3.99 \(\int \frac {\csc ^3(c+d x)}{(a+b \sin ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=153 \[ -\frac {b^{3/2} (5 a+4 b) \tanh ^{-1}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{2 a^3 d (a+b)^{3/2}}-\frac {(a-4 b) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac {b (a+2 b) \cos (c+d x)}{2 a^2 d (a+b) \left (a-b \cos ^2(c+d x)+b\right )}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d \left (a-b \cos ^2(c+d x)+b\right )} \]

[Out]

-1/2*(a-4*b)*arctanh(cos(d*x+c))/a^3/d-1/2*b^(3/2)*(5*a+4*b)*arctanh(cos(d*x+c)*b^(1/2)/(a+b)^(1/2))/a^3/(a+b)
^(3/2)/d-1/2*b*(a+2*b)*cos(d*x+c)/a^2/(a+b)/d/(a+b-b*cos(d*x+c)^2)-1/2*cot(d*x+c)*csc(d*x+c)/a/d/(a+b-b*cos(d*
x+c)^2)

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Rubi [A]  time = 0.24, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3186, 414, 527, 522, 206, 208} \[ -\frac {b^{3/2} (5 a+4 b) \tanh ^{-1}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{2 a^3 d (a+b)^{3/2}}-\frac {b (a+2 b) \cos (c+d x)}{2 a^2 d (a+b) \left (a-b \cos ^2(c+d x)+b\right )}-\frac {(a-4 b) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d \left (a-b \cos ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a - 4*b)*ArcTanh[Cos[c + d*x]])/(2*a^3*d) - (b^(3/2)*(5*a + 4*b)*ArcTanh[(Sqrt[b]*Cos[c + d*x])/Sqrt[a + b]
])/(2*a^3*(a + b)^(3/2)*d) - (b*(a + 2*b)*Cos[c + d*x])/(2*a^2*(a + b)*d*(a + b - b*Cos[c + d*x]^2)) - (Cot[c
+ d*x]*Csc[c + d*x])/(2*a*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 527

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

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 ^3(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 )^2 \left (a+b-b x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {\cot (c+d x) \csc (c+d x)}{2 a d \left (a+b-b \cos ^2(c+d x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {a-b-3 b x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{2 a d}\\ &=-\frac {b (a+2 b) \cos (c+d x)}{2 a^2 (a+b) d \left (a+b-b \cos ^2(c+d x)\right )}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d \left (a+b-b \cos ^2(c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {-2 \left (a^2-2 a b-2 b^2\right )+2 b (a+2 b) x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\cos (c+d x)\right )}{4 a^2 (a+b) d}\\ &=-\frac {b (a+2 b) \cos (c+d x)}{2 a^2 (a+b) d \left (a+b-b \cos ^2(c+d x)\right )}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d \left (a+b-b \cos ^2(c+d x)\right )}-\frac {(a-4 b) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{2 a^3 d}-\frac {\left (b^2 (5 a+4 b)\right ) \operatorname {Subst}\left (\int \frac {1}{a+b-b x^2} \, dx,x,\cos (c+d x)\right )}{2 a^3 (a+b) d}\\ &=-\frac {(a-4 b) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac {b^{3/2} (5 a+4 b) \tanh ^{-1}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{2 a^3 (a+b)^{3/2} d}-\frac {b (a+2 b) \cos (c+d x)}{2 a^2 (a+b) d \left (a+b-b \cos ^2(c+d x)\right )}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d \left (a+b-b \cos ^2(c+d x)\right )}\\ \end {align*}

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Mathematica [C]  time = 1.57, size = 390, normalized size = 2.55 \[ \frac {\csc ^3(c+d x) (-2 a+b \cos (2 (c+d x))-b) \left (\frac {4 b^{3/2} (5 a+4 b) \csc (c+d x) (2 a-b \cos (2 (c+d x))+b) \tan ^{-1}\left (\frac {\sqrt {b}-i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )}{(-a-b)^{3/2}}+\frac {4 b^{3/2} (5 a+4 b) \csc (c+d x) (2 a-b \cos (2 (c+d x))+b) \tan ^{-1}\left (\frac {\sqrt {b}+i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )}{(-a-b)^{3/2}}+\frac {8 a b^2 \cot (c+d x)}{a+b}+a \csc ^2\left (\frac {1}{2} (c+d x)\right ) \csc (c+d x) (2 a-b \cos (2 (c+d x))+b)+4 (a-4 b) \csc (c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) (2 a-b \cos (2 (c+d x))+b)-a \csc (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (2 a-b \cos (2 (c+d x))+b)-4 (a-4 b) \csc (c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) (2 a-b \cos (2 (c+d x))+b)\right )}{32 a^3 d \left (a \csc ^2(c+d x)+b\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a - b + b*Cos[2*(c + d*x)])*Csc[c + d*x]^3*((8*a*b^2*Cot[c + d*x])/(a + b) + (4*b^(3/2)*(5*a + 4*b)*ArcTa
n[(Sqrt[b] - I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b]]*(2*a + b - b*Cos[2*(c + d*x)])*Csc[c + d*x])/(-a - b)^(
3/2) + (4*b^(3/2)*(5*a + 4*b)*ArcTan[(Sqrt[b] + I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b]]*(2*a + b - b*Cos[2*(
c + d*x)])*Csc[c + d*x])/(-a - b)^(3/2) + a*(2*a + b - b*Cos[2*(c + d*x)])*Csc[(c + d*x)/2]^2*Csc[c + d*x] + 4
*(a - 4*b)*(2*a + b - b*Cos[2*(c + d*x)])*Csc[c + d*x]*Log[Cos[(c + d*x)/2]] - 4*(a - 4*b)*(2*a + b - b*Cos[2*
(c + d*x)])*Csc[c + d*x]*Log[Sin[(c + d*x)/2]] - a*(2*a + b - b*Cos[2*(c + d*x)])*Csc[c + d*x]*Sec[(c + d*x)/2
]^2))/(32*a^3*d*(b + a*Csc[c + d*x]^2)^2)

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fricas [B]  time = 0.57, size = 838, normalized size = 5.48 \[ \left [\frac {2 \, {\left (a^{2} b + 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left ({\left (5 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 5 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3} - {\left (5 \, a^{2} b + 14 \, a b^{2} + 8 \, b^{3}\right )} \cos \left (d x + c\right )^{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 (a^{3} + 2 \, a^{2} b + 2 \, a b^{2}\right )} \cos \left (d x + c\right ) - {\left ({\left (a^{2} b - 3 \, a b^{2} - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + a^{3} - 2 \, a^{2} b - 7 \, a b^{2} - 4 \, b^{3} - {\left (a^{3} - a^{2} b - 10 \, a b^{2} - 8 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{2} b - 3 \, a b^{2} - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + a^{3} - 2 \, a^{2} b - 7 \, a b^{2} - 4 \, b^{3} - {\left (a^{3} - a^{2} b - 10 \, a b^{2} - 8 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{4} b + a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{4} - {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} d\right )}}, \frac {2 \, {\left (a^{2} b + 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left ({\left (5 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 5 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3} - {\left (5 \, a^{2} b + 14 \, a b^{2} + 8 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-\frac {b}{a + b}} \arctan \left (\sqrt {-\frac {b}{a + b}} \cos \left (d x + c\right )\right ) - 2 \, {\left (a^{3} + 2 \, a^{2} b + 2 \, a b^{2}\right )} \cos \left (d x + c\right ) - {\left ({\left (a^{2} b - 3 \, a b^{2} - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + a^{3} - 2 \, a^{2} b - 7 \, a b^{2} - 4 \, b^{3} - {\left (a^{3} - a^{2} b - 10 \, a b^{2} - 8 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{2} b - 3 \, a b^{2} - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + a^{3} - 2 \, a^{2} b - 7 \, a b^{2} - 4 \, b^{3} - {\left (a^{3} - a^{2} b - 10 \, a b^{2} - 8 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{4} b + a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{4} - {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3/(a+b*sin(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

[1/4*(2*(a^2*b + 2*a*b^2)*cos(d*x + c)^3 + ((5*a*b^2 + 4*b^3)*cos(d*x + c)^4 + 5*a^2*b + 9*a*b^2 + 4*b^3 - (5*
a^2*b + 14*a*b^2 + 8*b^3)*cos(d*x + c)^2)*sqrt(b/(a + b))*log(-(b*cos(d*x + c)^2 - 2*(a + b)*sqrt(b/(a + b))*c
os(d*x + c) + a + b)/(b*cos(d*x + c)^2 - a - b)) - 2*(a^3 + 2*a^2*b + 2*a*b^2)*cos(d*x + c) - ((a^2*b - 3*a*b^
2 - 4*b^3)*cos(d*x + c)^4 + a^3 - 2*a^2*b - 7*a*b^2 - 4*b^3 - (a^3 - a^2*b - 10*a*b^2 - 8*b^3)*cos(d*x + c)^2)
*log(1/2*cos(d*x + c) + 1/2) + ((a^2*b - 3*a*b^2 - 4*b^3)*cos(d*x + c)^4 + a^3 - 2*a^2*b - 7*a*b^2 - 4*b^3 - (
a^3 - a^2*b - 10*a*b^2 - 8*b^3)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2))/((a^4*b + a^3*b^2)*d*cos(d*x + c
)^4 - (a^5 + 3*a^4*b + 2*a^3*b^2)*d*cos(d*x + c)^2 + (a^5 + 2*a^4*b + a^3*b^2)*d), 1/4*(2*(a^2*b + 2*a*b^2)*co
s(d*x + c)^3 + 2*((5*a*b^2 + 4*b^3)*cos(d*x + c)^4 + 5*a^2*b + 9*a*b^2 + 4*b^3 - (5*a^2*b + 14*a*b^2 + 8*b^3)*
cos(d*x + c)^2)*sqrt(-b/(a + b))*arctan(sqrt(-b/(a + b))*cos(d*x + c)) - 2*(a^3 + 2*a^2*b + 2*a*b^2)*cos(d*x +
c) - ((a^2*b - 3*a*b^2 - 4*b^3)*cos(d*x + c)^4 + a^3 - 2*a^2*b - 7*a*b^2 - 4*b^3 - (a^3 - a^2*b - 10*a*b^2 -
8*b^3)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) + ((a^2*b - 3*a*b^2 - 4*b^3)*cos(d*x + c)^4 + a^3 - 2*a^2*b
- 7*a*b^2 - 4*b^3 - (a^3 - a^2*b - 10*a*b^2 - 8*b^3)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2))/((a^4*b +
a^3*b^2)*d*cos(d*x + c)^4 - (a^5 + 3*a^4*b + 2*a^3*b^2)*d*cos(d*x + c)^2 + (a^5 + 2*a^4*b + a^3*b^2)*d)]

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giac [B]  time = 0.20, size = 512, normalized size = 3.35 \[ \frac {\frac {12 \, {\left (5 \, a b^{2} + 4 \, b^{3}\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^{4} + a^{3} b\right )} \sqrt {-a b - b^{2}}} + \frac {3 \, a^{3} + 3 \, a^{2} b - \frac {8 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {12 \, a^{2} b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {28 \, a b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {7 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {a^{2} b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {16 \, a b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {16 \, b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {2 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {6 \, a^{2} b {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {8 \, a b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{{\left (a^{4} + a^{3} b\right )} {\left (\frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {4 \, b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}} + \frac {6 \, {\left (a - 4 \, b\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{3}} - \frac {3 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{a^{2} {\left (\cos \left (d x + c\right ) + 1\right )}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3/(a+b*sin(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/24*(12*(5*a*b^2 + 4*b^3)*arctan((b*cos(d*x + c) + a + b)/(sqrt(-a*b - b^2)*cos(d*x + c) + sqrt(-a*b - b^2)))
/((a^4 + a^3*b)*sqrt(-a*b - b^2)) + (3*a^3 + 3*a^2*b - 8*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 12*a^2*b*
(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 28*a*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 7*a^3*(cos(d*x + c) -
1)^2/(cos(d*x + c) + 1)^2 - a^2*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 16*a*b^2*(cos(d*x + c) - 1)^2/(
cos(d*x + c) + 1)^2 + 16*b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 2*a^3*(cos(d*x + c) - 1)^3/(cos(d*x +
c) + 1)^3 + 6*a^2*b*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 8*a*b^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) +
1)^3)/((a^4 + a^3*b)*(a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2
- 4*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + a*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3)) + 6*(a - 4*b)*
log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^3 - 3*(cos(d*x + c) - 1)/(a^2*(cos(d*x + c) + 1)))/d

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maple [A]  time = 0.62, size = 226, normalized size = 1.48 \[ \frac {1}{4 d \,a^{2} \left (\cos \left (d x +c \right )-1\right )}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{4 d \,a^{2}}-\frac {\ln \left (\cos \left (d x +c \right )-1\right ) b}{d \,a^{3}}+\frac {b^{2} \cos \left (d x +c \right )}{2 d \,a^{2} \left (a +b \right ) \left (b \left (\cos ^{2}\left (d x +c \right )\right )-a -b \right )}-\frac {5 b^{2} \arctanh \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{2 d \,a^{2} \left (a +b \right ) \sqrt {\left (a +b \right ) b}}-\frac {2 b^{3} \arctanh \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{d \,a^{3} \left (a +b \right ) \sqrt {\left (a +b \right ) b}}+\frac {1}{4 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )}-\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {\ln \left (1+\cos \left (d x +c \right )\right ) b}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3/(a+b*sin(d*x+c)^2)^2,x)

[Out]

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

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maxima [A]  time = 0.48, size = 223, normalized size = 1.46 \[ \frac {\frac {{\left (5 \, a b^{2} + 4 \, b^{3}\right )} \log \left (\frac {b \cos \left (d x + c\right ) - \sqrt {{\left (a + b\right )} b}}{b \cos \left (d x + c\right ) + \sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{4} + a^{3} b\right )} \sqrt {{\left (a + b\right )} b}} + \frac {2 \, {\left ({\left (a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )\right )}}{{\left (a^{3} b + a^{2} b^{2}\right )} \cos \left (d x + c\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2} - {\left (a^{4} + 3 \, a^{3} b + 2 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}} - \frac {{\left (a - 4 \, b\right )} \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {{\left (a - 4 \, b\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3/(a+b*sin(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

1/4*((5*a*b^2 + 4*b^3)*log((b*cos(d*x + c) - sqrt((a + b)*b))/(b*cos(d*x + c) + sqrt((a + b)*b)))/((a^4 + a^3*
b)*sqrt((a + b)*b)) + 2*((a*b + 2*b^2)*cos(d*x + c)^3 - (a^2 + 2*a*b + 2*b^2)*cos(d*x + c))/((a^3*b + a^2*b^2)
*cos(d*x + c)^4 + a^4 + 2*a^3*b + a^2*b^2 - (a^4 + 3*a^3*b + 2*a^2*b^2)*cos(d*x + c)^2) - (a - 4*b)*log(cos(d*
x + c) + 1)/a^3 + (a - 4*b)*log(cos(d*x + c) - 1)/a^3)/d

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mupad [B]  time = 14.95, size = 2338, normalized size = 15.28 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^3*(a + b*sin(c + d*x)^2)^2),x)

[Out]

- ((cos(c + d*x)*(2*a*b + a^2 + 2*b^2))/(2*a^2*(a + b)) - (b*cos(c + d*x)^3*(a + 2*b))/(2*a^2*(a + b)))/(d*(a
+ b + b*cos(c + d*x)^4 - cos(c + d*x)^2*(a + 2*b))) - (atan((((a - 4*b)*((cos(c + d*x)*(64*a*b^6 + 32*b^7 + 26
*a^2*b^5 - 6*a^3*b^4 + a^4*b^3))/(2*(2*a^5*b + a^6 + a^4*b^2)) + (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^8*b^3 - 2*a^9*
b^2)/(2*a^7*b + a^8 + a^6*b^2) - (cos(c + d*x)*(a - 4*b)*(32*a^6*b^5 + 80*a^7*b^4 + 64*a^8*b^3 + 16*a^9*b^2))/
(8*a^3*(2*a^5*b + a^6 + a^4*b^2)))*(a - 4*b))/(4*a^3))*1i)/(4*a^3) + ((a - 4*b)*((cos(c + d*x)*(64*a*b^6 + 32*
b^7 + 26*a^2*b^5 - 6*a^3*b^4 + a^4*b^3))/(2*(2*a^5*b + a^6 + a^4*b^2)) - (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^8*b^3
- 2*a^9*b^2)/(2*a^7*b + a^8 + a^6*b^2) + (cos(c + d*x)*(a - 4*b)*(32*a^6*b^5 + 80*a^7*b^4 + 64*a^8*b^3 + 16*a^
9*b^2))/(8*a^3*(2*a^5*b + a^6 + a^4*b^2)))*(a - 4*b))/(4*a^3))*1i)/(4*a^3))/((12*a*b^6 + 8*b^7 + (3*a^2*b^5)/2
- (5*a^3*b^4)/4)/(2*a^7*b + a^8 + a^6*b^2) - ((a - 4*b)*((cos(c + d*x)*(64*a*b^6 + 32*b^7 + 26*a^2*b^5 - 6*a^
3*b^4 + a^4*b^3))/(2*(2*a^5*b + a^6 + a^4*b^2)) + (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^8*b^3 - 2*a^9*b^2)/(2*a^7*b +
a^8 + a^6*b^2) - (cos(c + d*x)*(a - 4*b)*(32*a^6*b^5 + 80*a^7*b^4 + 64*a^8*b^3 + 16*a^9*b^2))/(8*a^3*(2*a^5*b
+ a^6 + a^4*b^2)))*(a - 4*b))/(4*a^3)))/(4*a^3) + ((a - 4*b)*((cos(c + d*x)*(64*a*b^6 + 32*b^7 + 26*a^2*b^5 -
6*a^3*b^4 + a^4*b^3))/(2*(2*a^5*b + a^6 + a^4*b^2)) - (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^8*b^3 - 2*a^9*b^2)/(2*a^
7*b + a^8 + a^6*b^2) + (cos(c + d*x)*(a - 4*b)*(32*a^6*b^5 + 80*a^7*b^4 + 64*a^8*b^3 + 16*a^9*b^2))/(8*a^3*(2*
a^5*b + a^6 + a^4*b^2)))*(a - 4*b))/(4*a^3)))/(4*a^3)))*(a - 4*b)*1i)/(2*a^3*d) - (atan(((((5*a)/4 + b)*(b^3*(
a + b)^3)^(1/2)*((cos(c + d*x)*(64*a*b^6 + 32*b^7 + 26*a^2*b^5 - 6*a^3*b^4 + a^4*b^3))/(2*(2*a^5*b + a^6 + a^4
*b^2)) + (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^8*b^3 - 2*a^9*b^2)/(2*a^7*b + a^8 + a^6*b^2) - (cos(c + d*x)*((5*a)/4
+ b)*(b^3*(a + b)^3)^(1/2)*(32*a^6*b^5 + 80*a^7*b^4 + 64*a^8*b^3 + 16*a^9*b^2))/(2*(2*a^5*b + a^6 + a^4*b^2)*(
3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2)))*((5*a)/4 + b)*(b^3*(a + b)^3)^(1/2))/(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b
^2))*1i)/(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2) + (((5*a)/4 + b)*(b^3*(a + b)^3)^(1/2)*((cos(c + d*x)*(64*a*b^6
+ 32*b^7 + 26*a^2*b^5 - 6*a^3*b^4 + a^4*b^3))/(2*(2*a^5*b + a^6 + a^4*b^2)) - (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^
8*b^3 - 2*a^9*b^2)/(2*a^7*b + a^8 + a^6*b^2) + (cos(c + d*x)*((5*a)/4 + b)*(b^3*(a + b)^3)^(1/2)*(32*a^6*b^5 +
80*a^7*b^4 + 64*a^8*b^3 + 16*a^9*b^2))/(2*(2*a^5*b + a^6 + a^4*b^2)*(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2)))*(
(5*a)/4 + b)*(b^3*(a + b)^3)^(1/2))/(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2))*1i)/(3*a^5*b + a^6 + a^3*b^3 + 3*a^
4*b^2))/((12*a*b^6 + 8*b^7 + (3*a^2*b^5)/2 - (5*a^3*b^4)/4)/(2*a^7*b + a^8 + a^6*b^2) - (((5*a)/4 + b)*(b^3*(a
+ b)^3)^(1/2)*((cos(c + d*x)*(64*a*b^6 + 32*b^7 + 26*a^2*b^5 - 6*a^3*b^4 + a^4*b^3))/(2*(2*a^5*b + a^6 + a^4*
b^2)) + (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^8*b^3 - 2*a^9*b^2)/(2*a^7*b + a^8 + a^6*b^2) - (cos(c + d*x)*((5*a)/4 +
b)*(b^3*(a + b)^3)^(1/2)*(32*a^6*b^5 + 80*a^7*b^4 + 64*a^8*b^3 + 16*a^9*b^2))/(2*(2*a^5*b + a^6 + a^4*b^2)*(3
*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2)))*((5*a)/4 + b)*(b^3*(a + b)^3)^(1/2))/(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^
2)))/(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2) + (((5*a)/4 + b)*(b^3*(a + b)^3)^(1/2)*((cos(c + d*x)*(64*a*b^6 + 3
2*b^7 + 26*a^2*b^5 - 6*a^3*b^4 + a^4*b^3))/(2*(2*a^5*b + a^6 + a^4*b^2)) - (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^8*b^
3 - 2*a^9*b^2)/(2*a^7*b + a^8 + a^6*b^2) + (cos(c + d*x)*((5*a)/4 + b)*(b^3*(a + b)^3)^(1/2)*(32*a^6*b^5 + 80*
a^7*b^4 + 64*a^8*b^3 + 16*a^9*b^2))/(2*(2*a^5*b + a^6 + a^4*b^2)*(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2)))*((5*a
)/4 + b)*(b^3*(a + b)^3)^(1/2))/(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2)))/(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2))
)*((5*a)/4 + b)*(b^3*(a + b)^3)^(1/2)*2i)/(d*(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^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)**3/(a+b*sin(d*x+c)**2)**2,x)

[Out]

Timed out

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