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

Optimal. Leaf size=46 \[ \frac {a^2 \log (\sin (c+d x))}{d}+\frac {2 a b \sin (c+d x)}{d}+\frac {b^2 \sin ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2721, 43} \[ \frac {a^2 \log (\sin (c+d x))}{d}+\frac {2 a b \sin (c+d x)}{d}+\frac {b^2 \sin ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2721

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 46, normalized size = 1.00 \[ \frac {a^2 \log (\sin (c+d x))}{d}+\frac {2 a b \sin (c+d x)}{d}+\frac {b^2 \sin ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 42, normalized size = 0.91 \[ -\frac {b^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 4 \, a b \sin \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.42, size = 41, normalized size = 0.89 \[ \frac {b^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 4 \, a b \sin \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^2*sin(d*x + c)^2 + 2*a^2*log(abs(sin(d*x + c))) + 4*a*b*sin(d*x + c))/d

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maple [A]  time = 0.10, size = 45, normalized size = 0.98 \[ \frac {a^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}+\frac {2 a b \sin \left (d x +c \right )}{d}+\frac {b^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.48, size = 40, normalized size = 0.87 \[ \frac {b^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + 4 \, a b \sin \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^2*sin(d*x + c)^2 + 2*a^2*log(sin(d*x + c)) + 4*a*b*sin(d*x + c))/d

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mupad [B]  time = 6.44, size = 117, normalized size = 2.54 \[ \frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {2\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*log(tan(c/2 + (d*x)/2)))/d + (2*b^2*tan(c/2 + (d*x)/2)^2 + 4*a*b*tan(c/2 + (d*x)/2)^3 + 4*a*b*tan(c/2 + (
d*x)/2))/(d*(2*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 + 1)) - (a^2*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 \[ \int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \cot {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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