3.448 \(\int \frac {\cot ^7(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=121 \[ \frac {(a+b)^3 \log \left (a+b \sin ^2(c+d x)\right )}{2 a^4 d}-\frac {(a+b)^3 \log (\sin (c+d x))}{a^4 d}+\frac {(3 a+b) \csc ^4(c+d x)}{4 a^2 d}-\frac {\left (3 a^2+3 a b+b^2\right ) \csc ^2(c+d x)}{2 a^3 d}-\frac {\csc ^6(c+d x)}{6 a d} \]

[Out]

-1/2*(3*a^2+3*a*b+b^2)*csc(d*x+c)^2/a^3/d+1/4*(3*a+b)*csc(d*x+c)^4/a^2/d-1/6*csc(d*x+c)^6/a/d-(a+b)^3*ln(sin(d
*x+c))/a^4/d+1/2*(a+b)^3*ln(a+b*sin(d*x+c)^2)/a^4/d

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Rubi [A]  time = 0.11, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3194, 88} \[ -\frac {\left (3 a^2+3 a b+b^2\right ) \csc ^2(c+d x)}{2 a^3 d}+\frac {(3 a+b) \csc ^4(c+d x)}{4 a^2 d}+\frac {(a+b)^3 \log \left (a+b \sin ^2(c+d x)\right )}{2 a^4 d}-\frac {(a+b)^3 \log (\sin (c+d x))}{a^4 d}-\frac {\csc ^6(c+d x)}{6 a d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^7/(a + b*Sin[c + d*x]^2),x]

[Out]

-((3*a^2 + 3*a*b + b^2)*Csc[c + d*x]^2)/(2*a^3*d) + ((3*a + b)*Csc[c + d*x]^4)/(4*a^2*d) - Csc[c + d*x]^6/(6*a
*d) - ((a + b)^3*Log[Sin[c + d*x]])/(a^4*d) + ((a + b)^3*Log[a + b*Sin[c + d*x]^2])/(2*a^4*d)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 3194

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

Rubi steps

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

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Mathematica [A]  time = 0.26, size = 100, normalized size = 0.83 \[ -\frac {2 a^3 \csc ^6(c+d x)+6 a \left (3 a^2+3 a b+b^2\right ) \csc ^2(c+d x)-3 a^2 (3 a+b) \csc ^4(c+d x)-6 (a+b)^3 \log \left (a+b \sin ^2(c+d x)\right )+12 (a+b)^3 \log (\sin (c+d x))}{12 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^7/(a + b*Sin[c + d*x]^2),x]

[Out]

-1/12*(6*a*(3*a^2 + 3*a*b + b^2)*Csc[c + d*x]^2 - 3*a^2*(3*a + b)*Csc[c + d*x]^4 + 2*a^3*Csc[c + d*x]^6 + 12*(
a + b)^3*Log[Sin[c + d*x]] - 6*(a + b)^3*Log[a + b*Sin[c + d*x]^2])/(a^4*d)

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fricas [B]  time = 0.60, size = 371, normalized size = 3.07 \[ \frac {6 \, {\left (3 \, a^{3} + 3 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{4} + 11 \, a^{3} + 15 \, a^{2} b + 6 \, a b^{2} - 3 \, {\left (9 \, a^{3} + 11 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{4} - a^{3} - 3 \, a^{2} b - 3 \, a b^{2} - b^{3} + 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-b \cos \left (d x + c\right )^{2} + a + b\right ) - 12 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{4} - a^{3} - 3 \, a^{2} b - 3 \, a b^{2} - b^{3} + 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right )}{12 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} - 3 \, a^{4} d \cos \left (d x + c\right )^{4} + 3 \, a^{4} d \cos \left (d x + c\right )^{2} - a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^7/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

1/12*(6*(3*a^3 + 3*a^2*b + a*b^2)*cos(d*x + c)^4 + 11*a^3 + 15*a^2*b + 6*a*b^2 - 3*(9*a^3 + 11*a^2*b + 4*a*b^2
)*cos(d*x + c)^2 + 6*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d*x + c)^6 - 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d
*x + c)^4 - a^3 - 3*a^2*b - 3*a*b^2 - b^3 + 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d*x + c)^2)*log(-b*cos(d*x +
c)^2 + a + b) - 12*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d*x + c)^6 - 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d*
x + c)^4 - a^3 - 3*a^2*b - 3*a*b^2 - b^3 + 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d*x + c)^2)*log(1/2*sin(d*x +
c)))/(a^4*d*cos(d*x + c)^6 - 3*a^4*d*cos(d*x + c)^4 + 3*a^4*d*cos(d*x + c)^2 - a^4*d)

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giac [B]  time = 0.32, size = 353, normalized size = 2.92 \[ \frac {\frac {a^{2} {\left (\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} + \frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1}\right )}^{3} + 12 \, a^{2} {\left (\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} + \frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1}\right )}^{2} + 6 \, a b {\left (\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} + \frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1}\right )}^{2} + 84 \, a^{2} {\left (\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} + \frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1}\right )} + 120 \, a b {\left (\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} + \frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1}\right )} + 48 \, b^{2} {\left (\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} + \frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1}\right )}}{a^{3}} + \frac {192 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left ({\left | -a {\left (\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} + \frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1}\right )} + 2 \, a + 4 \, b \right |}\right )}{a^{4}}}{384 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^7/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

1/384*((a^2*((cos(d*x + c) + 1)/(cos(d*x + c) - 1) + (cos(d*x + c) - 1)/(cos(d*x + c) + 1))^3 + 12*a^2*((cos(d
*x + c) + 1)/(cos(d*x + c) - 1) + (cos(d*x + c) - 1)/(cos(d*x + c) + 1))^2 + 6*a*b*((cos(d*x + c) + 1)/(cos(d*
x + c) - 1) + (cos(d*x + c) - 1)/(cos(d*x + c) + 1))^2 + 84*a^2*((cos(d*x + c) + 1)/(cos(d*x + c) - 1) + (cos(
d*x + c) - 1)/(cos(d*x + c) + 1)) + 120*a*b*((cos(d*x + c) + 1)/(cos(d*x + c) - 1) + (cos(d*x + c) - 1)/(cos(d
*x + c) + 1)) + 48*b^2*((cos(d*x + c) + 1)/(cos(d*x + c) - 1) + (cos(d*x + c) - 1)/(cos(d*x + c) + 1)))/a^3 +
192*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*log(abs(-a*((cos(d*x + c) + 1)/(cos(d*x + c) - 1) + (cos(d*x + c) - 1)/(co
s(d*x + c) + 1)) + 2*a + 4*b))/a^4)/d

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maple [B]  time = 0.64, size = 489, normalized size = 4.04 \[ \frac {1}{48 d a \left (\cos \left (d x +c \right )-1\right )^{3}}+\frac {5}{32 d a \left (\cos \left (d x +c \right )-1\right )^{2}}+\frac {b}{16 d \,a^{2} \left (\cos \left (d x +c \right )-1\right )^{2}}+\frac {19}{32 d a \left (\cos \left (d x +c \right )-1\right )}+\frac {11 b}{16 d \,a^{2} \left (\cos \left (d x +c \right )-1\right )}+\frac {b^{2}}{4 d \,a^{3} \left (\cos \left (d x +c \right )-1\right )}-\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 d a}-\frac {3 \ln \left (\cos \left (d x +c \right )-1\right ) b}{2 d \,a^{2}}-\frac {3 \ln \left (\cos \left (d x +c \right )-1\right ) b^{2}}{2 d \,a^{3}}-\frac {\ln \left (\cos \left (d x +c \right )-1\right ) b^{3}}{2 d \,a^{4}}+\frac {\ln \left (b \left (\cos ^{2}\left (d x +c \right )\right )-a -b \right )}{2 d a}+\frac {3 \ln \left (b \left (\cos ^{2}\left (d x +c \right )\right )-a -b \right ) b}{2 d \,a^{2}}+\frac {3 \ln \left (b \left (\cos ^{2}\left (d x +c \right )\right )-a -b \right ) b^{2}}{2 d \,a^{3}}+\frac {\ln \left (b \left (\cos ^{2}\left (d x +c \right )\right )-a -b \right ) b^{3}}{2 d \,a^{4}}-\frac {1}{48 d a \left (1+\cos \left (d x +c \right )\right )^{3}}+\frac {5}{32 a d \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {b}{16 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {19}{32 a d \left (1+\cos \left (d x +c \right )\right )}-\frac {11 b}{16 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )}-\frac {b^{2}}{4 d \,a^{3} \left (1+\cos \left (d x +c \right )\right )}-\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{2 d a}-\frac {3 \ln \left (1+\cos \left (d x +c \right )\right ) b}{2 d \,a^{2}}-\frac {3 \ln \left (1+\cos \left (d x +c \right )\right ) b^{2}}{2 d \,a^{3}}-\frac {\ln \left (1+\cos \left (d x +c \right )\right ) b^{3}}{2 d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^7/(a+b*sin(d*x+c)^2),x)

[Out]

1/48/d/a/(cos(d*x+c)-1)^3+5/32/d/a/(cos(d*x+c)-1)^2+1/16/d/a^2/(cos(d*x+c)-1)^2*b+19/32/d/a/(cos(d*x+c)-1)+11/
16/d/a^2/(cos(d*x+c)-1)*b+1/4/d/a^3/(cos(d*x+c)-1)*b^2-1/2/d/a*ln(cos(d*x+c)-1)-3/2/d/a^2*ln(cos(d*x+c)-1)*b-3
/2/d/a^3*ln(cos(d*x+c)-1)*b^2-1/2/d/a^4*ln(cos(d*x+c)-1)*b^3+1/2/d/a*ln(b*cos(d*x+c)^2-a-b)+3/2/d/a^2*ln(b*cos
(d*x+c)^2-a-b)*b+3/2/d/a^3*ln(b*cos(d*x+c)^2-a-b)*b^2+1/2/d/a^4*ln(b*cos(d*x+c)^2-a-b)*b^3-1/48/d/a/(1+cos(d*x
+c))^3+5/32/a/d/(1+cos(d*x+c))^2+1/16/d/a^2/(1+cos(d*x+c))^2*b-19/32/a/d/(1+cos(d*x+c))-11/16/d/a^2/(1+cos(d*x
+c))*b-1/4/d/a^3/(1+cos(d*x+c))*b^2-1/2/d/a*ln(1+cos(d*x+c))-3/2/d/a^2*ln(1+cos(d*x+c))*b-3/2/d/a^3*ln(1+cos(d
*x+c))*b^2-1/2/d/a^4*ln(1+cos(d*x+c))*b^3

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maxima [A]  time = 0.32, size = 137, normalized size = 1.13 \[ \frac {\frac {6 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (b \sin \left (d x + c\right )^{2} + a\right )}{a^{4}} - \frac {6 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (\sin \left (d x + c\right )^{2}\right )}{a^{4}} - \frac {6 \, {\left (3 \, a^{2} + 3 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{4} - 3 \, {\left (3 \, a^{2} + a b\right )} \sin \left (d x + c\right )^{2} + 2 \, a^{2}}{a^{3} \sin \left (d x + c\right )^{6}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^7/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

1/12*(6*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*log(b*sin(d*x + c)^2 + a)/a^4 - 6*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*log(
sin(d*x + c)^2)/a^4 - (6*(3*a^2 + 3*a*b + b^2)*sin(d*x + c)^4 - 3*(3*a^2 + a*b)*sin(d*x + c)^2 + 2*a^2)/(a^3*s
in(d*x + c)^6))/d

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mupad [B]  time = 15.35, size = 138, normalized size = 1.14 \[ \frac {\ln \left (a+a\,{\mathrm {tan}\left (c+d\,x\right )}^2+b\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{2\,a^4\,d}-\frac {\frac {1}{6\,a}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a+b\right )}{4\,a^2}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,{\left (a+b\right )}^2}{2\,a^3}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^6}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^7/(a + b*sin(c + d*x)^2),x)

[Out]

(log(a + a*tan(c + d*x)^2 + b*tan(c + d*x)^2)*(3*a*b^2 + 3*a^2*b + a^3 + b^3))/(2*a^4*d) - (1/(6*a) - (tan(c +
d*x)^2*(a + b))/(4*a^2) + (tan(c + d*x)^4*(a + b)^2)/(2*a^3))/(d*tan(c + d*x)^6) - (log(tan(c + d*x))*(3*a*b^
2 + 3*a^2*b + a^3 + b^3))/(a^4*d)

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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(cot(d*x+c)**7/(a+b*sin(d*x+c)**2),x)

[Out]

Timed out

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