3.1203 \(\int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=86 \[ \frac {a \sin ^4(c+d x)}{4 d}-\frac {a \sin ^2(c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d}+\frac {b \sin ^5(c+d x)}{5 d}-\frac {2 b \sin ^3(c+d x)}{3 d}+\frac {b \sin (c+d x)}{d} \]

[Out]

a*ln(sin(d*x+c))/d+b*sin(d*x+c)/d-a*sin(d*x+c)^2/d-2/3*b*sin(d*x+c)^3/d+1/4*a*sin(d*x+c)^4/d+1/5*b*sin(d*x+c)^
5/d

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Rubi [A]  time = 0.08, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2837, 12, 766} \[ \frac {a \sin ^4(c+d x)}{4 d}-\frac {a \sin ^2(c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d}+\frac {b \sin ^5(c+d x)}{5 d}-\frac {2 b \sin ^3(c+d x)}{3 d}+\frac {b \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*Cot[c + d*x]*(a + b*Sin[c + d*x]),x]

[Out]

(a*Log[Sin[c + d*x]])/d + (b*Sin[c + d*x])/d - (a*Sin[c + d*x]^2)/d - (2*b*Sin[c + d*x]^3)/(3*d) + (a*Sin[c +
d*x]^4)/(4*d) + (b*Sin[c + d*x]^5)/(5*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 766

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(e*x
)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, m}, x] && IGtQ[p, 0]

Rule 2837

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 86, normalized size = 1.00 \[ \frac {a \sin ^4(c+d x)}{4 d}-\frac {a \sin ^2(c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d}+\frac {b \sin ^5(c+d x)}{5 d}-\frac {2 b \sin ^3(c+d x)}{3 d}+\frac {b \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*Cot[c + d*x]*(a + b*Sin[c + d*x]),x]

[Out]

(a*Log[Sin[c + d*x]])/d + (b*Sin[c + d*x])/d - (a*Sin[c + d*x]^2)/d - (2*b*Sin[c + d*x]^3)/(3*d) + (a*Sin[c +
d*x]^4)/(4*d) + (b*Sin[c + d*x]^5)/(5*d)

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fricas [A]  time = 0.75, size = 74, normalized size = 0.86 \[ \frac {15 \, a \cos \left (d x + c\right )^{4} + 30 \, a \cos \left (d x + c\right )^{2} + 60 \, a \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (3 \, b \cos \left (d x + c\right )^{4} + 4 \, b \cos \left (d x + c\right )^{2} + 8 \, b\right )} \sin \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/60*(15*a*cos(d*x + c)^4 + 30*a*cos(d*x + c)^2 + 60*a*log(1/2*sin(d*x + c)) + 4*(3*b*cos(d*x + c)^4 + 4*b*cos
(d*x + c)^2 + 8*b)*sin(d*x + c))/d

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giac [A]  time = 0.21, size = 70, normalized size = 0.81 \[ \frac {12 \, b \sin \left (d x + c\right )^{5} + 15 \, a \sin \left (d x + c\right )^{4} - 40 \, b \sin \left (d x + c\right )^{3} - 60 \, a \sin \left (d x + c\right )^{2} + 60 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 60 \, b \sin \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/60*(12*b*sin(d*x + c)^5 + 15*a*sin(d*x + c)^4 - 40*b*sin(d*x + c)^3 - 60*a*sin(d*x + c)^2 + 60*a*log(abs(sin
(d*x + c))) + 60*b*sin(d*x + c))/d

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maple [A]  time = 0.38, size = 94, normalized size = 1.09 \[ \frac {a \left (\cos ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a \left (\cos ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a \ln \left (\sin \left (d x +c \right )\right )}{d}+\frac {8 b \sin \left (d x +c \right )}{15 d}+\frac {\left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right ) b}{5 d}+\frac {4 b \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{15 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)*(a+b*sin(d*x+c)),x)

[Out]

1/4*a*cos(d*x+c)^4/d+1/2*a*cos(d*x+c)^2/d+a*ln(sin(d*x+c))/d+8/15*b*sin(d*x+c)/d+1/5/d*cos(d*x+c)^4*sin(d*x+c)
*b+4/15/d*b*sin(d*x+c)*cos(d*x+c)^2

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maxima [A]  time = 0.32, size = 69, normalized size = 0.80 \[ \frac {12 \, b \sin \left (d x + c\right )^{5} + 15 \, a \sin \left (d x + c\right )^{4} - 40 \, b \sin \left (d x + c\right )^{3} - 60 \, a \sin \left (d x + c\right )^{2} + 60 \, a \log \left (\sin \left (d x + c\right )\right ) + 60 \, b \sin \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/60*(12*b*sin(d*x + c)^5 + 15*a*sin(d*x + c)^4 - 40*b*sin(d*x + c)^3 - 60*a*sin(d*x + c)^2 + 60*a*log(sin(d*x
+ c)) + 60*b*sin(d*x + c))/d

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mupad [B]  time = 12.00, size = 126, normalized size = 1.47 \[ \frac {a\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {a\,{\cos \left (c+d\,x\right )}^2}{2\,d}+\frac {a\,{\cos \left (c+d\,x\right )}^4}{4\,d}+\frac {8\,b\,\sin \left (c+d\,x\right )}{15\,d}+\frac {4\,b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{15\,d}+\frac {b\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{5\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^5*(a + b*sin(c + d*x)))/sin(c + d*x),x)

[Out]

(a*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (a*log(1/cos(c/2 + (d*x)/2)^2))/d + (a*cos(c + d*x)^2)/(2*d
) + (a*cos(c + d*x)^4)/(4*d) + (8*b*sin(c + d*x))/(15*d) + (4*b*cos(c + d*x)^2*sin(c + d*x))/(15*d) + (b*cos(c
+ d*x)^4*sin(c + d*x))/(5*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (c + d x \right )}\right ) \cos ^{5}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)*(a+b*sin(d*x+c)),x)

[Out]

Integral((a + b*sin(c + d*x))*cos(c + d*x)**5*csc(c + d*x), x)

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