3.1469 \(\int \frac{\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=248 \[ -\frac{4 b^4 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{5/2}}-\frac{2 b^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a d \left (a^2-b^2\right )^{5/2}}-\frac{b^5 \cos (c+d x)}{a^2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac{2 b \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{\cot (c+d x)}{a^2 d}+\frac{\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)} \]

[Out]

(-2*b^4*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*(a^2 - b^2)^(5/2)*d) - (4*b^4*(2*a^2 - b^2)*ArcTa
n[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2)^(5/2)*d) + (2*b*ArcTanh[Cos[c + d*x]])/(a^3*d) -
Cot[c + d*x]/(a^2*d) + Cos[c + d*x]/(2*(a + b)^2*d*(1 - Sin[c + d*x])) - Cos[c + d*x]/(2*(a - b)^2*d*(1 + Sin
[c + d*x])) - (b^5*Cos[c + d*x])/(a^2*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x]))

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Rubi [A]  time = 0.372225, antiderivative size = 248, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 10, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.345, Rules used = {2897, 3770, 3767, 8, 2648, 2664, 12, 2660, 618, 204} \[ -\frac{4 b^4 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{5/2}}-\frac{2 b^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a d \left (a^2-b^2\right )^{5/2}}-\frac{b^5 \cos (c+d x)}{a^2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac{2 b \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{\cot (c+d x)}{a^2 d}+\frac{\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^4*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*(a^2 - b^2)^(5/2)*d) - (4*b^4*(2*a^2 - b^2)*ArcTa
n[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2)^(5/2)*d) + (2*b*ArcTanh[Cos[c + d*x]])/(a^3*d) -
Cot[c + d*x]/(a^2*d) + Cos[c + d*x]/(2*(a + b)^2*d*(1 - Sin[c + d*x])) - Cos[c + d*x]/(2*(a - b)^2*d*(1 + Sin
[c + d*x])) - (b^5*Cos[c + d*x])/(a^2*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x]))

Rule 2897

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x]
/; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && (LtQ[m, -1] || (EqQ[m, -1] && G
tQ[p, 0]))

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 3.23369, size = 254, normalized size = 1.02 \[ \frac{\frac{4 b^4 \left (2 b^2-5 a^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{5/2}}-\frac{2 b^5 \cos (c+d x)}{a^2 (a-b)^2 (a+b)^2 (a+b \sin (c+d x))}-\frac{4 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a^3}+\frac{4 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^3}+\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{a^2}-\frac{\cot \left (\frac{1}{2} (c+d x)\right )}{a^2}+\frac{2 \sin \left (\frac{1}{2} (c+d x)\right )}{(a+b)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{2 \sin \left (\frac{1}{2} (c+d x)\right )}{(a-b)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.159, size = 346, normalized size = 1.4 \begin{align*}{\frac{1}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-2\,{\frac{{b}^{6}\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}{a}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}-2\,{\frac{{b}^{5}}{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}-10\,{\frac{{b}^{4}}{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}a\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+4\,{\frac{{b}^{6}}{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}{a}^{3}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-{\frac{1}{d \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-2\,{\frac{b\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/a^2*tan(1/2*d*x+1/2*c)-1/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)-2/d*b^6/(a-b)^2/(a+b)^2/a^3/(tan(1/2*d*x+1/2*c
)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*tan(1/2*d*x+1/2*c)-2/d*b^5/(a-b)^2/(a+b)^2/a^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1
/2*d*x+1/2*c)*b+a)-10/d*b^4/(a-b)^2/(a+b)^2/a/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2
)^(1/2))+4/d*b^6/(a-b)^2/(a+b)^2/a^3/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-
1/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)-1/2/d/a^2/tan(1/2*d*x+1/2*c)-2/d/a^3*b*ln(tan(1/2*d*x+1/2*c))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 7.63576, size = 2925, normalized size = 11.79 \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)^2*sec(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[-1/2*(2*a^8 - 4*a^6*b^2 + 2*a^4*b^4 - 2*(2*a^8 - 5*a^6*b^2 + 4*a^4*b^4 - a^2*b^6)*cos(d*x + c)^2 - ((5*a^2*b^
5 - 2*b^7)*cos(d*x + c)^3 - (5*a^3*b^4 - 2*a*b^6)*cos(d*x + c)*sin(d*x + c) - (5*a^2*b^5 - 2*b^7)*cos(d*x + c)
)*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 + 2*(a*cos(d*x + c)*sin(
d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - 2*((a^6*
b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(d*x + c)^3 - (a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*cos(d*x + c)*sin(d
*x + c) - (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 2*((a^6*b^2 - 3*
a^4*b^4 + 3*a^2*b^6 - b^8)*cos(d*x + c)^3 - (a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*cos(d*x + c)*sin(d*x + c)
- (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(a^7*b - 2*a^5*b^3 +
a^3*b^5 + (2*a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - 2*a*b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^9*b - 3*a^7*b^3 + 3*a^
5*b^5 - a^3*b^7)*d*cos(d*x + c)^3 - (a^10 - 3*a^8*b^2 + 3*a^6*b^4 - a^4*b^6)*d*cos(d*x + c)*sin(d*x + c) - (a^
9*b - 3*a^7*b^3 + 3*a^5*b^5 - a^3*b^7)*d*cos(d*x + c)), -(a^8 - 2*a^6*b^2 + a^4*b^4 - (2*a^8 - 5*a^6*b^2 + 4*a
^4*b^4 - a^2*b^6)*cos(d*x + c)^2 - ((5*a^2*b^5 - 2*b^7)*cos(d*x + c)^3 - (5*a^3*b^4 - 2*a*b^6)*cos(d*x + c)*si
n(d*x + c) - (5*a^2*b^5 - 2*b^7)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*c
os(d*x + c))) - ((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(d*x + c)^3 - (a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b
^7)*cos(d*x + c)*sin(d*x + c) - (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1
/2) + ((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(d*x + c)^3 - (a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*cos(d*
x + c)*sin(d*x + c) - (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - (a^
7*b - 2*a^5*b^3 + a^3*b^5 + (2*a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - 2*a*b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^9*b
- 3*a^7*b^3 + 3*a^5*b^5 - a^3*b^7)*d*cos(d*x + c)^3 - (a^10 - 3*a^8*b^2 + 3*a^6*b^4 - a^4*b^6)*d*cos(d*x + c)*
sin(d*x + c) - (a^9*b - 3*a^7*b^3 + 3*a^5*b^5 - a^3*b^7)*d*cos(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)**2*sec(d*x+c)**2/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.32381, size = 706, normalized size = 2.85 \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)^2*sec(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/10*(20*(5*a^2*b^4 - 2*b^6)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/s
qrt(a^2 - b^2)))/((a^7 - 2*a^5*b^2 + a^3*b^4)*sqrt(a^2 - b^2)) - (4*a^5*b*tan(1/2*d*x + 1/2*c)^5 - 8*a^3*b^3*t
an(1/2*d*x + 1/2*c)^5 + 4*a*b^5*tan(1/2*d*x + 1/2*c)^5 - 25*a^6*tan(1/2*d*x + 1/2*c)^4 - 2*a^4*b^2*tan(1/2*d*x
+ 1/2*c)^4 - 21*a^2*b^4*tan(1/2*d*x + 1/2*c)^4 - 12*b^6*tan(1/2*d*x + 1/2*c)^4 - 10*a^5*b*tan(1/2*d*x + 1/2*c
)^3 - 20*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 30*a*b^5*tan(1/2*d*x + 1/2*c)^3 - 20*a^6*tan(1/2*d*x + 1/2*c)^2 + 52
*a^4*b^2*tan(1/2*d*x + 1/2*c)^2 + 16*a^2*b^4*tan(1/2*d*x + 1/2*c)^2 + 12*b^6*tan(1/2*d*x + 1/2*c)^2 + 46*a^5*b
*tan(1/2*d*x + 1/2*c) - 12*a^3*b^3*tan(1/2*d*x + 1/2*c) + 26*a*b^5*tan(1/2*d*x + 1/2*c) + 5*a^6 - 10*a^4*b^2 +
5*a^2*b^4)/((a^7 - 2*a^5*b^2 + a^3*b^4)*(a*tan(1/2*d*x + 1/2*c)^5 + 2*b*tan(1/2*d*x + 1/2*c)^4 - 2*b*tan(1/2*
d*x + 1/2*c)^2 - a*tan(1/2*d*x + 1/2*c))) + 20*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 5*tan(1/2*d*x + 1/2*c)/a
^2)/d