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

Optimal. Leaf size=162 \[ \frac{b^2 (6 a+5 b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{7/2} d (a+b)^{3/2}}-\frac{\left (2 a^2-a b-5 b^2\right ) \cot (c+d x)}{2 a^3 d (a+b)}-\frac{(2 a+5 b) \cot ^3(c+d x)}{6 a^2 d (a+b)}+\frac{b \csc ^3(c+d x) \sec (c+d x)}{2 a d (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )} \]

[Out]

(b^2*(6*a + 5*b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(2*a^(7/2)*(a + b)^(3/2)*d) - ((2*a^2 - a*b - 5*b
^2)*Cot[c + d*x])/(2*a^3*(a + b)*d) - ((2*a + 5*b)*Cot[c + d*x]^3)/(6*a^2*(a + b)*d) + (b*Csc[c + d*x]^3*Sec[c
+ d*x])/(2*a*(a + b)*d*(a + (a + b)*Tan[c + d*x]^2))

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Rubi [A]  time = 0.202876, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3187, 468, 570, 205} \[ \frac{b^2 (6 a+5 b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{7/2} d (a+b)^{3/2}}-\frac{\left (2 a^2-a b-5 b^2\right ) \cot (c+d x)}{2 a^3 d (a+b)}-\frac{(2 a+5 b) \cot ^3(c+d x)}{6 a^2 d (a+b)}+\frac{b \csc ^3(c+d x) \sec (c+d x)}{2 a d (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*(6*a + 5*b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(2*a^(7/2)*(a + b)^(3/2)*d) - ((2*a^2 - a*b - 5*b
^2)*Cot[c + d*x])/(2*a^3*(a + b)*d) - ((2*a + 5*b)*Cot[c + d*x]^3)/(6*a^2*(a + b)*d) + (b*Csc[c + d*x]^3*Sec[c
+ d*x])/(2*a*(a + b)*d*(a + (a + b)*Tan[c + d*x]^2))

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 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[((c*b -
a*d)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(a*b*e*n*(p + 1)), x] + Dist[1/(a*b*n*(p + 1)), I
nt[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c*b - a*d)*(m + 1)) + d*(c*b*n*(p
+ 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 570

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^
(r_.), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a,
b, c, d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]

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

Mathematica [A]  time = 1.24527, size = 202, normalized size = 1.25 \[ \frac{\csc ^4(c+d x) (-2 a+b \cos (2 (c+d x))-b) \left (2 a^{3/2} \cot (c+d x) \csc ^2(c+d x) (2 a-b \cos (2 (c+d x))+b)-\frac{3 \sqrt{a} b^3 \sin (2 (c+d x))}{a+b}+\frac{3 b^2 (6 a+5 b) (-2 a+b \cos (2 (c+d x))-b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{(a+b)^{3/2}}+4 \sqrt{a} (a-3 b) \cot (c+d x) (2 a-b \cos (2 (c+d x))+b)\right )}{24 a^{7/2} d \left (a \csc ^2(c+d x)+b\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a - b + b*Cos[2*(c + d*x)])*Csc[c + d*x]^4*((3*b^2*(6*a + 5*b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]]
*(-2*a - b + b*Cos[2*(c + d*x)]))/(a + b)^(3/2) + 4*Sqrt[a]*(a - 3*b)*(2*a + b - b*Cos[2*(c + d*x)])*Cot[c + d
*x] + 2*a^(3/2)*(2*a + b - b*Cos[2*(c + d*x)])*Cot[c + d*x]*Csc[c + d*x]^2 - (3*Sqrt[a]*b^3*Sin[2*(c + d*x)])/
(a + b)))/(24*a^(7/2)*d*(b + a*Csc[c + d*x]^2)^2)

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Maple [A]  time = 0.147, size = 179, normalized size = 1.1 \begin{align*}{\frac{{b}^{3}\tan \left ( dx+c \right ) }{2\,d{a}^{3} \left ( a+b \right ) \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) }}+3\,{\frac{{b}^{2}}{{a}^{2}d \left ( a+b \right ) \sqrt{a \left ( a+b \right ) }}\arctan \left ({\frac{ \left ( a+b \right ) \tan \left ( dx+c \right ) }{\sqrt{a \left ( a+b \right ) }}} \right ) }+{\frac{5\,{b}^{3}}{2\,d{a}^{3} \left ( a+b \right ) }\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\,{a}^{2}d \left ( \tan \left ( dx+c \right ) \right ) ^{3}}}-{\frac{1}{{a}^{2}d\tan \left ( dx+c \right ) }}+2\,{\frac{b}{d{a}^{3}\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)^2,x)

[Out]

1/2/d*b^3/a^3/(a+b)*tan(d*x+c)/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)+3/d/a^2*b^2/(a+b)/(a*(a+b))^(1/2)*arctan((a+b
)*tan(d*x+c)/(a*(a+b))^(1/2))+5/2/d*b^3/a^3/(a+b)/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))-1/3
/d/a^2/tan(d*x+c)^3-1/d/a^2/tan(d*x+c)+2/d/a^3/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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.07046, size = 1912, normalized size = 11.8 \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)^4/(a+b*sin(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

[-1/24*(4*(4*a^4*b - 4*a^3*b^2 - 23*a^2*b^3 - 15*a*b^4)*cos(d*x + c)^5 - 8*(2*a^5 + 3*a^4*b - 12*a^3*b^2 - 28*
a^2*b^3 - 15*a*b^4)*cos(d*x + c)^3 + 3*((6*a*b^3 + 5*b^4)*cos(d*x + c)^4 + 6*a^2*b^2 + 11*a*b^3 + 5*b^4 - (6*a
^2*b^2 + 17*a*b^3 + 10*b^4)*cos(d*x + c)^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*(2*a^5 + 2*a^4*b - 6*a^3*b^2 - 11*a^2*b^3 - 5*a*b^4)*cos(d*x + c))/(((a^6*b + 2*a^5*b^2 + a^4*b^3)*d
*cos(d*x + c)^4 - (a^7 + 4*a^6*b + 5*a^5*b^2 + 2*a^4*b^3)*d*cos(d*x + c)^2 + (a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*
b^3)*d)*sin(d*x + c)), -1/12*(2*(4*a^4*b - 4*a^3*b^2 - 23*a^2*b^3 - 15*a*b^4)*cos(d*x + c)^5 - 4*(2*a^5 + 3*a^
4*b - 12*a^3*b^2 - 28*a^2*b^3 - 15*a*b^4)*cos(d*x + c)^3 + 3*((6*a*b^3 + 5*b^4)*cos(d*x + c)^4 + 6*a^2*b^2 + 1
1*a*b^3 + 5*b^4 - (6*a^2*b^2 + 17*a*b^3 + 10*b^4)*cos(d*x + c)^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*(2*a^5 + 2*a^4*b - 6*a^3*b^2 -
11*a^2*b^3 - 5*a*b^4)*cos(d*x + c))/(((a^6*b + 2*a^5*b^2 + a^4*b^3)*d*cos(d*x + c)^4 - (a^7 + 4*a^6*b + 5*a^5
*b^2 + 2*a^4*b^3)*d*cos(d*x + c)^2 + (a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3)*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)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.19397, size = 235, normalized size = 1.45 \begin{align*} \frac{\frac{3 \, b^{3} \tan \left (d x + c\right )}{{\left (a^{4} + a^{3} b\right )}{\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}} + \frac{3 \,{\left (6 \, a b^{2} + 5 \, b^{3}\right )}{\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 )}}{{\left (a^{4} + a^{3} b\right )} \sqrt{a^{2} + a b}} - \frac{2 \,{\left (3 \, a \tan \left (d x + c\right )^{2} - 6 \, b \tan \left (d x + c\right )^{2} + a\right )}}{a^{3} \tan \left (d x + c\right )^{3}}}{6 \, 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)^2,x, algorithm="giac")

[Out]

1/6*(3*b^3*tan(d*x + c)/((a^4 + a^3*b)*(a*tan(d*x + c)^2 + b*tan(d*x + c)^2 + a)) + 3*(6*a*b^2 + 5*b^3)*(pi*fl
oor((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)))/((a^4 + a^
3*b)*sqrt(a^2 + a*b)) - 2*(3*a*tan(d*x + c)^2 - 6*b*tan(d*x + c)^2 + a)/(a^3*tan(d*x + c)^3))/d