3.98 \(\int \frac {\csc (c+d x)}{(a+b \sin ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=103 \[ \frac {\sqrt {b} (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{2 a^2 d (a+b)^{3/2}}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {b \cos (c+d x)}{2 a d (a+b) \left (a-b \cos ^2(c+d x)+b\right )} \]

[Out]

-arctanh(cos(d*x+c))/a^2/d+1/2*b*cos(d*x+c)/a/(a+b)/d/(a+b-b*cos(d*x+c)^2)+1/2*(3*a+2*b)*arctanh(cos(d*x+c)*b^
(1/2)/(a+b)^(1/2))*b^(1/2)/a^2/(a+b)^(3/2)/d

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Rubi [A]  time = 0.13, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3186, 414, 522, 206, 208} \[ \frac {\sqrt {b} (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{2 a^2 d (a+b)^{3/2}}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {b \cos (c+d x)}{2 a d (a+b) \left (a-b \cos ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Cos[c + d*x]]/(a^2*d)) + (Sqrt[b]*(3*a + 2*b)*ArcTanh[(Sqrt[b]*Cos[c + d*x])/Sqrt[a + b]])/(2*a^2*(a
+ b)^(3/2)*d) + (b*Cos[c + d*x])/(2*a*(a + b)*d*(a + b - b*Cos[c + d*x]^2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 208

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

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 522

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

Rule 3186

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

Rubi steps

\begin {align*} \int \frac {\csc (c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \left (a+b-b x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {b \cos (c+d x)}{2 a (a+b) d \left (a+b-b \cos ^2(c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {-2 a-b-b x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\cos (c+d x)\right )}{2 a (a+b) d}\\ &=\frac {b \cos (c+d x)}{2 a (a+b) d \left (a+b-b \cos ^2(c+d x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{a^2 d}+\frac {(b (3 a+2 b)) \operatorname {Subst}\left (\int \frac {1}{a+b-b x^2} \, dx,x,\cos (c+d x)\right )}{2 a^2 (a+b) d}\\ &=-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\sqrt {b} (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{2 a^2 (a+b)^{3/2} d}+\frac {b \cos (c+d x)}{2 a (a+b) d \left (a+b-b \cos ^2(c+d x)\right )}\\ \end {align*}

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Mathematica [C]  time = 0.80, size = 194, normalized size = 1.88 \[ \frac {\frac {\sqrt {b} (3 a+2 b) \tan ^{-1}\left (\frac {\sqrt {b}-i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )}{(-a-b)^{3/2}}+\frac {\sqrt {b} (3 a+2 b) \tan ^{-1}\left (\frac {\sqrt {b}+i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )}{(-a-b)^{3/2}}+2 \left (\frac {a b \cos (c+d x)}{(a+b) (2 a-b \cos (2 (c+d x))+b)}+\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[b]*(3*a + 2*b)*ArcTan[(Sqrt[b] - I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b]])/(-a - b)^(3/2) + (Sqrt[b]*(
3*a + 2*b)*ArcTan[(Sqrt[b] + I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b]])/(-a - b)^(3/2) + 2*((a*b*Cos[c + d*x])
/((a + b)*(2*a + b - b*Cos[2*(c + d*x)])) - Log[Cos[(c + d*x)/2]] + Log[Sin[(c + d*x)/2]]))/(2*a^2*d)

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fricas [B]  time = 0.54, size = 455, normalized size = 4.42 \[ \left [-\frac {2 \, a b \cos \left (d x + c\right ) - {\left ({\left (3 \, a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, a^{2} - 5 \, a b - 2 \, b^{2}\right )} \sqrt {\frac {b}{a + b}} \log \left (\frac {b \cos \left (d x + c\right )^{2} + 2 \, {\left (a + b\right )} \sqrt {\frac {b}{a + b}} \cos \left (d x + c\right ) + a + b}{b \cos \left (d x + c\right )^{2} - a - b}\right ) + 2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{3} b + a^{2} b^{2}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} d\right )}}, -\frac {a b \cos \left (d x + c\right ) + {\left ({\left (3 \, a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, a^{2} - 5 \, a b - 2 \, b^{2}\right )} \sqrt {-\frac {b}{a + b}} \arctan \left (\sqrt {-\frac {b}{a + b}} \cos \left (d x + c\right )\right ) + {\left ({\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left ({\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left ({\left (a^{3} b + a^{2} b^{2}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(2*a*b*cos(d*x + c) - ((3*a*b + 2*b^2)*cos(d*x + c)^2 - 3*a^2 - 5*a*b - 2*b^2)*sqrt(b/(a + b))*log((b*co
s(d*x + c)^2 + 2*(a + b)*sqrt(b/(a + b))*cos(d*x + c) + a + b)/(b*cos(d*x + c)^2 - a - b)) + 2*((a*b + b^2)*co
s(d*x + c)^2 - a^2 - 2*a*b - b^2)*log(1/2*cos(d*x + c) + 1/2) - 2*((a*b + b^2)*cos(d*x + c)^2 - a^2 - 2*a*b -
b^2)*log(-1/2*cos(d*x + c) + 1/2))/((a^3*b + a^2*b^2)*d*cos(d*x + c)^2 - (a^4 + 2*a^3*b + a^2*b^2)*d), -1/2*(a
*b*cos(d*x + c) + ((3*a*b + 2*b^2)*cos(d*x + c)^2 - 3*a^2 - 5*a*b - 2*b^2)*sqrt(-b/(a + b))*arctan(sqrt(-b/(a
+ b))*cos(d*x + c)) + ((a*b + b^2)*cos(d*x + c)^2 - a^2 - 2*a*b - b^2)*log(1/2*cos(d*x + c) + 1/2) - ((a*b + b
^2)*cos(d*x + c)^2 - a^2 - 2*a*b - b^2)*log(-1/2*cos(d*x + c) + 1/2))/((a^3*b + a^2*b^2)*d*cos(d*x + c)^2 - (a
^4 + 2*a^3*b + a^2*b^2)*d)]

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giac [B]  time = 0.19, size = 246, normalized size = 2.39 \[ -\frac {\frac {{\left (3 \, a b + 2 \, b^{2}\right )} \arctan \left (\frac {b \cos \left (d x + c\right ) + a + b}{\sqrt {-a b - b^{2}} \cos \left (d x + c\right ) + \sqrt {-a b - b^{2}}}\right )}{{\left (a^{3} + a^{2} b\right )} \sqrt {-a b - b^{2}}} - \frac {2 \, {\left (a b - \frac {a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {2 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{{\left (a^{3} + a^{2} b\right )} {\left (a - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {4 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}} - \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((3*a*b + 2*b^2)*arctan((b*cos(d*x + c) + a + b)/(sqrt(-a*b - b^2)*cos(d*x + c) + sqrt(-a*b - b^2)))/((a^
3 + a^2*b)*sqrt(-a*b - b^2)) - 2*(a*b - a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*b^2*(cos(d*x + c) - 1)/(
cos(d*x + c) + 1))/((a^3 + a^2*b)*(a - 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 4*b*(cos(d*x + c) - 1)/(cos
(d*x + c) + 1) + a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)) - log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) +
1))/a^2)/d

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maple [A]  time = 0.51, size = 150, normalized size = 1.46 \[ \frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 d \,a^{2}}-\frac {b \cos \left (d x +c \right )}{2 d a \left (a +b \right ) \left (b \left (\cos ^{2}\left (d x +c \right )\right )-a -b \right )}+\frac {3 b \arctanh \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{2 d a \left (a +b \right ) \sqrt {\left (a +b \right ) b}}+\frac {b^{2} \arctanh \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{d \,a^{2} \left (a +b \right ) \sqrt {\left (a +b \right ) b}}-\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{2 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/a^2*ln(cos(d*x+c)-1)-1/2/d/a*b/(a+b)*cos(d*x+c)/(b*cos(d*x+c)^2-a-b)+3/2/d/a*b/(a+b)/((a+b)*b)^(1/2)*arc
tanh(cos(d*x+c)*b/((a+b)*b)^(1/2))+1/d/a^2*b^2/(a+b)/((a+b)*b)^(1/2)*arctanh(cos(d*x+c)*b/((a+b)*b)^(1/2))-1/2
/d/a^2*ln(1+cos(d*x+c))

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maxima [A]  time = 0.57, size = 149, normalized size = 1.45 \[ \frac {\frac {2 \, b \cos \left (d x + c\right )}{a^{3} + 2 \, a^{2} b + a b^{2} - {\left (a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{2}} - \frac {{\left (3 \, a b + 2 \, b^{2}\right )} \log \left (\frac {b \cos \left (d x + c\right ) - \sqrt {{\left (a + b\right )} b}}{b \cos \left (d x + c\right ) + \sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{3} + a^{2} b\right )} \sqrt {{\left (a + b\right )} b}} - \frac {2 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {2 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 14.63, size = 2039, normalized size = 19.80 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan(((((cos(c + d*x)*(20*a*b^4 + 8*b^5 + 13*a^2*b^3))/(2*(2*a^3*b + a^4 + a^2*b^2)) + ((b*(a + b)^3)^(1/2)*(
(2*a^4*b^4 + 6*a^5*b^3 + 4*a^6*b^2)/(2*a^4*b + a^5 + a^3*b^2) - (cos(c + d*x)*(b*(a + b)^3)^(1/2)*(3*a + 2*b)*
(32*a^4*b^5 + 80*a^5*b^4 + 64*a^6*b^3 + 16*a^7*b^2))/(8*(2*a^3*b + a^4 + a^2*b^2)*(3*a^4*b + a^5 + a^2*b^3 + 3
*a^3*b^2)))*(3*a + 2*b))/(4*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)))*(b*(a + b)^3)^(1/2)*(3*a + 2*b)*1i)/(4*(3*
a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)) + (((cos(c + d*x)*(20*a*b^4 + 8*b^5 + 13*a^2*b^3))/(2*(2*a^3*b + a^4 + a^2
*b^2)) - ((b*(a + b)^3)^(1/2)*((2*a^4*b^4 + 6*a^5*b^3 + 4*a^6*b^2)/(2*a^4*b + a^5 + a^3*b^2) + (cos(c + d*x)*(
b*(a + b)^3)^(1/2)*(3*a + 2*b)*(32*a^4*b^5 + 80*a^5*b^4 + 64*a^6*b^3 + 16*a^7*b^2))/(8*(2*a^3*b + a^4 + a^2*b^
2)*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)))*(3*a + 2*b))/(4*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)))*(b*(a + b)^
3)^(1/2)*(3*a + 2*b)*1i)/(4*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)))/(((3*a*b^3)/2 + b^4)/(2*a^4*b + a^5 + a^3*
b^2) - (((cos(c + d*x)*(20*a*b^4 + 8*b^5 + 13*a^2*b^3))/(2*(2*a^3*b + a^4 + a^2*b^2)) + ((b*(a + b)^3)^(1/2)*(
(2*a^4*b^4 + 6*a^5*b^3 + 4*a^6*b^2)/(2*a^4*b + a^5 + a^3*b^2) - (cos(c + d*x)*(b*(a + b)^3)^(1/2)*(3*a + 2*b)*
(32*a^4*b^5 + 80*a^5*b^4 + 64*a^6*b^3 + 16*a^7*b^2))/(8*(2*a^3*b + a^4 + a^2*b^2)*(3*a^4*b + a^5 + a^2*b^3 + 3
*a^3*b^2)))*(3*a + 2*b))/(4*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)))*(b*(a + b)^3)^(1/2)*(3*a + 2*b))/(4*(3*a^4
*b + a^5 + a^2*b^3 + 3*a^3*b^2)) + (((cos(c + d*x)*(20*a*b^4 + 8*b^5 + 13*a^2*b^3))/(2*(2*a^3*b + a^4 + a^2*b^
2)) - ((b*(a + b)^3)^(1/2)*((2*a^4*b^4 + 6*a^5*b^3 + 4*a^6*b^2)/(2*a^4*b + a^5 + a^3*b^2) + (cos(c + d*x)*(b*(
a + b)^3)^(1/2)*(3*a + 2*b)*(32*a^4*b^5 + 80*a^5*b^4 + 64*a^6*b^3 + 16*a^7*b^2))/(8*(2*a^3*b + a^4 + a^2*b^2)*
(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)))*(3*a + 2*b))/(4*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)))*(b*(a + b)^3)^
(1/2)*(3*a + 2*b))/(4*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2))))*(b*(a + b)^3)^(1/2)*(3*a + 2*b)*1i)/(2*d*(3*a^4
*b + a^5 + a^2*b^3 + 3*a^3*b^2)) - (atan((((((2*a^4*b^4 + 6*a^5*b^3 + 4*a^6*b^2)/(2*(2*a^4*b + a^5 + a^3*b^2))
- (cos(c + d*x)*(32*a^4*b^5 + 80*a^5*b^4 + 64*a^6*b^3 + 16*a^7*b^2))/(8*a^2*(2*a^3*b + a^4 + a^2*b^2)))*1i)/(
2*a^2) + (cos(c + d*x)*(20*a*b^4 + 8*b^5 + 13*a^2*b^3)*1i)/(4*(2*a^3*b + a^4 + a^2*b^2)))/a^2 - ((((2*a^4*b^4
+ 6*a^5*b^3 + 4*a^6*b^2)/(2*(2*a^4*b + a^5 + a^3*b^2)) + (cos(c + d*x)*(32*a^4*b^5 + 80*a^5*b^4 + 64*a^6*b^3 +
16*a^7*b^2))/(8*a^2*(2*a^3*b + a^4 + a^2*b^2)))*1i)/(2*a^2) - (cos(c + d*x)*(20*a*b^4 + 8*b^5 + 13*a^2*b^3)*1
i)/(4*(2*a^3*b + a^4 + a^2*b^2)))/a^2)/((((2*a^4*b^4 + 6*a^5*b^3 + 4*a^6*b^2)/(2*(2*a^4*b + a^5 + a^3*b^2)) -
(cos(c + d*x)*(32*a^4*b^5 + 80*a^5*b^4 + 64*a^6*b^3 + 16*a^7*b^2))/(8*a^2*(2*a^3*b + a^4 + a^2*b^2)))/(2*a^2)
+ (cos(c + d*x)*(20*a*b^4 + 8*b^5 + 13*a^2*b^3))/(4*(2*a^3*b + a^4 + a^2*b^2)))/a^2 + (((2*a^4*b^4 + 6*a^5*b^3
+ 4*a^6*b^2)/(2*(2*a^4*b + a^5 + a^3*b^2)) + (cos(c + d*x)*(32*a^4*b^5 + 80*a^5*b^4 + 64*a^6*b^3 + 16*a^7*b^2
))/(8*a^2*(2*a^3*b + a^4 + a^2*b^2)))/(2*a^2) - (cos(c + d*x)*(20*a*b^4 + 8*b^5 + 13*a^2*b^3))/(4*(2*a^3*b + a
^4 + a^2*b^2)))/a^2 - ((3*a*b^3)/2 + b^4)/(2*a^4*b + a^5 + a^3*b^2)))*1i)/(a^2*d) + (b*cos(c + d*x))/(2*a*d*(a
+ b)*(a + b - b*cos(c + 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(csc(d*x+c)/(a+b*sin(d*x+c)**2)**2,x)

[Out]

Timed out

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