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

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

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.243339, antiderivative size = 153, 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} \[ -\frac{b^{3/2} (5 a+4 b) \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{2 a^3 d (a+b)^{3/2}}-\frac{b (a+2 b) \cos (c+d x)}{2 a^2 d (a+b) \left (a-b \cos ^2(c+d x)+b\right )}-\frac{(a-4 b) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d \left (a-b \cos ^2(c+d x)+b\right )} \]

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 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 ^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*}

Mathematica [C]  time = 1.63596, size = 390, normalized size = 2.55 \[ \frac{\csc ^3(c+d x) (-2 a+b \cos (2 (c+d x))-b) \left (\frac{8 a b^2 \cot (c+d x)}{a+b}+\frac{4 b^{3/2} (5 a+4 b) \csc (c+d x) (2 a-b \cos (2 (c+d x))+b) \tan ^{-1}\left (\frac{\sqrt{b}-i \sqrt{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a-b}}\right )}{(-a-b)^{3/2}}+\frac{4 b^{3/2} (5 a+4 b) \csc (c+d x) (2 a-b \cos (2 (c+d x))+b) \tan ^{-1}\left (\frac{\sqrt{b}+i \sqrt{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a-b}}\right )}{(-a-b)^{3/2}}+a \csc ^2\left (\frac{1}{2} (c+d x)\right ) \csc (c+d x) (2 a-b \cos (2 (c+d x))+b)+4 (a-4 b) \csc (c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right ) (2 a-b \cos (2 (c+d x))+b)-a \csc (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (2 a-b \cos (2 (c+d x))+b)-4 (a-4 b) \csc (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right ) (2 a-b \cos (2 (c+d x))+b)\right )}{32 a^3 d \left (a \csc ^2(c+d x)+b\right )^2} \]

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)

________________________________________________________________________________________

Maple [A]  time = 0.14, size = 226, normalized size = 1.5 \begin{align*}{\frac{1}{4\,{a}^{2}d \left ( -1+\cos \left ( dx+c \right ) \right ) }}-{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) b}{d{a}^{3}}}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{4\,{a}^{2}d}}+{\frac{1}{4\,{a}^{2}d \left ( 1+\cos \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( 1+\cos \left ( dx+c \right ) \right ) b}{d{a}^{3}}}-{\frac{\ln \left ( 1+\cos \left ( dx+c \right ) \right ) }{4\,{a}^{2}d}}+{\frac{\cos \left ( dx+c \right ){b}^{2}}{2\,{a}^{2}d \left ( a+b \right ) \left ( b \left ( \cos \left ( dx+c \right ) \right ) ^{2}-a-b \right ) }}-{\frac{5\,{b}^{2}}{2\,{a}^{2}d \left ( a+b \right ) }{\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}}}}-2\,{\frac{{b}^{3}}{d{a}^{3} \left ( a+b \right ) \sqrt{ \left ( a+b \right ) b}}{\it Artanh} \left ({\frac{b\cos \left ( dx+c \right ) }{\sqrt{ \left ( a+b \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.56541, size = 1928, normalized size = 12.6 \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)^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)]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.23108, size = 691, normalized size = 4.52 \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)^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