3.838 \(\int \frac{\csc ^3(c+d x) \sec ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=194 \[ -\frac{2 \tan ^7(c+d x)}{7 a^2 d}-\frac{8 \tan ^5(c+d x)}{5 a^2 d}-\frac{4 \tan ^3(c+d x)}{a^2 d}-\frac{8 \tan (c+d x)}{a^2 d}+\frac{2 \cot (c+d x)}{a^2 d}+\frac{11 \sec ^7(c+d x)}{14 a^2 d}+\frac{11 \sec ^5(c+d x)}{10 a^2 d}+\frac{11 \sec ^3(c+d x)}{6 a^2 d}+\frac{11 \sec (c+d x)}{2 a^2 d}-\frac{11 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac{\csc ^2(c+d x) \sec ^7(c+d x)}{2 a^2 d} \]

[Out]

(-11*ArcTanh[Cos[c + d*x]])/(2*a^2*d) + (2*Cot[c + d*x])/(a^2*d) + (11*Sec[c + d*x])/(2*a^2*d) + (11*Sec[c + d
*x]^3)/(6*a^2*d) + (11*Sec[c + d*x]^5)/(10*a^2*d) + (11*Sec[c + d*x]^7)/(14*a^2*d) - (Csc[c + d*x]^2*Sec[c + d
*x]^7)/(2*a^2*d) - (8*Tan[c + d*x])/(a^2*d) - (4*Tan[c + d*x]^3)/(a^2*d) - (8*Tan[c + d*x]^5)/(5*a^2*d) - (2*T
an[c + d*x]^7)/(7*a^2*d)

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Rubi [A]  time = 0.365604, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2875, 2873, 2622, 302, 207, 2620, 270, 288} \[ -\frac{2 \tan ^7(c+d x)}{7 a^2 d}-\frac{8 \tan ^5(c+d x)}{5 a^2 d}-\frac{4 \tan ^3(c+d x)}{a^2 d}-\frac{8 \tan (c+d x)}{a^2 d}+\frac{2 \cot (c+d x)}{a^2 d}+\frac{11 \sec ^7(c+d x)}{14 a^2 d}+\frac{11 \sec ^5(c+d x)}{10 a^2 d}+\frac{11 \sec ^3(c+d x)}{6 a^2 d}+\frac{11 \sec (c+d x)}{2 a^2 d}-\frac{11 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac{\csc ^2(c+d x) \sec ^7(c+d x)}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^3*Sec[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]

[Out]

(-11*ArcTanh[Cos[c + d*x]])/(2*a^2*d) + (2*Cot[c + d*x])/(a^2*d) + (11*Sec[c + d*x])/(2*a^2*d) + (11*Sec[c + d
*x]^3)/(6*a^2*d) + (11*Sec[c + d*x]^5)/(10*a^2*d) + (11*Sec[c + d*x]^7)/(14*a^2*d) - (Csc[c + d*x]^2*Sec[c + d
*x]^7)/(2*a^2*d) - (8*Tan[c + d*x])/(a^2*d) - (4*Tan[c + d*x]^3)/(a^2*d) - (8*Tan[c + d*x]^5)/(5*a^2*d) - (2*T
an[c + d*x]^7)/(7*a^2*d)

Rule 2875

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(a/g)^(2*m), Int[((g*Cos[e + f*x])^(2*m + p)*(d*Sin[e + f*x])^n)/(a - b*Sin[e +
f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]

Rule 2873

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

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 207

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

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

\begin{align*} \int \frac{\csc ^3(c+d x) \sec ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\int \csc ^3(c+d x) \sec ^8(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac{\int \left (a^2 \csc (c+d x) \sec ^8(c+d x)-2 a^2 \csc ^2(c+d x) \sec ^8(c+d x)+a^2 \csc ^3(c+d x) \sec ^8(c+d x)\right ) \, dx}{a^4}\\ &=\frac{\int \csc (c+d x) \sec ^8(c+d x) \, dx}{a^2}+\frac{\int \csc ^3(c+d x) \sec ^8(c+d x) \, dx}{a^2}-\frac{2 \int \csc ^2(c+d x) \sec ^8(c+d x) \, dx}{a^2}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^{10}}{\left (-1+x^2\right )^2} \, dx,x,\sec (c+d x)\right )}{a^2 d}+\frac{\operatorname{Subst}\left (\int \frac{x^8}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^2 d}-\frac{2 \operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^4}{x^2} \, dx,x,\tan (c+d x)\right )}{a^2 d}\\ &=-\frac{\csc ^2(c+d x) \sec ^7(c+d x)}{2 a^2 d}+\frac{\operatorname{Subst}\left (\int \left (1+x^2+x^4+x^6+\frac{1}{-1+x^2}\right ) \, dx,x,\sec (c+d x)\right )}{a^2 d}-\frac{2 \operatorname{Subst}\left (\int \left (4+\frac{1}{x^2}+6 x^2+4 x^4+x^6\right ) \, dx,x,\tan (c+d x)\right )}{a^2 d}+\frac{9 \operatorname{Subst}\left (\int \frac{x^8}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{2 a^2 d}\\ &=\frac{2 \cot (c+d x)}{a^2 d}+\frac{\sec (c+d x)}{a^2 d}+\frac{\sec ^3(c+d x)}{3 a^2 d}+\frac{\sec ^5(c+d x)}{5 a^2 d}+\frac{\sec ^7(c+d x)}{7 a^2 d}-\frac{\csc ^2(c+d x) \sec ^7(c+d x)}{2 a^2 d}-\frac{8 \tan (c+d x)}{a^2 d}-\frac{4 \tan ^3(c+d x)}{a^2 d}-\frac{8 \tan ^5(c+d x)}{5 a^2 d}-\frac{2 \tan ^7(c+d x)}{7 a^2 d}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^2 d}+\frac{9 \operatorname{Subst}\left (\int \left (1+x^2+x^4+x^6+\frac{1}{-1+x^2}\right ) \, dx,x,\sec (c+d x)\right )}{2 a^2 d}\\ &=-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac{2 \cot (c+d x)}{a^2 d}+\frac{11 \sec (c+d x)}{2 a^2 d}+\frac{11 \sec ^3(c+d x)}{6 a^2 d}+\frac{11 \sec ^5(c+d x)}{10 a^2 d}+\frac{11 \sec ^7(c+d x)}{14 a^2 d}-\frac{\csc ^2(c+d x) \sec ^7(c+d x)}{2 a^2 d}-\frac{8 \tan (c+d x)}{a^2 d}-\frac{4 \tan ^3(c+d x)}{a^2 d}-\frac{8 \tan ^5(c+d x)}{5 a^2 d}-\frac{2 \tan ^7(c+d x)}{7 a^2 d}+\frac{9 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{2 a^2 d}\\ &=-\frac{11 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}+\frac{2 \cot (c+d x)}{a^2 d}+\frac{11 \sec (c+d x)}{2 a^2 d}+\frac{11 \sec ^3(c+d x)}{6 a^2 d}+\frac{11 \sec ^5(c+d x)}{10 a^2 d}+\frac{11 \sec ^7(c+d x)}{14 a^2 d}-\frac{\csc ^2(c+d x) \sec ^7(c+d x)}{2 a^2 d}-\frac{8 \tan (c+d x)}{a^2 d}-\frac{4 \tan ^3(c+d x)}{a^2 d}-\frac{8 \tan ^5(c+d x)}{5 a^2 d}-\frac{2 \tan ^7(c+d x)}{7 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.582447, size = 277, normalized size = 1.43 \[ \frac{-36960 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^7+36960 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^7+\frac{\csc ^2(c+d x) (4488 \sin (c+d x)-7536 \sin (2 (c+d x))+3836 \sin (3 (c+d x))-780 \sin (5 (c+d x))+2512 \sin (6 (c+d x))-768 \sin (7 (c+d x))-6908 \cos (c+d x)-563 \cos (2 (c+d x))+4396 \cos (3 (c+d x))-5390 \cos (4 (c+d x))+3140 \cos (5 (c+d x))-1917 \cos (6 (c+d x))-628 \cos (7 (c+d x))+4510)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}}{6720 d (a \sin (c+d x)+a)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^3*Sec[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]

[Out]

(-36960*Log[Cos[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^7 + 36960*Log[Sin[(c + d*x)/2]]*(Cos[(c +
d*x)/2] + Sin[(c + d*x)/2])^7 + (Csc[c + d*x]^2*(4510 - 6908*Cos[c + d*x] - 563*Cos[2*(c + d*x)] + 4396*Cos[3*
(c + d*x)] - 5390*Cos[4*(c + d*x)] + 3140*Cos[5*(c + d*x)] - 1917*Cos[6*(c + d*x)] - 628*Cos[7*(c + d*x)] + 44
88*Sin[c + d*x] - 7536*Sin[2*(c + d*x)] + 3836*Sin[3*(c + d*x)] - 780*Sin[5*(c + d*x)] + 2512*Sin[6*(c + d*x)]
- 768*Sin[7*(c + d*x)]))/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3)/(6720*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2
])^3*(a + a*Sin[c + d*x])^2)

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Maple [A]  time = 0.162, size = 303, normalized size = 1.6 \begin{align*}{\frac{1}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{1}{d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{12\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{3}{4\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{4}{7\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-7}}-2\,{\frac{1}{d{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{6}}}+{\frac{26}{5\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}-8\,{\frac{1}{d{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{4}}}+{\frac{139}{12\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{83}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{67}{4\,d{a}^{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{1}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}+{\frac{11}{2\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3*sec(d*x+c)^4/(a+a*sin(d*x+c))^2,x)

[Out]

1/8/d/a^2*tan(1/2*d*x+1/2*c)^2-1/d/a^2*tan(1/2*d*x+1/2*c)-1/12/d/a^2/(tan(1/2*d*x+1/2*c)-1)^3-1/8/d/a^2/(tan(1
/2*d*x+1/2*c)-1)^2-3/4/d/a^2/(tan(1/2*d*x+1/2*c)-1)+4/7/d/a^2/(tan(1/2*d*x+1/2*c)+1)^7-2/d/a^2/(tan(1/2*d*x+1/
2*c)+1)^6+26/5/d/a^2/(tan(1/2*d*x+1/2*c)+1)^5-8/d/a^2/(tan(1/2*d*x+1/2*c)+1)^4+139/12/d/a^2/(tan(1/2*d*x+1/2*c
)+1)^3-83/8/d/a^2/(tan(1/2*d*x+1/2*c)+1)^2+67/4/d/a^2/(tan(1/2*d*x+1/2*c)+1)-1/8/d/a^2/tan(1/2*d*x+1/2*c)^2+1/
d/a^2/tan(1/2*d*x+1/2*c)+11/2/d/a^2*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.15984, size = 710, normalized size = 3.66 \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)^4/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/840*((420*sin(d*x + c)/(cos(d*x + c) + 1) + 15173*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 38432*sin(d*x + c)^3
/(cos(d*x + c) + 1)^3 + 894*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 95344*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 -
77182*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 61992*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 101115*sin(d*x + c)^8/
(cos(d*x + c) + 1)^8 + 11340*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 33495*sin(d*x + c)^10/(cos(d*x + c) + 1)^10
- 14280*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 105)/(a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 4*a^2*sin(d*x
+ c)^3/(cos(d*x + c) + 1)^3 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 8*a^2*sin(d*x + c)^5/(cos(d*x + c)
+ 1)^5 - 14*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 14*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 8*a^2*sin(d
*x + c)^9/(cos(d*x + c) + 1)^9 - 3*a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 4*a^2*sin(d*x + c)^11/(cos(d*x
+ c) + 1)^11 - a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12) - 105*(8*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x +
c)^2/(cos(d*x + c) + 1)^2)/a^2 + 4620*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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Fricas [A]  time = 1.90948, size = 786, normalized size = 4.05 \begin{align*} -\frac{3834 \, \cos \left (d x + c\right )^{6} - 3056 \, \cos \left (d x + c\right )^{4} - 468 \, \cos \left (d x + c\right )^{2} + 1155 \,{\left (\cos \left (d x + c\right )^{7} - 3 \, \cos \left (d x + c\right )^{5} + 2 \, \cos \left (d x + c\right )^{3} - 2 \,{\left (\cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 1155 \,{\left (\cos \left (d x + c\right )^{7} - 3 \, \cos \left (d x + c\right )^{5} + 2 \, \cos \left (d x + c\right )^{3} - 2 \,{\left (\cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 4 \,{\left (768 \, \cos \left (d x + c\right )^{6} - 765 \, \cos \left (d x + c\right )^{4} - 98 \, \cos \left (d x + c\right )^{2} - 10\right )} \sin \left (d x + c\right ) - 100}{420 \,{\left (a^{2} d \cos \left (d x + c\right )^{7} - 3 \, a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} - 2 \,{\left (a^{2} d \cos \left (d x + c\right )^{5} - a^{2} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/420*(3834*cos(d*x + c)^6 - 3056*cos(d*x + c)^4 - 468*cos(d*x + c)^2 + 1155*(cos(d*x + c)^7 - 3*cos(d*x + c)
^5 + 2*cos(d*x + c)^3 - 2*(cos(d*x + c)^5 - cos(d*x + c)^3)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 1155*(
cos(d*x + c)^7 - 3*cos(d*x + c)^5 + 2*cos(d*x + c)^3 - 2*(cos(d*x + c)^5 - cos(d*x + c)^3)*sin(d*x + c))*log(-
1/2*cos(d*x + c) + 1/2) + 4*(768*cos(d*x + c)^6 - 765*cos(d*x + c)^4 - 98*cos(d*x + c)^2 - 10)*sin(d*x + c) -
100)/(a^2*d*cos(d*x + c)^7 - 3*a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^3 - 2*(a^2*d*cos(d*x + c)^5 - a^2*d
*cos(d*x + c)^3)*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)**4/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.33592, size = 321, normalized size = 1.65 \begin{align*} \frac{\frac{4620 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac{105 \,{\left (a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a^{4}} - \frac{105 \,{\left (66 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}}{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}} - \frac{35 \,{\left (18 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 33 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 17\right )}}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} + \frac{14070 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 75705 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 177205 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 226450 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 166488 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 66661 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 11533}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{7}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/840*(4620*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 105*(a^2*tan(1/2*d*x + 1/2*c)^2 - 8*a^2*tan(1/2*d*x + 1/2*c))
/a^4 - 105*(66*tan(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) + 1)/(a^2*tan(1/2*d*x + 1/2*c)^2) - 35*(18*tan(
1/2*d*x + 1/2*c)^2 - 33*tan(1/2*d*x + 1/2*c) + 17)/(a^2*(tan(1/2*d*x + 1/2*c) - 1)^3) + (14070*tan(1/2*d*x + 1
/2*c)^6 + 75705*tan(1/2*d*x + 1/2*c)^5 + 177205*tan(1/2*d*x + 1/2*c)^4 + 226450*tan(1/2*d*x + 1/2*c)^3 + 16648
8*tan(1/2*d*x + 1/2*c)^2 + 66661*tan(1/2*d*x + 1/2*c) + 11533)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^7))/d