3.1218 \(\int \cos ^3(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=125 \[ \frac {\left (a^2-2 b^2\right ) \sin ^3(c+d x)}{3 d}-\frac {\left (2 a^2-b^2\right ) \sin (c+d x)}{d}-\frac {a^2 \csc (c+d x)}{d}+\frac {a b \sin ^4(c+d x)}{2 d}-\frac {2 a b \sin ^2(c+d x)}{d}+\frac {2 a b \log (\sin (c+d x))}{d}+\frac {b^2 \sin ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2837, 12, 948} \[ \frac {\left (a^2-2 b^2\right ) \sin ^3(c+d x)}{3 d}-\frac {\left (2 a^2-b^2\right ) \sin (c+d x)}{d}-\frac {a^2 \csc (c+d x)}{d}+\frac {a b \sin ^4(c+d x)}{2 d}-\frac {2 a b \sin ^2(c+d x)}{d}+\frac {2 a b \log (\sin (c+d x))}{d}+\frac {b^2 \sin ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*Cot[c + d*x]^2*(a + b*Sin[c + d*x])^2,x]

[Out]

-((a^2*Csc[c + d*x])/d) + (2*a*b*Log[Sin[c + d*x]])/d - ((2*a^2 - b^2)*Sin[c + d*x])/d - (2*a*b*Sin[c + d*x]^2
)/d + ((a^2 - 2*b^2)*Sin[c + d*x]^3)/(3*d) + (a*b*Sin[c + d*x]^4)/(2*d) + (b^2*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 948

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 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 ^3(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {b^2 (a+x)^2 \left (b^2-x^2\right )^2}{x^2} \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(a+x)^2 \left (b^2-x^2\right )^2}{x^2} \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-2 a^2 b^2 \left (1-\frac {b^2}{2 a^2}\right )+\frac {a^2 b^4}{x^2}+\frac {2 a b^4}{x}-4 a b^2 x+\left (a^2-2 b^2\right ) x^2+2 a x^3+x^4\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=-\frac {a^2 \csc (c+d x)}{d}+\frac {2 a b \log (\sin (c+d x))}{d}-\frac {\left (2 a^2-b^2\right ) \sin (c+d x)}{d}-\frac {2 a b \sin ^2(c+d x)}{d}+\frac {\left (a^2-2 b^2\right ) \sin ^3(c+d x)}{3 d}+\frac {a b \sin ^4(c+d x)}{2 d}+\frac {b^2 \sin ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 142, normalized size = 1.14 \[ \frac {a^2 \sin ^3(c+d x)}{3 d}-\frac {2 a^2 \sin (c+d x)}{d}-\frac {a^2 \csc (c+d x)}{d}+\frac {a b \sin ^4(c+d x)}{2 d}-\frac {2 a b \sin ^2(c+d x)}{d}+\frac {2 a b \log (\sin (c+d x))}{d}+\frac {b^2 \sin ^5(c+d x)}{5 d}-\frac {2 b^2 \sin ^3(c+d x)}{3 d}+\frac {b^2 \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*Cot[c + d*x]^2*(a + b*Sin[c + d*x])^2,x]

[Out]

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

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fricas [A]  time = 0.66, size = 135, normalized size = 1.08 \[ -\frac {48 \, b^{2} \cos \left (d x + c\right )^{6} - 16 \, {\left (5 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} - 480 \, a b \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 64 \, {\left (5 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 640 \, a^{2} - 128 \, b^{2} - 15 \, {\left (8 \, a b \cos \left (d x + c\right )^{4} + 16 \, a b \cos \left (d x + c\right )^{2} - 11 \, a b\right )} \sin \left (d x + c\right )}{240 \, d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/240*(48*b^2*cos(d*x + c)^6 - 16*(5*a^2 - b^2)*cos(d*x + c)^4 - 480*a*b*log(1/2*sin(d*x + c))*sin(d*x + c) -
64*(5*a^2 - b^2)*cos(d*x + c)^2 + 640*a^2 - 128*b^2 - 15*(8*a*b*cos(d*x + c)^4 + 16*a*b*cos(d*x + c)^2 - 11*a
*b)*sin(d*x + c))/(d*sin(d*x + c))

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giac [A]  time = 0.24, size = 127, normalized size = 1.02 \[ \frac {6 \, b^{2} \sin \left (d x + c\right )^{5} + 15 \, a b \sin \left (d x + c\right )^{4} + 10 \, a^{2} \sin \left (d x + c\right )^{3} - 20 \, b^{2} \sin \left (d x + c\right )^{3} - 60 \, a b \sin \left (d x + c\right )^{2} + 60 \, a b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 60 \, a^{2} \sin \left (d x + c\right ) + 30 \, b^{2} \sin \left (d x + c\right ) - \frac {30 \, {\left (2 \, a b \sin \left (d x + c\right ) + a^{2}\right )}}{\sin \left (d x + c\right )}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/30*(6*b^2*sin(d*x + c)^5 + 15*a*b*sin(d*x + c)^4 + 10*a^2*sin(d*x + c)^3 - 20*b^2*sin(d*x + c)^3 - 60*a*b*si
n(d*x + c)^2 + 60*a*b*log(abs(sin(d*x + c))) - 60*a^2*sin(d*x + c) + 30*b^2*sin(d*x + c) - 30*(2*a*b*sin(d*x +
c) + a^2)/sin(d*x + c))/d

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maple [A]  time = 0.45, size = 185, normalized size = 1.48 \[ -\frac {a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {8 a^{2} \sin \left (d x +c \right )}{3 d}-\frac {a^{2} \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{d}-\frac {4 a^{2} \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a b \left (\cos ^{4}\left (d x +c \right )\right )}{2 d}+\frac {a b \left (\cos ^{2}\left (d x +c \right )\right )}{d}+\frac {2 a b \ln \left (\sin \left (d x +c \right )\right )}{d}+\frac {8 b^{2} \sin \left (d x +c \right )}{15 d}+\frac {\sin \left (d x +c \right ) b^{2} \left (\cos ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 \sin \left (d x +c \right ) b^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{15 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^2*(a+b*sin(d*x+c))^2,x)

[Out]

-1/d*a^2/sin(d*x+c)*cos(d*x+c)^6-8/3*a^2*sin(d*x+c)/d-1/d*a^2*sin(d*x+c)*cos(d*x+c)^4-4/3/d*a^2*sin(d*x+c)*cos
(d*x+c)^2+1/2/d*a*b*cos(d*x+c)^4+1/d*a*b*cos(d*x+c)^2+2*a*b*ln(sin(d*x+c))/d+8/15*b^2*sin(d*x+c)/d+1/5/d*sin(d
*x+c)*b^2*cos(d*x+c)^4+4/15/d*sin(d*x+c)*b^2*cos(d*x+c)^2

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maxima [A]  time = 0.33, size = 105, normalized size = 0.84 \[ \frac {6 \, b^{2} \sin \left (d x + c\right )^{5} + 15 \, a b \sin \left (d x + c\right )^{4} - 60 \, a b \sin \left (d x + c\right )^{2} + 10 \, {\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{3} + 60 \, a b \log \left (\sin \left (d x + c\right )\right ) - 30 \, {\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right ) - \frac {30 \, a^{2}}{\sin \left (d x + c\right )}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^2*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/30*(6*b^2*sin(d*x + c)^5 + 15*a*b*sin(d*x + c)^4 - 60*a*b*sin(d*x + c)^2 + 10*(a^2 - 2*b^2)*sin(d*x + c)^3 +
60*a*b*log(sin(d*x + c)) - 30*(2*a^2 - b^2)*sin(d*x + c) - 30*a^2/sin(d*x + c))/d

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mupad [B]  time = 12.02, size = 445, normalized size = 3.56 \[ \frac {16\,a\,b\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d}-\frac {8\,a\,b\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{d}-\frac {16\,a\,b\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d}+\frac {8\,a\,b\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d}+\frac {20\,a^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {16\,a^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {8\,a^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {22\,b^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {256\,b^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {368\,b^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{15\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {96\,b^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{5\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {32\,b^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{5\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {2\,a\,b\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {2\,a\,b\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {9\,a^2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {2\,b^2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^5*(a + b*sin(c + d*x))^2)/sin(c + d*x)^2,x)

[Out]

(16*a*b*cos(c/2 + (d*x)/2)^4)/d - (8*a*b*cos(c/2 + (d*x)/2)^2)/d - (16*a*b*cos(c/2 + (d*x)/2)^6)/d + (8*a*b*co
s(c/2 + (d*x)/2)^8)/d + (20*a^2*cos(c/2 + (d*x)/2)^3)/(3*d*sin(c/2 + (d*x)/2)) - (16*a^2*cos(c/2 + (d*x)/2)^5)
/(3*d*sin(c/2 + (d*x)/2)) + (8*a^2*cos(c/2 + (d*x)/2)^7)/(3*d*sin(c/2 + (d*x)/2)) - (22*b^2*cos(c/2 + (d*x)/2)
^3)/(3*d*sin(c/2 + (d*x)/2)) + (256*b^2*cos(c/2 + (d*x)/2)^5)/(15*d*sin(c/2 + (d*x)/2)) - (368*b^2*cos(c/2 + (
d*x)/2)^7)/(15*d*sin(c/2 + (d*x)/2)) + (96*b^2*cos(c/2 + (d*x)/2)^9)/(5*d*sin(c/2 + (d*x)/2)) - (32*b^2*cos(c/
2 + (d*x)/2)^11)/(5*d*sin(c/2 + (d*x)/2)) - (2*a*b*log(1/cos(c/2 + (d*x)/2)^2))/d + (2*a*b*log(sin(c/2 + (d*x)
/2)/cos(c/2 + (d*x)/2)))/d - (9*a^2*cos(c/2 + (d*x)/2))/(2*d*sin(c/2 + (d*x)/2)) - (a^2*sin(c/2 + (d*x)/2))/(2
*d*cos(c/2 + (d*x)/2)) + (2*b^2*cos(c/2 + (d*x)/2))/(d*sin(c/2 + (d*x)/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**2*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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