d*x)/2))/4) + (75*a^3*b*tan(c/2 + (d*x)/2))/(2*(15*a*b^3
- (75*a^3*b)/2 + (225*a^2*b^2*tan(c/2 + (d*x)/2))/4))))/(2*d) + (a*b*tan(c/2 + (d*x)/2))/d

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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)**3*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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