3.210 \(\int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=62 \[ \frac {a^3 \sin ^2(c+d x)}{2 d}+\frac {3 a^3 \sin (c+d x)}{d}-\frac {a^3 \csc (c+d x)}{d}+\frac {3 a^3 \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 62, 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 = {2833, 12, 43} \[ \frac {a^3 \sin ^2(c+d x)}{2 d}+\frac {3 a^3 \sin (c+d x)}{d}-\frac {a^3 \csc (c+d x)}{d}+\frac {3 a^3 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*Csc[c + d*x])/d) + (3*a^3*Log[Sin[c + d*x]])/d + (3*a^3*Sin[c + d*x])/d + (a^3*Sin[c + d*x]^2)/(2*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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2833

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 62, normalized size = 1.00 \[ \frac {a^3 \sin ^2(c+d x)}{2 d}+\frac {3 a^3 \sin (c+d x)}{d}-\frac {a^3 \csc (c+d x)}{d}+\frac {3 a^3 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.50, size = 78, normalized size = 1.26 \[ -\frac {12 \, a^{3} \cos \left (d x + c\right )^{2} - 12 \, a^{3} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 8 \, a^{3} + {\left (2 \, a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 55, normalized size = 0.89 \[ \frac {a^{3} \sin \left (d x + c\right )^{2} + 6 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 6 \, a^{3} \sin \left (d x + c\right ) - \frac {2 \, a^{3}}{\sin \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.14, size = 63, normalized size = 1.02 \[ \frac {a^{3} \left (\sin ^{2}\left (d x +c \right )\right )}{2 d}+\frac {3 a^{3} \sin \left (d x +c \right )}{d}-\frac {a^{3}}{d \sin \left (d x +c \right )}+\frac {3 a^{3} \ln \left (\sin \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 54, normalized size = 0.87 \[ \frac {a^{3} \sin \left (d x + c\right )^{2} + 6 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 6 \, a^{3} \sin \left (d x + c\right ) - \frac {2 \, a^{3}}{\sin \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.59, size = 156, normalized size = 2.52 \[ \frac {3\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {11\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+10\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {3\,a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*(a + a*sin(c + d*x))^3)/sin(c + d*x)^2,x)

[Out]

(3*a^3*log(tan(c/2 + (d*x)/2)))/d + (10*a^3*tan(c/2 + (d*x)/2)^2 + 4*a^3*tan(c/2 + (d*x)/2)^3 + 11*a^3*tan(c/2
+ (d*x)/2)^4 - a^3)/(d*(2*tan(c/2 + (d*x)/2) + 4*tan(c/2 + (d*x)/2)^3 + 2*tan(c/2 + (d*x)/2)^5)) - (a^3*tan(c
/2 + (d*x)/2))/(2*d) - (3*a^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int \cos {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int 3 \sin {\left (c + d x \right )} \cos {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*csc(d*x+c)**2*(a+a*sin(d*x+c))**3,x)

[Out]

a**3*(Integral(cos(c + d*x)*csc(c + d*x)**2, x) + Integral(3*sin(c + d*x)*cos(c + d*x)*csc(c + d*x)**2, x) + I
ntegral(3*sin(c + d*x)**2*cos(c + d*x)*csc(c + d*x)**2, x) + Integral(sin(c + d*x)**3*cos(c + d*x)*csc(c + d*x
)**2, x))

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