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

Optimal. Leaf size=85 \[ -\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{a^2 d \sqrt {a+b}}-\frac {(a-2 b) \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3186, 414, 522, 206, 208} \[ -\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{a^2 d \sqrt {a+b}}-\frac {(a-2 b) \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 2.21, size = 224, normalized size = 2.64 \[ -\frac {\csc ^2(c+d x) (2 a-b \cos (2 (c+d x))+b) \left (-8 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b}-i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )-8 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b}+i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )+\sqrt {-a-b} \left (4 (a-2 b) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+a \csc ^2\left (\frac {1}{2} (c+d x)\right )-a \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )}{16 a^2 d \sqrt {-a-b} \left (a \csc ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*((2*a + b - b*Cos[2*(c + d*x)])*Csc[c + d*x]^2*(-8*b^(3/2)*ArcTan[(Sqrt[b] - I*Sqrt[a]*Tan[(c + d*x)/2])
/Sqrt[-a - b]] - 8*b^(3/2)*ArcTan[(Sqrt[b] + I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b]] + Sqrt[-a - b]*(a*Csc[(
c + d*x)/2]^2 + 4*(a - 2*b)*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]]) - a*Sec[(c + d*x)/2]^2)))/(a^2*Sqr
t[-a - b]*d*(b + a*Csc[c + d*x]^2))

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fricas [A]  time = 0.49, size = 327, normalized size = 3.85 \[ \left [\frac {2 \, {\left (b \cos \left (d x + c\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}} \log \left (-\frac {b \cos \left (d x + c\right )^{2} - 2 \, {\left (a + b\right )} \sqrt {\frac {b}{a + b}} \cos \left (d x + c\right ) + a + b}{b \cos \left (d x + c\right )^{2} - a - b}\right ) + 2 \, a \cos \left (d x + c\right ) - {\left ({\left (a - 2 \, b\right )} \cos \left (d x + c\right )^{2} - a + 2 \, b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (a - 2 \, b\right )} \cos \left (d x + c\right )^{2} - a + 2 \, b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}}, \frac {4 \, {\left (b \cos \left (d x + c\right )^{2} - b\right )} \sqrt {-\frac {b}{a + b}} \arctan \left (\sqrt {-\frac {b}{a + b}} \cos \left (d x + c\right )\right ) + 2 \, a \cos \left (d x + c\right ) - {\left ({\left (a - 2 \, b\right )} \cos \left (d x + c\right )^{2} - a + 2 \, b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (a - 2 \, b\right )} \cos \left (d x + c\right )^{2} - a + 2 \, b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.21, size = 196, normalized size = 2.31 \[ \frac {\frac {8 \, b^{2} \arctan \left (\frac {b \cos \left (d x + c\right ) + a + b}{\sqrt {-a b - b^{2}} \cos \left (d x + c\right ) + \sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} a^{2}} + \frac {2 \, {\left (a - 2 \, b\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2}} + \frac {{\left (a - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {4 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}} - \frac {\cos \left (d x + c\right ) - 1}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*b^2*arctan((b*cos(d*x + c) + a + b)/(sqrt(-a*b - b^2)*cos(d*x + c) + sqrt(-a*b - b^2)))/(sqrt(-a*b - b^
2)*a^2) + 2*(a - 2*b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^2 + (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))*(cos(d*x + c) + 1)/(a^2*(cos(d*x + c) - 1)) - (cos
(d*x + c) - 1)/(a*(cos(d*x + c) + 1)))/d

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maple [A]  time = 0.55, size = 142, normalized size = 1.67 \[ \frac {1}{4 a d \left (\cos \left (d x +c \right )-1\right )}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{4 a d}-\frac {\ln \left (\cos \left (d x +c \right )-1\right ) b}{2 d \,a^{2}}-\frac {b^{2} \arctanh \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{d \,a^{2} \sqrt {\left (a +b \right ) b}}+\frac {1}{4 a d \left (1+\cos \left (d x +c \right )\right )}-\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{4 a d}+\frac {\ln \left (1+\cos \left (d x +c \right )\right ) b}{2 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/a/d/(cos(d*x+c)-1)+1/4/a/d*ln(cos(d*x+c)-1)-1/2/d/a^2*ln(cos(d*x+c)-1)*b-1/d*b^2/a^2/((a+b)*b)^(1/2)*arcta
nh(cos(d*x+c)*b/((a+b)*b)^(1/2))+1/4/a/d/(1+cos(d*x+c))-1/4/a/d*ln(1+cos(d*x+c))+1/2/d/a^2*ln(1+cos(d*x+c))*b

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maxima [A]  time = 0.43, size = 120, normalized size = 1.41 \[ \frac {\frac {2 \, b^{2} \log \left (\frac {b \cos \left (d x + c\right ) - \sqrt {{\left (a + b\right )} b}}{b \cos \left (d x + c\right ) + \sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{2}} + \frac {2 \, \cos \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a} - \frac {{\left (a - 2 \, b\right )} \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {{\left (a - 2 \, b\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 13.91, size = 592, normalized size = 6.96 \[ -\frac {a\,\left (b\,\cos \left (c+d\,x\right )-b\,\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )+b\,{\cos \left (c+d\,x\right )}^2\,\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )\right )+a^2\,\left (\cos \left (c+d\,x\right )+\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )-{\cos \left (c+d\,x\right )}^2\,\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )\right )-2\,b^2\,\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )+\mathrm {atan}\left (\frac {-a\,\cos \left (c+d\,x\right )\,{\left (b^4+a\,b^3\right )}^{3/2}\,4{}\mathrm {i}-b\,\cos \left (c+d\,x\right )\,{\left (b^4+a\,b^3\right )}^{3/2}\,8{}\mathrm {i}+b^5\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,8{}\mathrm {i}+a^2\,b^3\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,1{}\mathrm {i}-a^3\,b^2\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,2{}\mathrm {i}+a\,b^4\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,12{}\mathrm {i}+a^4\,b\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,1{}\mathrm {i}}{-a^5\,b^2+a^4\,b^3+5\,a^3\,b^4+3\,a^2\,b^5}\right )\,\sqrt {b^4+a\,b^3}\,2{}\mathrm {i}+2\,b^2\,{\cos \left (c+d\,x\right )}^2\,\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )-{\cos \left (c+d\,x\right )}^2\,\mathrm {atan}\left (\frac {-a\,\cos \left (c+d\,x\right )\,{\left (b^4+a\,b^3\right )}^{3/2}\,4{}\mathrm {i}-b\,\cos \left (c+d\,x\right )\,{\left (b^4+a\,b^3\right )}^{3/2}\,8{}\mathrm {i}+b^5\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,8{}\mathrm {i}+a^2\,b^3\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,1{}\mathrm {i}-a^3\,b^2\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,2{}\mathrm {i}+a\,b^4\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,12{}\mathrm {i}+a^4\,b\,\cos \left (c+d\,x\right )\,\sqrt {b^4+a\,b^3}\,1{}\mathrm {i}}{-a^5\,b^2+a^4\,b^3+5\,a^3\,b^4+3\,a^2\,b^5}\right )\,\sqrt {b^4+a\,b^3}\,2{}\mathrm {i}}{d\,\left (-2\,a^3\,{\cos \left (c+d\,x\right )}^2+2\,a^3-2\,b\,a^2\,{\cos \left (c+d\,x\right )}^2+2\,b\,a^2\right )} \]

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)),x)

[Out]

-(a*(b*cos(c + d*x) - b*atanh(cos(c + d*x)) + b*cos(c + d*x)^2*atanh(cos(c + d*x))) + a^2*(cos(c + d*x) + atan
h(cos(c + d*x)) - cos(c + d*x)^2*atanh(cos(c + d*x))) - 2*b^2*atanh(cos(c + d*x)) + atan((b^5*cos(c + d*x)*(a*
b^3 + b^4)^(1/2)*8i - b*cos(c + d*x)*(a*b^3 + b^4)^(3/2)*8i - a*cos(c + d*x)*(a*b^3 + b^4)^(3/2)*4i + a^2*b^3*
cos(c + d*x)*(a*b^3 + b^4)^(1/2)*1i - a^3*b^2*cos(c + d*x)*(a*b^3 + b^4)^(1/2)*2i + a*b^4*cos(c + d*x)*(a*b^3
+ b^4)^(1/2)*12i + a^4*b*cos(c + d*x)*(a*b^3 + b^4)^(1/2)*1i)/(3*a^2*b^5 + 5*a^3*b^4 + a^4*b^3 - a^5*b^2))*(a*
b^3 + b^4)^(1/2)*2i + 2*b^2*cos(c + d*x)^2*atanh(cos(c + d*x)) - cos(c + d*x)^2*atan((b^5*cos(c + d*x)*(a*b^3
+ b^4)^(1/2)*8i - b*cos(c + d*x)*(a*b^3 + b^4)^(3/2)*8i - a*cos(c + d*x)*(a*b^3 + b^4)^(3/2)*4i + a^2*b^3*cos(
c + d*x)*(a*b^3 + b^4)^(1/2)*1i - a^3*b^2*cos(c + d*x)*(a*b^3 + b^4)^(1/2)*2i + a*b^4*cos(c + d*x)*(a*b^3 + b^
4)^(1/2)*12i + a^4*b*cos(c + d*x)*(a*b^3 + b^4)^(1/2)*1i)/(3*a^2*b^5 + 5*a^3*b^4 + a^4*b^3 - a^5*b^2))*(a*b^3
+ b^4)^(1/2)*2i)/(d*(2*a^2*b + 2*a^3 - 2*a^3*cos(c + d*x)^2 - 2*a^2*b*cos(c + d*x)^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{3}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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