3.93 \(\int \frac{\csc ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=140 \[ \frac{b^4 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{a^{9/2} d \sqrt{a+b}}-\frac{\left (3 a^2-2 a b+b^2\right ) \cot ^3(c+d x)}{3 a^3 d}-\frac{(a-b) \left (a^2+b^2\right ) \cot (c+d x)}{a^4 d}-\frac{(3 a-b) \cot ^5(c+d x)}{5 a^2 d}-\frac{\cot ^7(c+d x)}{7 a d} \]

[Out]

(b^4*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(9/2)*Sqrt[a + b]*d) - ((a - b)*(a^2 + b^2)*Cot[c + d*x])/
(a^4*d) - ((3*a^2 - 2*a*b + b^2)*Cot[c + d*x]^3)/(3*a^3*d) - ((3*a - b)*Cot[c + d*x]^5)/(5*a^2*d) - Cot[c + d*
x]^7/(7*a*d)

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Rubi [A]  time = 0.155528, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3187, 461, 205} \[ \frac{b^4 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{a^{9/2} d \sqrt{a+b}}-\frac{\left (3 a^2-2 a b+b^2\right ) \cot ^3(c+d x)}{3 a^3 d}-\frac{(a-b) \left (a^2+b^2\right ) \cot (c+d x)}{a^4 d}-\frac{(3 a-b) \cot ^5(c+d x)}{5 a^2 d}-\frac{\cot ^7(c+d x)}{7 a d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^8/(a + b*Sin[c + d*x]^2),x]

[Out]

(b^4*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(9/2)*Sqrt[a + b]*d) - ((a - b)*(a^2 + b^2)*Cot[c + d*x])/
(a^4*d) - ((3*a^2 - 2*a*b + b^2)*Cot[c + d*x]^3)/(3*a^3*d) - ((3*a - b)*Cot[c + d*x]^5)/(5*a^2*d) - Cot[c + d*
x]^7/(7*a*d)

Rule 3187

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

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\csc ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^4}{x^8 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{a x^8}+\frac{3 a-b}{a^2 x^6}+\frac{3 a^2-2 a b+b^2}{a^3 x^4}+\frac{(a-b) \left (a^2+b^2\right )}{a^4 x^2}+\frac{b^4}{a^4 \left (a+(a+b) x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{(a-b) \left (a^2+b^2\right ) \cot (c+d x)}{a^4 d}-\frac{\left (3 a^2-2 a b+b^2\right ) \cot ^3(c+d x)}{3 a^3 d}-\frac{(3 a-b) \cot ^5(c+d x)}{5 a^2 d}-\frac{\cot ^7(c+d x)}{7 a d}+\frac{b^4 \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{a^4 d}\\ &=\frac{b^4 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{a^{9/2} \sqrt{a+b} d}-\frac{(a-b) \left (a^2+b^2\right ) \cot (c+d x)}{a^4 d}-\frac{\left (3 a^2-2 a b+b^2\right ) \cot ^3(c+d x)}{3 a^3 d}-\frac{(3 a-b) \cot ^5(c+d x)}{5 a^2 d}-\frac{\cot ^7(c+d x)}{7 a d}\\ \end{align*}

Mathematica [A]  time = 1.72937, size = 137, normalized size = 0.98 \[ \frac{b^4 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{a^{9/2} d \sqrt{a+b}}-\frac{\cot (c+d x) \left (a \left (24 a^2-28 a b+35 b^2\right ) \csc ^2(c+d x)+3 a^2 (6 a-7 b) \csc ^4(c+d x)-56 a^2 b+15 a^3 \csc ^6(c+d x)+48 a^3+70 a b^2-105 b^3\right )}{105 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^8/(a + b*Sin[c + d*x]^2),x]

[Out]

(b^4*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(9/2)*Sqrt[a + b]*d) - (Cot[c + d*x]*(48*a^3 - 56*a^2*b +
70*a*b^2 - 105*b^3 + a*(24*a^2 - 28*a*b + 35*b^2)*Csc[c + d*x]^2 + 3*a^2*(6*a - 7*b)*Csc[c + d*x]^4 + 15*a^3*C
sc[c + d*x]^6))/(105*a^4*d)

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Maple [A]  time = 0.158, size = 207, normalized size = 1.5 \begin{align*}{\frac{{b}^{4}}{d{a}^{4}}\arctan \left ({ \left ( a+b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}}-{\frac{1}{7\,da \left ( \tan \left ( dx+c \right ) \right ) ^{7}}}-{\frac{3}{5\,da \left ( \tan \left ( dx+c \right ) \right ) ^{5}}}+{\frac{b}{5\,{a}^{2}d \left ( \tan \left ( dx+c \right ) \right ) ^{5}}}-{\frac{1}{da \left ( \tan \left ( dx+c \right ) \right ) ^{3}}}+{\frac{2\,b}{3\,{a}^{2}d \left ( \tan \left ( dx+c \right ) \right ) ^{3}}}-{\frac{{b}^{2}}{3\,d{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}}-{\frac{1}{da\tan \left ( dx+c \right ) }}+{\frac{b}{{a}^{2}d\tan \left ( dx+c \right ) }}-{\frac{{b}^{2}}{d{a}^{3}\tan \left ( dx+c \right ) }}+{\frac{{b}^{3}}{d{a}^{4}\tan \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^8/(a+sin(d*x+c)^2*b),x)

[Out]

1/d*b^4/a^4/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))-1/7/d/a/tan(d*x+c)^7-3/5/d/a/tan(d*x+c)^5
+1/5/d/a^2/tan(d*x+c)^5*b-1/d/a/tan(d*x+c)^3+2/3/d/a^2/tan(d*x+c)^3*b-1/3/d/a^3/tan(d*x+c)^3*b^2-1/d/a/tan(d*x
+c)+1/d/a^2/tan(d*x+c)*b-1/d/a^3/tan(d*x+c)*b^2+1/d/a^4/tan(d*x+c)*b^3

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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)^8/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.01844, size = 1839, normalized size = 13.14 \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)^8/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

[-1/420*(4*(48*a^5 - 8*a^4*b + 14*a^3*b^2 - 35*a^2*b^3 - 105*a*b^4)*cos(d*x + c)^7 - 28*(24*a^5 - 4*a^4*b + 7*
a^3*b^2 - 10*a^2*b^3 - 45*a*b^4)*cos(d*x + c)^5 + 140*(6*a^5 - a^4*b + a^3*b^2 - a^2*b^3 - 9*a*b^4)*cos(d*x +
c)^3 + 105*(b^4*cos(d*x + c)^6 - 3*b^4*cos(d*x + c)^4 + 3*b^4*cos(d*x + c)^2 - b^4)*sqrt(-a^2 - a*b)*log(((8*a
^2 + 8*a*b + b^2)*cos(d*x + c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c)^2 + 4*((2*a + b)*cos(d*x + c)^3 - (a +
b)*cos(d*x + c))*sqrt(-a^2 - a*b)*sin(d*x + c) + a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d
*x + c)^2 + a^2 + 2*a*b + b^2))*sin(d*x + c) - 420*(a^5 - a*b^4)*cos(d*x + c))/(((a^6 + a^5*b)*d*cos(d*x + c)^
6 - 3*(a^6 + a^5*b)*d*cos(d*x + c)^4 + 3*(a^6 + a^5*b)*d*cos(d*x + c)^2 - (a^6 + a^5*b)*d)*sin(d*x + c)), -1/2
10*(2*(48*a^5 - 8*a^4*b + 14*a^3*b^2 - 35*a^2*b^3 - 105*a*b^4)*cos(d*x + c)^7 - 14*(24*a^5 - 4*a^4*b + 7*a^3*b
^2 - 10*a^2*b^3 - 45*a*b^4)*cos(d*x + c)^5 + 70*(6*a^5 - a^4*b + a^3*b^2 - a^2*b^3 - 9*a*b^4)*cos(d*x + c)^3 +
105*(b^4*cos(d*x + c)^6 - 3*b^4*cos(d*x + c)^4 + 3*b^4*cos(d*x + c)^2 - b^4)*sqrt(a^2 + a*b)*arctan(1/2*((2*a
+ b)*cos(d*x + c)^2 - a - b)/(sqrt(a^2 + a*b)*cos(d*x + c)*sin(d*x + c)))*sin(d*x + c) - 210*(a^5 - a*b^4)*co
s(d*x + c))/(((a^6 + a^5*b)*d*cos(d*x + c)^6 - 3*(a^6 + a^5*b)*d*cos(d*x + c)^4 + 3*(a^6 + a^5*b)*d*cos(d*x +
c)^2 - (a^6 + a^5*b)*d)*sin(d*x + c))]

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

[Out]

Timed out

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Giac [A]  time = 1.19436, size = 290, normalized size = 2.07 \begin{align*} \frac{\frac{105 \,{\left (\pi \left \lfloor \frac{d x + c}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac{a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt{a^{2} + a b}}\right )\right )} b^{4}}{\sqrt{a^{2} + a b} a^{4}} - \frac{105 \, a^{3} \tan \left (d x + c\right )^{6} - 105 \, a^{2} b \tan \left (d x + c\right )^{6} + 105 \, a b^{2} \tan \left (d x + c\right )^{6} - 105 \, b^{3} \tan \left (d x + c\right )^{6} + 105 \, a^{3} \tan \left (d x + c\right )^{4} - 70 \, a^{2} b \tan \left (d x + c\right )^{4} + 35 \, a b^{2} \tan \left (d x + c\right )^{4} + 63 \, a^{3} \tan \left (d x + c\right )^{2} - 21 \, a^{2} b \tan \left (d x + c\right )^{2} + 15 \, a^{3}}{a^{4} \tan \left (d x + c\right )^{7}}}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^8/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

1/105*(105*(pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 +
a*b)))*b^4/(sqrt(a^2 + a*b)*a^4) - (105*a^3*tan(d*x + c)^6 - 105*a^2*b*tan(d*x + c)^6 + 105*a*b^2*tan(d*x + c)
^6 - 105*b^3*tan(d*x + c)^6 + 105*a^3*tan(d*x + c)^4 - 70*a^2*b*tan(d*x + c)^4 + 35*a*b^2*tan(d*x + c)^4 + 63*
a^3*tan(d*x + c)^2 - 21*a^2*b*tan(d*x + c)^2 + 15*a^3)/(a^4*tan(d*x + c)^7))/d