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

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

[Out]

1/2*b^2*(6*a+5*b)*arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))/a^(7/2)/(a+b)^(3/2)/d-1/2*(2*a^2-a*b-5*b^2)*cot(d*x+c
)/a^3/(a+b)/d-1/6*(2*a+5*b)*cot(d*x+c)^3/a^2/(a+b)/d+1/2*b*csc(d*x+c)^3*sec(d*x+c)/a/(a+b)/d/(a+(a+b)*tan(d*x+
c)^2)

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 162, normalized size of antiderivative = 1.00, 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 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]

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 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]

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 = 1.30, 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)

________________________________________________________________________________________

fricas [B]  time = 0.50, size = 843, normalized size = 5.20 \[ \left [-\frac {4 \, {\left (4 \, a^{4} b - 4 \, a^{3} b^{2} - 23 \, a^{2} b^{3} - 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} - 8 \, {\left (2 \, a^{5} + 3 \, a^{4} b - 12 \, a^{3} b^{2} - 28 \, a^{2} b^{3} - 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left ({\left (6 \, a b^{3} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 6 \, a^{2} b^{2} + 11 \, a b^{3} + 5 \, b^{4} - {\left (6 \, a^{2} b^{2} + 17 \, a b^{3} + 10 \, b^{4}\right )} \cos \left (d x + c\right )^{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 (2 \, a^{5} + 2 \, a^{4} b - 6 \, a^{3} b^{2} - 11 \, a^{2} b^{3} - 5 \, a b^{4}\right )} \cos \left (d x + c\right )}{24 \, {\left ({\left (a^{6} b + 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} d \cos \left (d x + c\right )^{4} - {\left (a^{7} + 4 \, a^{6} b + 5 \, a^{5} b^{2} + 2 \, a^{4} b^{3}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} d\right )} \sin \left (d x + c\right )}, -\frac {2 \, {\left (4 \, a^{4} b - 4 \, a^{3} b^{2} - 23 \, a^{2} b^{3} - 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} - 4 \, {\left (2 \, a^{5} + 3 \, a^{4} b - 12 \, a^{3} b^{2} - 28 \, a^{2} b^{3} - 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left ({\left (6 \, a b^{3} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 6 \, a^{2} b^{2} + 11 \, a b^{3} + 5 \, b^{4} - {\left (6 \, a^{2} b^{2} + 17 \, a b^{3} + 10 \, b^{4}\right )} \cos \left (d x + c\right )^{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 (2 \, a^{5} + 2 \, a^{4} b - 6 \, a^{3} b^{2} - 11 \, a^{2} b^{3} - 5 \, a b^{4}\right )} \cos \left (d x + c\right )}{12 \, {\left ({\left (a^{6} b + 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} d \cos \left (d x + c\right )^{4} - {\left (a^{7} + 4 \, a^{6} b + 5 \, a^{5} b^{2} + 2 \, a^{4} b^{3}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} d\right )} \sin \left (d x + c\right )}\right ] \]

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))]

________________________________________________________________________________________

giac [A]  time = 0.19, size = 174, normalized size = 1.07 \[ \frac {\frac {3 \, b^{3} \tan \left (d x + c\right )}{{\left (a^{4} + a^{3} b\right )} {\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}} + \frac {3 \, {\left (6 \, a b^{2} + 5 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )}}{{\left (a^{4} + a^{3} b\right )} \sqrt {a^{2} + a b}} - \frac {2 \, {\left (3 \, a \tan \left (d x + c\right )^{2} - 6 \, b \tan \left (d x + c\right )^{2} + a\right )}}{a^{3} \tan \left (d x + c\right )^{3}}}{6 \, d} \]

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

________________________________________________________________________________________

maple [A]  time = 0.56, size = 179, normalized size = 1.10 \[ \frac {b^{3} \tan \left (d x +c \right )}{2 d \,a^{3} \left (a +b \right ) \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}+\frac {3 b^{2} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{d \,a^{2} \left (a +b \right ) \sqrt {a \left (a +b \right )}}+\frac {5 b^{3} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 d \,a^{3} \left (a +b \right ) \sqrt {a \left (a +b \right )}}-\frac {1}{3 d \,a^{2} \tan \left (d x +c \right )^{3}}-\frac {1}{d \,a^{2} \tan \left (d x +c \right )}+\frac {2 b}{d \,a^{3} \tan \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

maxima [A]  time = 0.56, size = 172, normalized size = 1.06 \[ \frac {\frac {3 \, {\left (6 \, a b^{2} + 5 \, b^{3}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{4} + a^{3} b\right )} \sqrt {{\left (a + b\right )} a}} - \frac {3 \, {\left (2 \, a^{3} - 6 \, a b^{2} - 5 \, b^{3}\right )} \tan \left (d x + c\right )^{4} + 2 \, a^{3} + 2 \, a^{2} b + 2 \, {\left (4 \, a^{3} - a^{2} b - 5 \, a b^{2}\right )} \tan \left (d x + c\right )^{2}}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \tan \left (d x + c\right )^{5} + {\left (a^{5} + a^{4} b\right )} \tan \left (d x + c\right )^{3}}}{6 \, d} \]

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]

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

________________________________________________________________________________________

mupad [B]  time = 14.40, size = 164, normalized size = 1.01 \[ \frac {b^2\,\mathrm {atan}\left (\frac {b^2\,\mathrm {tan}\left (c+d\,x\right )\,\left (a^4+b\,a^3\right )\,\left (6\,a+5\,b\right )}{a^{7/2}\,\left (5\,b^3+6\,a\,b^2\right )\,\sqrt {a+b}}\right )\,\left (6\,a+5\,b\right )}{2\,a^{7/2}\,d\,{\left (a+b\right )}^{3/2}}-\frac {\frac {1}{3\,a}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (4\,a-5\,b\right )}{3\,a^2}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (-2\,a^3+6\,a\,b^2+5\,b^3\right )}{2\,a^3\,\left (a+b\right )}}{d\,\left (\left (a+b\right )\,{\mathrm {tan}\left (c+d\,x\right )}^5+a\,{\mathrm {tan}\left (c+d\,x\right )}^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^4*(a + b*sin(c + d*x)^2)^2),x)

[Out]

(b^2*atan((b^2*tan(c + d*x)*(a^3*b + a^4)*(6*a + 5*b))/(a^(7/2)*(6*a*b^2 + 5*b^3)*(a + b)^(1/2)))*(6*a + 5*b))
/(2*a^(7/2)*d*(a + b)^(3/2)) - (1/(3*a) + (tan(c + d*x)^2*(4*a - 5*b))/(3*a^2) - (tan(c + d*x)^4*(6*a*b^2 - 2*
a^3 + 5*b^3))/(2*a^3*(a + b)))/(d*(tan(c + d*x)^5*(a + b) + a*tan(c + d*x)^3))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

________________________________________________________________________________________