3.91 \(\int \frac{\csc ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=77 \[ \frac{b^2 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{a^{5/2} d \sqrt{a+b}}-\frac{(a-b) \cot (c+d x)}{a^2 d}-\frac{\cot ^3(c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.105693, antiderivative size = 77, 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} \[ \frac{b^2 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{a^{5/2} d \sqrt{a+b}}-\frac{(a-b) \cot (c+d x)}{a^2 d}-\frac{\cot ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4/(a + b*Sin[c + d*x]^2),x]

[Out]

(b^2*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(5/2)*Sqrt[a + b]*d) - ((a - b)*Cot[c + d*x])/(a^2*d) - Co
t[c + d*x]^3/(3*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 ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x^4 \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^4}+\frac{a-b}{a^2 x^2}+\frac{b^2}{a^2 \left (a+(a+b) x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{(a-b) \cot (c+d x)}{a^2 d}-\frac{\cot ^3(c+d x)}{3 a d}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{a^2 d}\\ &=\frac{b^2 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{a^{5/2} \sqrt{a+b} d}-\frac{(a-b) \cot (c+d x)}{a^2 d}-\frac{\cot ^3(c+d x)}{3 a d}\\ \end{align*}

Mathematica [A]  time = 0.654628, size = 119, normalized size = 1.55 \[ -\frac{\csc ^2(c+d x) (2 a-b \cos (2 (c+d x))+b) \left (\sqrt{a} \sqrt{a+b} \cot (c+d x) \left (a \csc ^2(c+d x)+2 a-3 b\right )-3 b^2 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )\right )}{6 a^{5/2} d \sqrt{a+b} \left (a \csc ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4/(a + b*Sin[c + d*x]^2),x]

[Out]

-((2*a + b - b*Cos[2*(c + d*x)])*Csc[c + d*x]^2*(-3*b^2*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]] + Sqrt[a]*S
qrt[a + b]*Cot[c + d*x]*(2*a - 3*b + a*Csc[c + d*x]^2)))/(6*a^(5/2)*Sqrt[a + b]*d*(b + a*Csc[c + d*x]^2))

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Maple [A]  time = 0.116, size = 85, normalized size = 1.1 \begin{align*}{\frac{{b}^{2}}{{a}^{2}d}\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}{3\,da \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 ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^4/(a+sin(d*x+c)^2*b),x)

[Out]

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

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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)^4/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.85966, size = 1046, normalized size = 13.58 \begin{align*} \left [-\frac{4 \,{\left (2 \, a^{3} - a^{2} b - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (b^{2} \cos \left (d x + c\right )^{2} - b^{2}\right )} \sqrt{-a^{2} - a b} \log \left (\frac{{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{3} -{\left (a + b\right )} \cos \left (d x + c\right )\right )} \sqrt{-a^{2} - a b} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \,{\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) \sin \left (d x + c\right ) - 12 \,{\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )}{12 \,{\left ({\left (a^{4} + a^{3} b\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{4} + a^{3} b\right )} d\right )} \sin \left (d x + c\right )}, -\frac{2 \,{\left (2 \, a^{3} - a^{2} b - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (b^{2} \cos \left (d x + c\right )^{2} - b^{2}\right )} \sqrt{a^{2} + a b} \arctan \left (\frac{{\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b}{2 \, \sqrt{a^{2} + a b} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 6 \,{\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )}{6 \,{\left ({\left (a^{4} + a^{3} b\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{4} + a^{3} b\right )} d\right )} \sin \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

[-1/12*(4*(2*a^3 - a^2*b - 3*a*b^2)*cos(d*x + c)^3 + 3*(b^2*cos(d*x + c)^2 - b^2)*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) - 12*(a^3 - a*b^2)*cos(d*x + c))/(((a^4 + a^3*b)*d*cos(d*x + c)^2 -
(a^4 + a^3*b)*d)*sin(d*x + c)), -1/6*(2*(2*a^3 - a^2*b - 3*a*b^2)*cos(d*x + c)^3 + 3*(b^2*cos(d*x + c)^2 - b^
2)*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) - 6*(a^3 - a*b^2)*cos(d*x + c))/(((a^4 + a^3*b)*d*cos(d*x + c)^2 - (a^4 + a^3*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)**4/(a+b*sin(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.15074, size = 150, normalized size = 1.95 \begin{align*} \frac{\frac{3 \,{\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^{2}}{\sqrt{a^{2} + a b} a^{2}} - \frac{3 \, a \tan \left (d x + c\right )^{2} - 3 \, b \tan \left (d x + c\right )^{2} + a}{a^{2} \tan \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

1/3*(3*(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^2/(sqrt(a^2 + a*b)*a^2) - (3*a*tan(d*x + c)^2 - 3*b*tan(d*x + c)^2 + a)/(a^2*tan(d*x + c)^3))/d