3.1062 \(\int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=90 \[ \frac {\left (2 a^2-b^2\right ) \cos (c+d x)}{3 d}-\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a b \sin (c+d x) \cos (c+d x)}{3 d}+\frac {\cos (c+d x) (a+b \sin (c+d x))^2}{3 d}+a b x \]

[Out]

a*b*x-a^2*arctanh(cos(d*x+c))/d+1/3*(2*a^2-b^2)*cos(d*x+c)/d+1/3*a*b*cos(d*x+c)*sin(d*x+c)/d+1/3*cos(d*x+c)*(a
+b*sin(d*x+c))^2/d

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Rubi [A]  time = 0.25, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2889, 3050, 3033, 3023, 2735, 3770} \[ \frac {\left (2 a^2-b^2\right ) \cos (c+d x)}{3 d}-\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a b \sin (c+d x) \cos (c+d x)}{3 d}+\frac {\cos (c+d x) (a+b \sin (c+d x))^2}{3 d}+a b x \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*b*x - (a^2*ArcTanh[Cos[c + d*x]])/d + ((2*a^2 - b^2)*Cos[c + d*x])/(3*d) + (a*b*Cos[c + d*x]*Sin[c + d*x])/(
3*d) + (Cos[c + d*x]*(a + b*Sin[c + d*x])^2)/(3*d)

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2889

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f,
m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3033

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)
*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n
+ 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n
*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0
] && NeQ[c, 0])))

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 91, normalized size = 1.01 \[ \frac {3 \left (4 a^2-b^2\right ) \cos (c+d x)+6 a \left (2 a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+b \sin (2 (c+d x))+2 b c+2 b d x\right )+b^2 (-\cos (3 (c+d x)))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(4*a^2 - b^2)*Cos[c + d*x] - b^2*Cos[3*(c + d*x)] + 6*a*(2*b*c + 2*b*d*x - 2*a*Log[Cos[(c + d*x)/2]] + 2*a*
Log[Sin[(c + d*x)/2]] + b*Sin[2*(c + d*x)]))/(12*d)

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fricas [A]  time = 0.77, size = 84, normalized size = 0.93 \[ -\frac {2 \, b^{2} \cos \left (d x + c\right )^{3} - 6 \, a b d x - 6 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, a^{2} \cos \left (d x + c\right ) + 3 \, a^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*b^2*cos(d*x + c)^3 - 6*a*b*d*x - 6*a*b*cos(d*x + c)*sin(d*x + c) - 6*a^2*cos(d*x + c) + 3*a^2*log(1/2*
cos(d*x + c) + 1/2) - 3*a^2*log(-1/2*cos(d*x + c) + 1/2))/d

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giac [A]  time = 0.20, size = 133, normalized size = 1.48 \[ \frac {3 \, {\left (d x + c\right )} a b + 3 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2} + b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(d*x + c)*a*b + 3*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(3*a*b*tan(1/2*d*x + 1/2*c)^5 - 3*a^2*tan(1/2*
d*x + 1/2*c)^4 + 3*b^2*tan(1/2*d*x + 1/2*c)^4 - 6*a^2*tan(1/2*d*x + 1/2*c)^2 - 3*a*b*tan(1/2*d*x + 1/2*c) - 3*
a^2 + b^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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maple [A]  time = 0.40, size = 83, normalized size = 0.92 \[ \frac {a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {a^{2} \cos \left (d x +c \right )}{d}+\frac {a b \cos \left (d x +c \right ) \sin \left (d x +c \right )}{d}+a b x +\frac {a b c}{d}-\frac {b^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 74, normalized size = 0.82 \[ -\frac {2 \, b^{2} \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b - 3 \, a^{2} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*b^2*cos(d*x + c)^3 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*a*b - 3*a^2*(2*cos(d*x + c) - log(cos(d*x + c)
+ 1) + log(cos(d*x + c) - 1)))/d

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mupad [B]  time = 9.87, size = 225, normalized size = 2.50 \[ \frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^2-2\,b^2\right )+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^2-\frac {2\,b^2}{3}-2\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {2\,a\,b\,\mathrm {atan}\left (\frac {4\,a^2\,b^2}{4\,a^3\,b-4\,a^2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {4\,a^3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^3\,b-4\,a^2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*log(tan(c/2 + (d*x)/2)))/d + (tan(c/2 + (d*x)/2)^4*(2*a^2 - 2*b^2) + 4*a^2*tan(c/2 + (d*x)/2)^2 + 2*a^2 -
(2*b^2)/3 - 2*a*b*tan(c/2 + (d*x)/2)^5 + 2*a*b*tan(c/2 + (d*x)/2))/(d*(3*tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (
d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) + (2*a*b*atan((4*a^2*b^2)/(4*a^3*b - 4*a^2*b^2*tan(c/2 + (d*x)/2)) + (4
*a^3*b*tan(c/2 + (d*x)/2))/(4*a^3*b - 4*a^2*b^2*tan(c/2 + (d*x)/2))))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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