3.84 \(\int \frac{\csc ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=125 \[ -\frac{\left (3 a^2-4 a b+8 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}+\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{a^3 d \sqrt{a+b}}-\frac{(3 a-4 b) \cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d} \]

[Out]

-((3*a^2 - 4*a*b + 8*b^2)*ArcTanh[Cos[c + d*x]])/(8*a^3*d) + (b^(5/2)*ArcTanh[(Sqrt[b]*Cos[c + d*x])/Sqrt[a +
b]])/(a^3*Sqrt[a + b]*d) - ((3*a - 4*b)*Cot[c + d*x]*Csc[c + d*x])/(8*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]^3)/(
4*a*d)

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Rubi [A]  time = 0.188042, antiderivative size = 125, normalized size of antiderivative = 1., 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} \[ -\frac{\left (3 a^2-4 a b+8 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}+\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{a^3 d \sqrt{a+b}}-\frac{(3 a-4 b) \cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^5/(a + b*Sin[c + d*x]^2),x]

[Out]

-((3*a^2 - 4*a*b + 8*b^2)*ArcTanh[Cos[c + d*x]])/(8*a^3*d) + (b^(5/2)*ArcTanh[(Sqrt[b]*Cos[c + d*x])/Sqrt[a +
b]])/(a^3*Sqrt[a + b]*d) - ((3*a - 4*b)*Cot[c + d*x]*Csc[c + d*x])/(8*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]^3)/(
4*a*d)

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]

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\csc ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^3 \left (a+b-b x^2\right )} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac{\operatorname{Subst}\left (\int \frac{3 a-b-3 b x^2}{\left (1-x^2\right )^2 \left (a+b-b x^2\right )} \, dx,x,\cos (c+d x)\right )}{4 a d}\\ &=-\frac{(3 a-4 b) \cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac{\operatorname{Subst}\left (\int \frac{3 a^2-a b+4 b^2-(3 a-4 b) b x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\cos (c+d x)\right )}{8 a^2 d}\\ &=-\frac{(3 a-4 b) \cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{a+b-b x^2} \, dx,x,\cos (c+d x)\right )}{a^3 d}-\frac{\left (3 a^2-4 a b+8 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{8 a^3 d}\\ &=-\frac{\left (3 a^2-4 a b+8 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}+\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{a^3 \sqrt{a+b} d}-\frac{(3 a-4 b) \cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d}\\ \end{align*}

Mathematica [C]  time = 6.29067, size = 657, normalized size = 5.26 \[ \frac{\left (3 a^2-4 a b+8 b^2\right ) \csc ^2(c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right ) (-2 a+b \cos (2 (c+d x))-b)}{16 a^3 d \left (a \csc ^2(c+d x)+b\right )}+\frac{\left (-3 a^2+4 a b-8 b^2\right ) \csc ^2(c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right ) (-2 a+b \cos (2 (c+d x))-b)}{16 a^3 d \left (a \csc ^2(c+d x)+b\right )}+\frac{b^{5/2} \csc ^2(c+d x) (-2 a+b \cos (2 (c+d x))-b) \tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (\sqrt{b} \cos \left (\frac{1}{2} (c+d x)\right )-i \sqrt{a} \sin \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{-a-b}}\right )}{2 a^3 d \sqrt{-a-b} \left (a \csc ^2(c+d x)+b\right )}+\frac{b^{5/2} \csc ^2(c+d x) (-2 a+b \cos (2 (c+d x))-b) \tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (\sqrt{b} \cos \left (\frac{1}{2} (c+d x)\right )+i \sqrt{a} \sin \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{-a-b}}\right )}{2 a^3 d \sqrt{-a-b} \left (a \csc ^2(c+d x)+b\right )}+\frac{(3 a-4 b) \csc ^2(c+d x) \csc ^2\left (\frac{1}{2} (c+d x)\right ) (-2 a+b \cos (2 (c+d x))-b)}{64 a^2 d \left (a \csc ^2(c+d x)+b\right )}+\frac{(4 b-3 a) \csc ^2(c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (-2 a+b \cos (2 (c+d x))-b)}{64 a^2 d \left (a \csc ^2(c+d x)+b\right )}+\frac{\csc ^2(c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right ) (-2 a+b \cos (2 (c+d x))-b)}{128 a d \left (a \csc ^2(c+d x)+b\right )}-\frac{\csc ^2(c+d x) \sec ^4\left (\frac{1}{2} (c+d x)\right ) (-2 a+b \cos (2 (c+d x))-b)}{128 a d \left (a \csc ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^5/(a + b*Sin[c + d*x]^2),x]

[Out]

(b^(5/2)*ArcTan[(Sec[(c + d*x)/2]*(Sqrt[b]*Cos[(c + d*x)/2] - I*Sqrt[a]*Sin[(c + d*x)/2]))/Sqrt[-a - b]]*(-2*a
- b + b*Cos[2*(c + d*x)])*Csc[c + d*x]^2)/(2*a^3*Sqrt[-a - b]*d*(b + a*Csc[c + d*x]^2)) + (b^(5/2)*ArcTan[(Se
c[(c + d*x)/2]*(Sqrt[b]*Cos[(c + d*x)/2] + I*Sqrt[a]*Sin[(c + d*x)/2]))/Sqrt[-a - b]]*(-2*a - b + b*Cos[2*(c +
d*x)])*Csc[c + d*x]^2)/(2*a^3*Sqrt[-a - b]*d*(b + a*Csc[c + d*x]^2)) + ((3*a - 4*b)*(-2*a - b + b*Cos[2*(c +
d*x)])*Csc[(c + d*x)/2]^2*Csc[c + d*x]^2)/(64*a^2*d*(b + a*Csc[c + d*x]^2)) + ((-2*a - b + b*Cos[2*(c + d*x)])
*Csc[(c + d*x)/2]^4*Csc[c + d*x]^2)/(128*a*d*(b + a*Csc[c + d*x]^2)) + ((3*a^2 - 4*a*b + 8*b^2)*(-2*a - b + b*
Cos[2*(c + d*x)])*Csc[c + d*x]^2*Log[Cos[(c + d*x)/2]])/(16*a^3*d*(b + a*Csc[c + d*x]^2)) + ((-3*a^2 + 4*a*b -
8*b^2)*(-2*a - b + b*Cos[2*(c + d*x)])*Csc[c + d*x]^2*Log[Sin[(c + d*x)/2]])/(16*a^3*d*(b + a*Csc[c + d*x]^2)
) + ((-3*a + 4*b)*(-2*a - b + b*Cos[2*(c + d*x)])*Csc[c + d*x]^2*Sec[(c + d*x)/2]^2)/(64*a^2*d*(b + a*Csc[c +
d*x]^2)) - ((-2*a - b + b*Cos[2*(c + d*x)])*Csc[c + d*x]^2*Sec[(c + d*x)/2]^4)/(128*a*d*(b + a*Csc[c + d*x]^2)
)

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Maple [B]  time = 0.122, size = 255, normalized size = 2. \begin{align*} -{\frac{1}{16\,da \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3}{16\,da \left ( -1+\cos \left ( dx+c \right ) \right ) }}-{\frac{b}{4\,{a}^{2}d \left ( -1+\cos \left ( dx+c \right ) \right ) }}+{\frac{3\,\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{16\,da}}-{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) b}{4\,{a}^{2}d}}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ){b}^{2}}{2\,d{a}^{3}}}+{\frac{1}{16\,da \left ( 1+\cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3}{16\,da \left ( 1+\cos \left ( dx+c \right ) \right ) }}-{\frac{b}{4\,{a}^{2}d \left ( 1+\cos \left ( dx+c \right ) \right ) }}-{\frac{3\,\ln \left ( 1+\cos \left ( dx+c \right ) \right ) }{16\,da}}+{\frac{\ln \left ( 1+\cos \left ( dx+c \right ) \right ) b}{4\,{a}^{2}d}}-{\frac{\ln \left ( 1+\cos \left ( dx+c \right ) \right ){b}^{2}}{2\,d{a}^{3}}}+{\frac{{b}^{3}}{d{a}^{3}}{\it Artanh} \left ({b\cos \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^5/(a+sin(d*x+c)^2*b),x)

[Out]

-1/16/d/a/(-1+cos(d*x+c))^2+3/16/d/a/(-1+cos(d*x+c))-1/4/d/a^2/(-1+cos(d*x+c))*b+3/16/d/a*ln(-1+cos(d*x+c))-1/
4/d/a^2*ln(-1+cos(d*x+c))*b+1/2/d/a^3*ln(-1+cos(d*x+c))*b^2+1/16/a/d/(1+cos(d*x+c))^2+3/16/a/d/(1+cos(d*x+c))-
1/4/d/a^2/(1+cos(d*x+c))*b-3/16/d/a*ln(1+cos(d*x+c))+1/4/d/a^2*ln(1+cos(d*x+c))*b-1/2/d/a^3*ln(1+cos(d*x+c))*b
^2+1/d*b^3/a^3/((a+b)*b)^(1/2)*arctanh(cos(d*x+c)*b/((a+b)*b)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.11396, size = 1482, normalized size = 11.86 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/16*(2*(3*a^2 - 4*a*b)*cos(d*x + c)^3 + 8*(b^2*cos(d*x + c)^4 - 2*b^2*cos(d*x + c)^2 + b^2)*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*(5*a^2
- 4*a*b)*cos(d*x + c) - ((3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^4 - 2*(3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^2 +
3*a^2 - 4*a*b + 8*b^2)*log(1/2*cos(d*x + c) + 1/2) + ((3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^4 - 2*(3*a^2 - 4*a*
b + 8*b^2)*cos(d*x + c)^2 + 3*a^2 - 4*a*b + 8*b^2)*log(-1/2*cos(d*x + c) + 1/2))/(a^3*d*cos(d*x + c)^4 - 2*a^3
*d*cos(d*x + c)^2 + a^3*d), 1/16*(2*(3*a^2 - 4*a*b)*cos(d*x + c)^3 - 16*(b^2*cos(d*x + c)^4 - 2*b^2*cos(d*x +
c)^2 + b^2)*sqrt(-b/(a + b))*arctan(sqrt(-b/(a + b))*cos(d*x + c)) - 2*(5*a^2 - 4*a*b)*cos(d*x + c) - ((3*a^2
- 4*a*b + 8*b^2)*cos(d*x + c)^4 - 2*(3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^2 + 3*a^2 - 4*a*b + 8*b^2)*log(1/2*co
s(d*x + c) + 1/2) + ((3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^4 - 2*(3*a^2 - 4*a*b + 8*b^2)*cos(d*x + c)^2 + 3*a^2
- 4*a*b + 8*b^2)*log(-1/2*cos(d*x + c) + 1/2))/(a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**5/(a+b*sin(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.26124, size = 451, normalized size = 3.61 \begin{align*} -\frac{\frac{64 \, b^{3} \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^{3}} + \frac{\frac{8 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{8 \, 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}}}{a^{2}} - \frac{4 \,{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\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{{\left (a^{2} - \frac{8 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{8 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{18 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{24 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{48 \, b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

-1/64*(64*b^3*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^3) + (8*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 8*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - a*(cos(d
*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/a^2 - 4*(3*a^2 - 4*a*b + 8*b^2)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x +
c) + 1))/a^3 + (a^2 - 8*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 8*a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1
) + 18*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 24*a*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 48*b
^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)^2/(a^3*(cos(d*x + c) - 1)^2))/d