3.11.96 \(\int \cos ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx\) [1096]

Optimal. Leaf size=89 \[ \frac {3 b x}{8}-\frac {a \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}+\frac {3 b \cos (c+d x) \sin (c+d x)}{8 d}+\frac {b \cos ^3(c+d x) \sin (c+d x)}{4 d} \]

[Out]

3/8*b*x-a*arctanh(cos(d*x+c))/d+a*cos(d*x+c)/d+1/3*a*cos(d*x+c)^3/d+3/8*b*cos(d*x+c)*sin(d*x+c)/d+1/4*b*cos(d*
x+c)^3*sin(d*x+c)/d

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Rubi [A]
time = 0.06, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2917, 2672, 308, 212, 2715, 8} \begin {gather*} \frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos (c+d x)}{d}-\frac {a \tanh ^{-1}(\cos (c+d x))}{d}+\frac {b \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 b \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 b x}{8} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*x)/8 - (a*ArcTanh[Cos[c + d*x]])/d + (a*Cos[c + d*x])/d + (a*Cos[c + d*x]^3)/(3*d) + (3*b*Cos[c + d*x]*Si
n[c + d*x])/(8*d) + (b*Cos[c + d*x]^3*Sin[c + d*x])/(4*d)

Rule 8

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

Rule 212

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

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2672

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

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2917

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

Rubi steps

\begin {align*} \int \cos ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx &=a \int \cos ^3(c+d x) \cot (c+d x) \, dx+b \int \cos ^4(c+d x) \, dx\\ &=\frac {b \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{4} (3 b) \int \cos ^2(c+d x) \, dx-\frac {a \text {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {3 b \cos (c+d x) \sin (c+d x)}{8 d}+\frac {b \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{8} (3 b) \int 1 \, dx-\frac {a \text {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {3 b x}{8}+\frac {a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}+\frac {3 b \cos (c+d x) \sin (c+d x)}{8 d}+\frac {b \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {a \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {3 b x}{8}-\frac {a \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}+\frac {3 b \cos (c+d x) \sin (c+d x)}{8 d}+\frac {b \cos ^3(c+d x) \sin (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 109, normalized size = 1.22 \begin {gather*} \frac {3 b (c+d x)}{8 d}+\frac {5 a \cos (c+d x)}{4 d}+\frac {a \cos (3 (c+d x))}{12 d}-\frac {a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {b \sin (2 (c+d x))}{4 d}+\frac {b \sin (4 (c+d x))}{32 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*(c + d*x))/(8*d) + (5*a*Cos[c + d*x])/(4*d) + (a*Cos[3*(c + d*x)])/(12*d) - (a*Log[Cos[(c + d*x)/2]])/d +
(a*Log[Sin[(c + d*x)/2]])/d + (b*Sin[2*(c + d*x)])/(4*d) + (b*Sin[4*(c + d*x)])/(32*d)

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Maple [A]
time = 0.14, size = 76, normalized size = 0.85

method	result	size
derivativedivides	\(\frac {a \left (\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(76\)
default	\(\frac {a \left (\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(76\)
risch	\(\frac {3 b x}{8}+\frac {5 a \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {5 a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {b \sin \left (4 d x +4 c \right )}{32 d}+\frac {a \cos \left (3 d x +3 c \right )}{12 d}+\frac {b \sin \left (2 d x +2 c \right )}{4 d}\)	\(116\)
norman	\(\frac {\frac {3 b x}{8}+\frac {8 a}{3 d}+\frac {5 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {3 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {3 b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {5 b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {3 b x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {9 b x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {3 b x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {3 b x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {4 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {20 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(221\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)*(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(a*(1/3*cos(d*x+c)^3+cos(d*x+c)+ln(csc(d*x+c)-cot(d*x+c)))+b*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)
+3/8*d*x+3/8*c))

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Maxima [A]
time = 0.29, size = 81, normalized size = 0.91 \begin {gather*} \frac {16 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b}{96 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/96*(16*(2*cos(d*x + c)^3 + 6*cos(d*x + c) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1))*a + 3*(12*d*x
+ 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*b)/d

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Fricas [A]
time = 0.40, size = 88, normalized size = 0.99 \begin {gather*} \frac {8 \, a \cos \left (d x + c\right )^{3} + 9 \, b d x + 24 \, a \cos \left (d x + c\right ) - 12 \, a \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 12 \, a \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (2 \, b \cos \left (d x + c\right )^{3} + 3 \, b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/24*(8*a*cos(d*x + c)^3 + 9*b*d*x + 24*a*cos(d*x + c) - 12*a*log(1/2*cos(d*x + c) + 1/2) + 12*a*log(-1/2*cos(
d*x + c) + 1/2) + 3*(2*b*cos(d*x + c)^3 + 3*b*cos(d*x + c))*sin(d*x + c))/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin {\left (c + d x \right )}\right ) \cos ^{4}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)*(a+b*sin(d*x+c)),x)

[Out]

Integral((a + b*sin(c + d*x))*cos(c + d*x)**4*csc(c + d*x), x)

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Giac [A]
time = 0.46, size = 145, normalized size = 1.63 \begin {gather*} \frac {9 \, {\left (d x + c\right )} b + 24 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (15 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 9 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 96 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 9 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 32 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/24*(9*(d*x + c)*b + 24*a*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(15*b*tan(1/2*d*x + 1/2*c)^7 - 48*a*tan(1/2*d*x
+ 1/2*c)^6 - 9*b*tan(1/2*d*x + 1/2*c)^5 - 96*a*tan(1/2*d*x + 1/2*c)^4 + 9*b*tan(1/2*d*x + 1/2*c)^3 - 80*a*tan(
1/2*d*x + 1/2*c)^2 - 15*b*tan(1/2*d*x + 1/2*c) - 32*a)/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d

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Mupad [B]
time = 10.82, size = 242, normalized size = 2.72 \begin {gather*} \frac {3\,b\,\mathrm {atan}\left (\frac {9\,b^2}{16\,\left (\frac {3\,a\,b}{2}-\frac {9\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}+\frac {3\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {3\,a\,b}{2}-\frac {9\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}\right )}{4\,d}+\frac {-\frac {5\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {20\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {5\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {8\,a}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\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+1\right )}+\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*(a + b*sin(c + d*x)))/sin(c + d*x),x)

[Out]

(3*b*atan((9*b^2)/(16*((3*a*b)/2 - (9*b^2*tan(c/2 + (d*x)/2))/16)) + (3*a*b*tan(c/2 + (d*x)/2))/(2*((3*a*b)/2
- (9*b^2*tan(c/2 + (d*x)/2))/16))))/(4*d) + ((8*a)/3 + (5*b*tan(c/2 + (d*x)/2))/4 + (20*a*tan(c/2 + (d*x)/2)^2
)/3 + 8*a*tan(c/2 + (d*x)/2)^4 + 4*a*tan(c/2 + (d*x)/2)^6 - (3*b*tan(c/2 + (d*x)/2)^3)/4 + (3*b*tan(c/2 + (d*x
)/2)^5)/4 - (5*b*tan(c/2 + (d*x)/2)^7)/4)/(d*(4*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d
*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (a*log(tan(c/2 + (d*x)/2)))/d

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