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

[Out]

-2*b^4*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a/(a^2-b^2)^(5/2)/d-4*b^4*(2*a^2-b^2)*arctan((b+a*tan(
1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^3/(a^2-b^2)^(5/2)/d+2*b*arctanh(cos(d*x+c))/a^3/d-cot(d*x+c)/a^2/d+1/2*cos(
d*x+c)/(a+b)^2/d/(1-sin(d*x+c))-1/2*cos(d*x+c)/(a-b)^2/d/(1+sin(d*x+c))-b^5*cos(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*
sin(d*x+c))

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Rubi [A]  time = 0.37, antiderivative size = 248, normalized size of antiderivative = 1.00, 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 8

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

Rule 12

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

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])

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 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 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 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 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 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 3770

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

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*}

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Mathematica [A]  time = 3.25, size = 254, normalized size = 1.02 \[ \frac {-\frac {4 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}+\frac {4 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}-\frac {2 b^5 \cos (c+d x)}{a^2 (a-b)^2 (a+b)^2 (a+b \sin (c+d x))}+\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{a^2}-\frac {\cot \left (\frac {1}{2} (c+d x)\right )}{a^2}+\frac {4 b^4 \left (2 b^2-5 a^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{5/2}}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^2 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}}{2 d} \]

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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fricas [B]  time = 1.27, size = 1355, normalized size = 5.46 \[ \text {result too large to display} \]

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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giac [B]  time = 0.26, size = 523, normalized size = 2.11 \[ -\frac {\frac {20 \, {\left (5 \, a^{2} b^{4} - 2 \, b^{6}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {4 \, a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 8 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 25 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 21 \, a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 10 \, a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 20 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 20 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 52 \, a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 16 \, a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 46 \, a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 26 \, a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, a^{6} - 10 \, a^{4} b^{2} + 5 \, a^{2} b^{4}}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}} + \frac {20 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}}}{10 \, d} \]

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

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maple [A]  time = 0.62, size = 346, normalized size = 1.40 \[ -\frac {1}{d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {1}{2 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}-\frac {1}{d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 b^{6} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (a -b \right )^{2} \left (a +b \right )^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}-\frac {2 b^{5}}{d \,a^{2} \left (a -b \right )^{2} \left (a +b \right )^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}-\frac {10 b^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d a \left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}+\frac {4 b^{6} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,a^{3} \left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}} \]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 13.33, size = 2151, normalized size = 8.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a + (2*tan(c/2 + (d*x)/2)*(5*a^4*b + 3*b^5 - 2*a^2*b^3))/(a^4 + b^4 - 2*a^2*b^2) + (4*tan(c/2 + (d*x)/2)^2*(b
^6 - a^6 + 3*a^4*b^2))/(a*(a^2 - b^2)^2) - (2*b*tan(c/2 + (d*x)/2)^3*(a^4 + 3*b^4 + 2*a^2*b^2))/(a^4 + b^4 - 2
*a^2*b^2) - (tan(c/2 + (d*x)/2)^4*(5*a^6 + 4*b^6 + a^2*b^4 + 2*a^4*b^2))/(a*(a^4 + b^4 - 2*a^2*b^2)))/(d*(2*a^
3*tan(c/2 + (d*x)/2)^5 - 2*a^3*tan(c/2 + (d*x)/2) - 4*a^2*b*tan(c/2 + (d*x)/2)^2 + 4*a^2*b*tan(c/2 + (d*x)/2)^
4)) + tan(c/2 + (d*x)/2)/(2*a^2*d) - (2*b*log(tan(c/2 + (d*x)/2)))/(a^3*d) + (b^4*atan(((b^4*(5*a^2 - 2*b^2)*(
-(a + b)^5*(a - b)^5)^(1/2)*(tan(c/2 + (d*x)/2)*(4*a^18*b - 16*a^4*b^15 + 104*a^6*b^13 - 272*a^8*b^11 + 372*a^
10*b^9 - 288*a^12*b^7 + 128*a^14*b^5 - 32*a^16*b^3) - 8*a^5*b^14 + 50*a^7*b^12 - 124*a^9*b^10 + 156*a^11*b^8 -
104*a^13*b^6 + 34*a^15*b^4 - 4*a^17*b^2 + (b^4*(5*a^2 - 2*b^2)*(-(a + b)^5*(a - b)^5)^(1/2)*(2*a^20*b - tan(c
/2 + (d*x)/2)*(6*a^21 - 8*a^7*b^14 + 54*a^9*b^12 - 156*a^11*b^10 + 250*a^13*b^8 - 240*a^15*b^6 + 138*a^17*b^4
- 44*a^19*b^2) + 2*a^8*b^13 - 12*a^10*b^11 + 30*a^12*b^9 - 40*a^14*b^7 + 30*a^16*b^5 - 12*a^18*b^3))/(a^13 - a
^3*b^10 + 5*a^5*b^8 - 10*a^7*b^6 + 10*a^9*b^4 - 5*a^11*b^2))*1i)/(a^13 - a^3*b^10 + 5*a^5*b^8 - 10*a^7*b^6 + 1
0*a^9*b^4 - 5*a^11*b^2) - (b^4*(5*a^2 - 2*b^2)*(-(a + b)^5*(a - b)^5)^(1/2)*(8*a^5*b^14 - tan(c/2 + (d*x)/2)*(
4*a^18*b - 16*a^4*b^15 + 104*a^6*b^13 - 272*a^8*b^11 + 372*a^10*b^9 - 288*a^12*b^7 + 128*a^14*b^5 - 32*a^16*b^
3) - 50*a^7*b^12 + 124*a^9*b^10 - 156*a^11*b^8 + 104*a^13*b^6 - 34*a^15*b^4 + 4*a^17*b^2 + (b^4*(5*a^2 - 2*b^2
)*(-(a + b)^5*(a - b)^5)^(1/2)*(2*a^20*b - tan(c/2 + (d*x)/2)*(6*a^21 - 8*a^7*b^14 + 54*a^9*b^12 - 156*a^11*b^
10 + 250*a^13*b^8 - 240*a^15*b^6 + 138*a^17*b^4 - 44*a^19*b^2) + 2*a^8*b^13 - 12*a^10*b^11 + 30*a^12*b^9 - 40*
a^14*b^7 + 30*a^16*b^5 - 12*a^18*b^3))/(a^13 - a^3*b^10 + 5*a^5*b^8 - 10*a^7*b^6 + 10*a^9*b^4 - 5*a^11*b^2))*1
i)/(a^13 - a^3*b^10 + 5*a^5*b^8 - 10*a^7*b^6 + 10*a^9*b^4 - 5*a^11*b^2))/(16*a^2*b^15 - 104*a^4*b^13 + 256*a^6
*b^11 - 304*a^8*b^9 + 176*a^10*b^7 - 40*a^12*b^5 - 2*tan(c/2 + (d*x)/2)*(20*a^5*b^12 - 8*a^3*b^14 + 24*a^7*b^1
0 - 76*a^9*b^8 + 40*a^11*b^6) + (b^4*(5*a^2 - 2*b^2)*(-(a + b)^5*(a - b)^5)^(1/2)*(tan(c/2 + (d*x)/2)*(4*a^18*
b - 16*a^4*b^15 + 104*a^6*b^13 - 272*a^8*b^11 + 372*a^10*b^9 - 288*a^12*b^7 + 128*a^14*b^5 - 32*a^16*b^3) - 8*
a^5*b^14 + 50*a^7*b^12 - 124*a^9*b^10 + 156*a^11*b^8 - 104*a^13*b^6 + 34*a^15*b^4 - 4*a^17*b^2 + (b^4*(5*a^2 -
2*b^2)*(-(a + b)^5*(a - b)^5)^(1/2)*(2*a^20*b - tan(c/2 + (d*x)/2)*(6*a^21 - 8*a^7*b^14 + 54*a^9*b^12 - 156*a
^11*b^10 + 250*a^13*b^8 - 240*a^15*b^6 + 138*a^17*b^4 - 44*a^19*b^2) + 2*a^8*b^13 - 12*a^10*b^11 + 30*a^12*b^9
- 40*a^14*b^7 + 30*a^16*b^5 - 12*a^18*b^3))/(a^13 - a^3*b^10 + 5*a^5*b^8 - 10*a^7*b^6 + 10*a^9*b^4 - 5*a^11*b
^2)))/(a^13 - a^3*b^10 + 5*a^5*b^8 - 10*a^7*b^6 + 10*a^9*b^4 - 5*a^11*b^2) + (b^4*(5*a^2 - 2*b^2)*(-(a + b)^5*
(a - b)^5)^(1/2)*(8*a^5*b^14 - tan(c/2 + (d*x)/2)*(4*a^18*b - 16*a^4*b^15 + 104*a^6*b^13 - 272*a^8*b^11 + 372*
a^10*b^9 - 288*a^12*b^7 + 128*a^14*b^5 - 32*a^16*b^3) - 50*a^7*b^12 + 124*a^9*b^10 - 156*a^11*b^8 + 104*a^13*b
^6 - 34*a^15*b^4 + 4*a^17*b^2 + (b^4*(5*a^2 - 2*b^2)*(-(a + b)^5*(a - b)^5)^(1/2)*(2*a^20*b - tan(c/2 + (d*x)/
2)*(6*a^21 - 8*a^7*b^14 + 54*a^9*b^12 - 156*a^11*b^10 + 250*a^13*b^8 - 240*a^15*b^6 + 138*a^17*b^4 - 44*a^19*b
^2) + 2*a^8*b^13 - 12*a^10*b^11 + 30*a^12*b^9 - 40*a^14*b^7 + 30*a^16*b^5 - 12*a^18*b^3))/(a^13 - a^3*b^10 + 5
*a^5*b^8 - 10*a^7*b^6 + 10*a^9*b^4 - 5*a^11*b^2)))/(a^13 - a^3*b^10 + 5*a^5*b^8 - 10*a^7*b^6 + 10*a^9*b^4 - 5*
a^11*b^2)))*(5*a^2 - 2*b^2)*(-(a + b)^5*(a - b)^5)^(1/2)*2i)/(d*(a^13 - a^3*b^10 + 5*a^5*b^8 - 10*a^7*b^6 + 10
*a^9*b^4 - 5*a^11*b^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]

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]

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

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