3.1470 \(\int \frac{\csc ^3(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=295 \[ \frac{2 b^5 \left (5 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 d \left (a^2-b^2\right )^{5/2}}+\frac{2 b^5 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{5/2}}+\frac{b^6 \cos (c+d x)}{a^3 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac{\left (a^2+3 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{2 b \cot (c+d x)}{a^3 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}+\frac{\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)} \]

[Out]

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

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Rubi [A]  time = 0.39499, antiderivative size = 295, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 11, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.379, Rules used = {2897, 3770, 3767, 8, 3768, 2648, 2664, 12, 2660, 618, 204} \[ \frac{2 b^5 \left (5 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 d \left (a^2-b^2\right )^{5/2}}+\frac{2 b^5 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{5/2}}+\frac{b^6 \cos (c+d x)}{a^3 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac{\left (a^2+3 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{2 b \cot (c+d x)}{a^3 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}+\frac{\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^3*Sec[c + d*x]^2)/(a + b*Sin[c + d*x])^2,x]

[Out]

(2*b^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^(5/2)*d) + (2*b^5*(5*a^2 - 3*b^2)*Ar
cTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^4*(a^2 - b^2)^(5/2)*d) - ArcTanh[Cos[c + d*x]]/(2*a^2*d) -
((a^2 + 3*b^2)*ArcTanh[Cos[c + d*x]])/(a^4*d) + (2*b*Cot[c + d*x])/(a^3*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*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^
6*Cos[c + d*x])/(a^3*(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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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

Mathematica [A]  time = 6.49093, size = 356, normalized size = 1.21 \[ \frac{3 \left (a^2+2 b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^4 d}-\frac{3 \left (a^2+2 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^4 d}+\frac{b^6 \cos (c+d x)}{a^3 d (a-b)^2 (a+b)^2 (a+b \sin (c+d x))}+\frac{6 b^5 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (a \sin \left (\frac{1}{2} (c+d x)\right )+b \cos \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2-b^2}}\right )}{a^4 d \left (a^2-b^2\right )^{5/2}}-\frac{b \tan \left (\frac{1}{2} (c+d x)\right )}{a^3 d}+\frac{b \cot \left (\frac{1}{2} (c+d x)\right )}{a^3 d}-\frac{\csc ^2\left (\frac{1}{2} (c+d x)\right )}{8 a^2 d}+\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right )}{8 a^2 d}+\frac{\sin \left (\frac{1}{2} (c+d x)\right )}{d (a+b)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{\sin \left (\frac{1}{2} (c+d x)\right )}{d (a-b)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^3*Sec[c + d*x]^2)/(a + b*Sin[c + d*x])^2,x]

[Out]

(6*b^5*(2*a^2 - b^2)*ArcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] + a*Sin[(c + d*x)/2]))/Sqrt[a^2 - b^2]])/(a^
4*(a^2 - b^2)^(5/2)*d) + (b*Cot[(c + d*x)/2])/(a^3*d) - Csc[(c + d*x)/2]^2/(8*a^2*d) - (3*(a^2 + 2*b^2)*Log[Co
s[(c + d*x)/2]])/(2*a^4*d) + (3*(a^2 + 2*b^2)*Log[Sin[(c + d*x)/2]])/(2*a^4*d) + Sec[(c + d*x)/2]^2/(8*a^2*d)
+ Sin[(c + d*x)/2]/((a + b)^2*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - Sin[(c + d*x)/2]/((a - b)^2*d*(Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2])) + (b^6*Cos[c + d*x])/(a^3*(a - b)^2*(a + b)^2*d*(a + b*Sin[c + d*x])) - (b*Tan
[(c + d*x)/2])/(a^3*d)

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Maple [A]  time = 0.167, size = 404, normalized size = 1.4 \begin{align*}{\frac{1}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{b}{d{a}^{3}}\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}^{7}\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{4} \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{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 ) }}+2\,{\frac{{b}^{6}}{d{a}^{3} \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{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 ) }}+12\,{\frac{{b}^{5}}{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}{a}^{2}\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 ) }-6\,{\frac{{b}^{7}}{d{a}^{4} \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}\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}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{3}{2\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){b}^{2}}{d{a}^{4}}}+{\frac{b}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3*sec(d*x+c)^2/(a+b*sin(d*x+c))^2,x)

[Out]

1/8/d/a^2*tan(1/2*d*x+1/2*c)^2-1/d/a^3*tan(1/2*d*x+1/2*c)*b-1/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)+2/d/a^4*b^7/(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^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)+12/d*b^5/(a-b)^2/(a+b)^2/a^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))-6/d/a^4*b^7/(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))+1/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)-1/8/d/a^2/tan(1/2*d*x+1/2*c)^2+3/2/d/a^2*l
n(tan(1/2*d*x+1/2*c))+3/d/a^4*ln(tan(1/2*d*x+1/2*c))*b^2+1/d*b/a^3/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)^3*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 = 14.6489, size = 3976, normalized size = 13.48 \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)^3*sec(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[-1/4*(4*a^9 - 8*a^7*b^2 + 4*a^5*b^4 - 4*(4*a^7*b^2 - 8*a^5*b^4 + 7*a^3*b^6 - 3*a*b^8)*cos(d*x + c)^4 - 6*(a^9
- 5*a^7*b^2 + 7*a^5*b^4 - 5*a^3*b^6 + 2*a*b^8)*cos(d*x + c)^2 - 6*((2*a^3*b^5 - a*b^7)*cos(d*x + c)^3 - (2*a^
3*b^5 - a*b^7)*cos(d*x + c) + ((2*a^2*b^6 - b^8)*cos(d*x + c)^3 - (2*a^2*b^6 - b^8)*cos(d*x + c))*sin(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)) + 3*((a^9
- a^7*b^2 - 3*a^5*b^4 + 5*a^3*b^6 - 2*a*b^8)*cos(d*x + c)^3 - (a^9 - a^7*b^2 - 3*a^5*b^4 + 5*a^3*b^6 - 2*a*b^
8)*cos(d*x + c) + ((a^8*b - a^6*b^3 - 3*a^4*b^5 + 5*a^2*b^7 - 2*b^9)*cos(d*x + c)^3 - (a^8*b - a^6*b^3 - 3*a^4
*b^5 + 5*a^2*b^7 - 2*b^9)*cos(d*x + c))*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 3*((a^9 - a^7*b^2 - 3*a^5*
b^4 + 5*a^3*b^6 - 2*a*b^8)*cos(d*x + c)^3 - (a^9 - a^7*b^2 - 3*a^5*b^4 + 5*a^3*b^6 - 2*a*b^8)*cos(d*x + c) + (
(a^8*b - a^6*b^3 - 3*a^4*b^5 + 5*a^2*b^7 - 2*b^9)*cos(d*x + c)^3 - (a^8*b - a^6*b^3 - 3*a^4*b^5 + 5*a^2*b^7 -
2*b^9)*cos(d*x + c))*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(2*a^8*b - 4*a^6*b^3 + 2*a^4*b^5 - (5*a^8*
b - 13*a^6*b^3 + 11*a^4*b^5 - 3*a^2*b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^
6)*d*cos(d*x + c)^3 - (a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cos(d*x + c) + ((a^10*b - 3*a^8*b^3 + 3*a^6*b
^5 - a^4*b^7)*d*cos(d*x + c)^3 - (a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*cos(d*x + c))*sin(d*x + c)), -1/
4*(4*a^9 - 8*a^7*b^2 + 4*a^5*b^4 - 4*(4*a^7*b^2 - 8*a^5*b^4 + 7*a^3*b^6 - 3*a*b^8)*cos(d*x + c)^4 - 6*(a^9 - 5
*a^7*b^2 + 7*a^5*b^4 - 5*a^3*b^6 + 2*a*b^8)*cos(d*x + c)^2 + 12*((2*a^3*b^5 - a*b^7)*cos(d*x + c)^3 - (2*a^3*b
^5 - a*b^7)*cos(d*x + c) + ((2*a^2*b^6 - b^8)*cos(d*x + c)^3 - (2*a^2*b^6 - b^8)*cos(d*x + c))*sin(d*x + c))*s
qrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + 3*((a^9 - a^7*b^2 - 3*a^5*b^4 +
5*a^3*b^6 - 2*a*b^8)*cos(d*x + c)^3 - (a^9 - a^7*b^2 - 3*a^5*b^4 + 5*a^3*b^6 - 2*a*b^8)*cos(d*x + c) + ((a^8*b
- a^6*b^3 - 3*a^4*b^5 + 5*a^2*b^7 - 2*b^9)*cos(d*x + c)^3 - (a^8*b - a^6*b^3 - 3*a^4*b^5 + 5*a^2*b^7 - 2*b^9)
*cos(d*x + c))*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 3*((a^9 - a^7*b^2 - 3*a^5*b^4 + 5*a^3*b^6 - 2*a*b^8
)*cos(d*x + c)^3 - (a^9 - a^7*b^2 - 3*a^5*b^4 + 5*a^3*b^6 - 2*a*b^8)*cos(d*x + c) + ((a^8*b - a^6*b^3 - 3*a^4*
b^5 + 5*a^2*b^7 - 2*b^9)*cos(d*x + c)^3 - (a^8*b - a^6*b^3 - 3*a^4*b^5 + 5*a^2*b^7 - 2*b^9)*cos(d*x + c))*sin(
d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(2*a^8*b - 4*a^6*b^3 + 2*a^4*b^5 - (5*a^8*b - 13*a^6*b^3 + 11*a^4*b
^5 - 3*a^2*b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cos(d*x + c)^3 - (a^
11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cos(d*x + c) + ((a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*cos(d*x +
c)^3 - (a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*cos(d*x + c))*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)**3*sec(d*x+c)**2/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.25932, size = 571, normalized size = 1.94 \begin{align*} \frac{\frac{48 \,{\left (2 \, a^{2} b^{5} - b^{7}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \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^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \sqrt{a^{2} - b^{2}}} + \frac{16 \,{\left (2 \, a^{6} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + b^{7} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a^{7} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{5} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a b^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, a^{4} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b^{7} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{7} - a^{5} b^{2} - a b^{6}\right )}}{{\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )}{\left (a \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 )^{3} - 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\right )}} + \frac{12 \,{\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{4}} + \frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{4}} - \frac{18 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 36 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2}}{a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/8*(48*(2*a^2*b^5 - b^7)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(
a^2 - b^2)))/((a^8 - 2*a^6*b^2 + a^4*b^4)*sqrt(a^2 - b^2)) + 16*(2*a^6*b*tan(1/2*d*x + 1/2*c)^3 + b^7*tan(1/2*
d*x + 1/2*c)^3 - a^7*tan(1/2*d*x + 1/2*c)^2 + 3*a^5*b^2*tan(1/2*d*x + 1/2*c)^2 + a*b^6*tan(1/2*d*x + 1/2*c)^2
- 2*a^4*b^3*tan(1/2*d*x + 1/2*c) - b^7*tan(1/2*d*x + 1/2*c) - a^7 - a^5*b^2 - a*b^6)/((a^8 - 2*a^6*b^2 + a^4*b
^4)*(a*tan(1/2*d*x + 1/2*c)^4 + 2*b*tan(1/2*d*x + 1/2*c)^3 - 2*b*tan(1/2*d*x + 1/2*c) - a)) + 12*(a^2 + 2*b^2)
*log(abs(tan(1/2*d*x + 1/2*c)))/a^4 + (a^2*tan(1/2*d*x + 1/2*c)^2 - 8*a*b*tan(1/2*d*x + 1/2*c))/a^4 - (18*a^2*
tan(1/2*d*x + 1/2*c)^2 + 36*b^2*tan(1/2*d*x + 1/2*c)^2 - 8*a*b*tan(1/2*d*x + 1/2*c) + a^2)/(a^4*tan(1/2*d*x +
1/2*c)^2))/d