3.5.55 \(\int \frac {\cot ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) [455]

Optimal. Leaf size=96 \[ -\frac {(a+b)^{5/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{7/2} d}-\frac {(a+b)^2 \cot (c+d x)}{a^3 d}+\frac {(a+b) \cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a d} \]

[Out]

-(a+b)^(5/2)*arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))/a^(7/2)/d-(a+b)^2*cot(d*x+c)/a^3/d+1/3*(a+b)*cot(d*x+c)^3/
a^2/d-1/5*cot(d*x+c)^5/a/d

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Rubi [A]
time = 0.07, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3274, 331, 211} \begin {gather*} -\frac {(a+b)^{5/2} \text {ArcTan}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{7/2} d}-\frac {(a+b)^2 \cot (c+d x)}{a^3 d}+\frac {(a+b) \cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6/(a + b*Sin[c + d*x]^2),x]

[Out]

-(((a + b)^(5/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(7/2)*d)) - ((a + b)^2*Cot[c + d*x])/(a^3*d) +
((a + b)*Cot[c + d*x]^3)/(3*a^2*d) - Cot[c + d*x]^5/(5*a*d)

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 3274

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[{ff
= FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(d*ff*x)^m*((a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(p
+ 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m}, x] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\cot ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{x^6 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {\cot ^5(c+d x)}{5 a d}-\frac {(a+b) \text {Subst}\left (\int \frac {1}{x^4 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{a d}\\ &=\frac {(a+b) \cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a d}+\frac {(a+b)^2 \text {Subst}\left (\int \frac {1}{x^2 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{a^2 d}\\ &=-\frac {(a+b)^2 \cot (c+d x)}{a^3 d}+\frac {(a+b) \cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a d}-\frac {(a+b)^3 \text {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{a^3 d}\\ &=-\frac {(a+b)^{5/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{7/2} d}-\frac {(a+b)^2 \cot (c+d x)}{a^3 d}+\frac {(a+b) \cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a d}\\ \end {align*}

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Mathematica [A]
time = 0.62, size = 101, normalized size = 1.05 \begin {gather*} \frac {-15 (a+b)^{5/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )-\sqrt {a} \cot (c+d x) \left (23 a^2+35 a b+15 b^2-a (11 a+5 b) \csc ^2(c+d x)+3 a^2 \csc ^4(c+d x)\right )}{15 a^{7/2} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6/(a + b*Sin[c + d*x]^2),x]

[Out]

(-15*(a + b)^(5/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]] - Sqrt[a]*Cot[c + d*x]*(23*a^2 + 35*a*b + 15*b^2
- a*(11*a + 5*b)*Csc[c + d*x]^2 + 3*a^2*Csc[c + d*x]^4))/(15*a^(7/2)*d)

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Maple [A]
time = 0.57, size = 115, normalized size = 1.20

method	result	size
derivativedivides	\(\frac {-\frac {1}{5 a \tan \left (d x +c \right )^{5}}-\frac {a^{2}+2 a b +b^{2}}{a^{3} \tan \left (d x +c \right )}-\frac {-a -b}{3 a^{2} \tan \left (d x +c \right )^{3}}+\frac {\left (-a^{3}-3 a^{2} b -3 a \,b^{2}-b^{3}\right ) \arctan \left (\frac {\tan \left (d x +c \right ) \left (a +b \right )}{\sqrt {a \left (a +b \right )}}\right )}{a^{3} \sqrt {a \left (a +b \right )}}}{d}\)	\(115\)
default	\(\frac {-\frac {1}{5 a \tan \left (d x +c \right )^{5}}-\frac {a^{2}+2 a b +b^{2}}{a^{3} \tan \left (d x +c \right )}-\frac {-a -b}{3 a^{2} \tan \left (d x +c \right )^{3}}+\frac {\left (-a^{3}-3 a^{2} b -3 a \,b^{2}-b^{3}\right ) \arctan \left (\frac {\tan \left (d x +c \right ) \left (a +b \right )}{\sqrt {a \left (a +b \right )}}\right )}{a^{3} \sqrt {a \left (a +b \right )}}}{d}\)	\(115\)
risch	\(-\frac {2 i \left (45 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}+45 b \,{\mathrm e}^{8 i \left (d x +c \right )} a +15 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-90 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-150 a b \,{\mathrm e}^{6 i \left (d x +c \right )}-60 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+140 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+200 a b \,{\mathrm e}^{4 i \left (d x +c \right )}+90 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-70 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-130 b \,{\mathrm e}^{2 i \left (d x +c \right )} a -60 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+23 a^{2}+35 a b +15 b^{2}\right )}{15 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 a^{2} d}-\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right ) b}{a^{3} d}-\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right ) b^{2}}{2 a^{4} d}+\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 a^{2} d}+\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right ) b}{a^{3} d}+\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right ) b^{2}}{2 a^{4} d}\)	\(501\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^6/(a+sin(d*x+c)^2*b),x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/5/a/tan(d*x+c)^5-1/a^3*(a^2+2*a*b+b^2)/tan(d*x+c)-1/3*(-a-b)/a^2/tan(d*x+c)^3+1/a^3*(-a^3-3*a^2*b-3*a*
b^2-b^3)/(a*(a+b))^(1/2)*arctan(tan(d*x+c)*(a+b)/(a*(a+b))^(1/2)))

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Maxima [A]
time = 0.51, size = 111, normalized size = 1.16 \begin {gather*} -\frac {\frac {15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} a^{3}} + \frac {15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (d x + c\right )^{4} - 5 \, {\left (a^{2} + a b\right )} \tan \left (d x + c\right )^{2} + 3 \, a^{2}}{a^{3} \tan \left (d x + c\right )^{5}}}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/15*(15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/(sqrt((a + b)*a)*a^3) +
(15*(a^2 + 2*a*b + b^2)*tan(d*x + c)^4 - 5*(a^2 + a*b)*tan(d*x + c)^2 + 3*a^2)/(a^3*tan(d*x + c)^5))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (84) = 168\).
time = 0.43, size = 576, normalized size = 6.00 \begin {gather*} \left [-\frac {4 \, {\left (23 \, a^{2} + 35 \, a b + 15 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 20 \, {\left (7 \, a^{2} + 13 \, a b + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{3} - 15 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-\frac {a + b}{a}} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (2 \, a^{2} + a b\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {a + b}{a}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) \sin \left (d x + c\right ) + 60 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )} \sin \left (d x + c\right )}, -\frac {2 \, {\left (23 \, a^{2} + 35 \, a b + 15 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 10 \, {\left (7 \, a^{2} + 13 \, a b + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{3} - 15 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {\frac {a + b}{a}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {a + b}{a}}}{2 \, {\left (a + b\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 30 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )} \sin \left (d x + c\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

[-1/60*(4*(23*a^2 + 35*a*b + 15*b^2)*cos(d*x + c)^5 - 20*(7*a^2 + 13*a*b + 6*b^2)*cos(d*x + c)^3 - 15*((a^2 +
2*a*b + b^2)*cos(d*x + c)^4 - 2*(a^2 + 2*a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)*sqrt(-(a + b)/a)*log((
(8*a^2 + 8*a*b + b^2)*cos(d*x + c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c)^2 + 4*((2*a^2 + a*b)*cos(d*x + c)^
3 - (a^2 + a*b)*cos(d*x + c))*sqrt(-(a + b)/a)*sin(d*x + c) + a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b
+ b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2))*sin(d*x + c) + 60*(a^2 + 2*a*b + b^2)*cos(d*x + c))/((a^3*d*cos(d*
x + c)^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d)*sin(d*x + c)), -1/30*(2*(23*a^2 + 35*a*b + 15*b^2)*cos(d*x + c)^5 -
10*(7*a^2 + 13*a*b + 6*b^2)*cos(d*x + c)^3 - 15*((a^2 + 2*a*b + b^2)*cos(d*x + c)^4 - 2*(a^2 + 2*a*b + b^2)*c
os(d*x + c)^2 + a^2 + 2*a*b + b^2)*sqrt((a + b)/a)*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)*sqrt((a + b)/
a)/((a + b)*cos(d*x + c)*sin(d*x + c)))*sin(d*x + c) + 30*(a^2 + 2*a*b + b^2)*cos(d*x + c))/((a^3*d*cos(d*x +
c)^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d)*sin(d*x + c))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot ^{6}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**6/(a+b*sin(d*x+c)**2),x)

[Out]

Integral(cot(c + d*x)**6/(a + b*sin(c + d*x)**2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (84) = 168\).
time = 0.53, size = 171, normalized size = 1.78 \begin {gather*} -\frac {\frac {15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )}}{\sqrt {a^{2} + a b} a^{3}} + \frac {15 \, a^{2} \tan \left (d x + c\right )^{4} + 30 \, a b \tan \left (d x + c\right )^{4} + 15 \, b^{2} \tan \left (d x + c\right )^{4} - 5 \, a^{2} \tan \left (d x + c\right )^{2} - 5 \, a b \tan \left (d x + c\right )^{2} + 3 \, a^{2}}{a^{3} \tan \left (d x + c\right )^{5}}}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

-1/15*(15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c
) + b*tan(d*x + c))/sqrt(a^2 + a*b)))/(sqrt(a^2 + a*b)*a^3) + (15*a^2*tan(d*x + c)^4 + 30*a*b*tan(d*x + c)^4 +
15*b^2*tan(d*x + c)^4 - 5*a^2*tan(d*x + c)^2 - 5*a*b*tan(d*x + c)^2 + 3*a^2)/(a^3*tan(d*x + c)^5))/d

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Mupad [B]
time = 16.15, size = 82, normalized size = 0.85 \begin {gather*} -\frac {\frac {1}{5\,a}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a+b\right )}{3\,a^2}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,{\left (a+b\right )}^2}{a^3}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^5}-\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a+b}}{\sqrt {a}}\right )\,{\left (a+b\right )}^{5/2}}{a^{7/2}\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^6/(a + b*sin(c + d*x)^2),x)

[Out]

- (1/(5*a) - (tan(c + d*x)^2*(a + b))/(3*a^2) + (tan(c + d*x)^4*(a + b)^2)/a^3)/(d*tan(c + d*x)^5) - (atan((ta
n(c + d*x)*(a + b)^(1/2))/a^(1/2))*(a + b)^(5/2))/(a^(7/2)*d)

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