3.1358 \(\int \frac{\csc ^3(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=332 \[ -\frac{2 b^7 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{5/2}}+\frac{b^2 \sec ^3(c+d x)}{3 a^3 d}+\frac{b^2 \sec (c+d x)}{a^3 d}-\frac{b^2 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{b^3 \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a^3 d \left (a^2-b^2\right )}-\frac{b^3 \sec (c+d x) \left (a \left (2 a^2-5 b^2\right ) \sin (c+d x)+3 b^3\right )}{3 a^3 d \left (a^2-b^2\right )^2}-\frac{b \tan ^3(c+d x)}{3 a^2 d}-\frac{2 b \tan (c+d x)}{a^2 d}+\frac{b \cot (c+d x)}{a^2 d}+\frac{5 \sec ^3(c+d x)}{6 a d}+\frac{5 \sec (c+d x)}{2 a d}-\frac{5 \tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac{\csc ^2(c+d x) \sec ^3(c+d x)}{2 a d} \]

[Out]

(-2*b^7*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2)^(5/2)*d) - (5*ArcTanh[Cos[c + d*x]]
)/(2*a*d) - (b^2*ArcTanh[Cos[c + d*x]])/(a^3*d) + (b*Cot[c + d*x])/(a^2*d) + (5*Sec[c + d*x])/(2*a*d) + (b^2*S
ec[c + d*x])/(a^3*d) + (5*Sec[c + d*x]^3)/(6*a*d) + (b^2*Sec[c + d*x]^3)/(3*a^3*d) - (Csc[c + d*x]^2*Sec[c + d
*x]^3)/(2*a*d) + (b^3*Sec[c + d*x]^3*(b - a*Sin[c + d*x]))/(3*a^3*(a^2 - b^2)*d) - (b^3*Sec[c + d*x]*(3*b^3 +
a*(2*a^2 - 5*b^2)*Sin[c + d*x]))/(3*a^3*(a^2 - b^2)^2*d) - (2*b*Tan[c + d*x])/(a^2*d) - (b*Tan[c + d*x]^3)/(3*
a^2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.547913, antiderivative size = 332, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 13, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.448, Rules used = {2898, 2622, 302, 207, 2620, 270, 288, 2696, 2866, 12, 2660, 618, 204} \[ -\frac{2 b^7 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{5/2}}+\frac{b^2 \sec ^3(c+d x)}{3 a^3 d}+\frac{b^2 \sec (c+d x)}{a^3 d}-\frac{b^2 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{b^3 \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a^3 d \left (a^2-b^2\right )}-\frac{b^3 \sec (c+d x) \left (a \left (2 a^2-5 b^2\right ) \sin (c+d x)+3 b^3\right )}{3 a^3 d \left (a^2-b^2\right )^2}-\frac{b \tan ^3(c+d x)}{3 a^2 d}-\frac{2 b \tan (c+d x)}{a^2 d}+\frac{b \cot (c+d x)}{a^2 d}+\frac{5 \sec ^3(c+d x)}{6 a d}+\frac{5 \sec (c+d x)}{2 a d}-\frac{5 \tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac{\csc ^2(c+d x) \sec ^3(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^7*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2)^(5/2)*d) - (5*ArcTanh[Cos[c + d*x]]
)/(2*a*d) - (b^2*ArcTanh[Cos[c + d*x]])/(a^3*d) + (b*Cot[c + d*x])/(a^2*d) + (5*Sec[c + d*x])/(2*a*d) + (b^2*S
ec[c + d*x])/(a^3*d) + (5*Sec[c + d*x]^3)/(6*a*d) + (b^2*Sec[c + d*x]^3)/(3*a^3*d) - (Csc[c + d*x]^2*Sec[c + d
*x]^3)/(2*a*d) + (b^3*Sec[c + d*x]^3*(b - a*Sin[c + d*x]))/(3*a^3*(a^2 - b^2)*d) - (b^3*Sec[c + d*x]*(3*b^3 +
a*(2*a^2 - 5*b^2)*Sin[c + d*x]))/(3*a^3*(a^2 - b^2)^2*d) - (2*b*Tan[c + d*x])/(a^2*d) - (b*Tan[c + d*x]^3)/(3*
a^2*d)

Rule 2898

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])
, x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b,
e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/2, 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]

Rule 2696

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((g*Co
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b - a*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 2866

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

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

Mathematica [B]  time = 6.23072, size = 947, normalized size = 2.85 \[ 16 \left (-\frac{\tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (b \cos \left (\frac{1}{2} (c+d x)\right )+a \sin \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2-b^2}}\right ) \csc (c+d x) (a+b \sin (c+d x)) b^7}{8 a^3 \left (a^2-b^2\right )^{5/2} d (b+a \csc (c+d x))}+\frac{\cot \left (\frac{1}{2} (c+d x)\right ) \csc (c+d x) (a+b \sin (c+d x)) b}{32 a^2 d (b+a \csc (c+d x))}-\frac{\csc (c+d x) (a+b \sin (c+d x)) \tan \left (\frac{1}{2} (c+d x)\right ) b}{32 a^2 d (b+a \csc (c+d x))}+\frac{\csc (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a+b \sin (c+d x))}{128 a d (b+a \csc (c+d x))}+\frac{\left (-5 a^2-2 b^2\right ) \csc (c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \sin (c+d x))}{32 a^3 d (b+a \csc (c+d x))}+\frac{\left (5 a^2+2 b^2\right ) \csc (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \sin (c+d x))}{32 a^3 d (b+a \csc (c+d x))}+\frac{\csc (c+d x) \sin \left (\frac{1}{2} (c+d x)\right ) (a+b \sin (c+d x))}{96 (a+b) d (b+a \csc (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{\csc (c+d x) \left (16 b \sin \left (\frac{1}{2} (c+d x)\right )-13 a \sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \sin (c+d x))}{96 (a-b)^2 d (b+a \csc (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{\csc (c+d x) \left (13 a \sin \left (\frac{1}{2} (c+d x)\right )+16 b \sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \sin (c+d x))}{96 (a+b)^2 d (b+a \csc (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{\csc ^2\left (\frac{1}{2} (c+d x)\right ) \csc (c+d x) (a+b \sin (c+d x))}{128 a d (b+a \csc (c+d x))}+\frac{a \left (13 a^2-19 b^2\right ) \csc (c+d x) (a+b \sin (c+d x))}{96 \left (a^2-b^2\right )^2 d (b+a \csc (c+d x))}+\frac{\csc (c+d x) (a+b \sin (c+d x))}{192 (a+b) d (b+a \csc (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{\csc (c+d x) (a+b \sin (c+d x))}{192 (a-b) d (b+a \csc (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{\csc (c+d x) \sin \left (\frac{1}{2} (c+d x)\right ) (a+b \sin (c+d x))}{96 (a-b) d (b+a \csc (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

16*((a*(13*a^2 - 19*b^2)*Csc[c + d*x]*(a + b*Sin[c + d*x]))/(96*(a^2 - b^2)^2*d*(b + a*Csc[c + d*x])) - (b^7*A
rcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] + a*Sin[(c + d*x)/2]))/Sqrt[a^2 - b^2]]*Csc[c + d*x]*(a + b*Sin[c
+ d*x]))/(8*a^3*(a^2 - b^2)^(5/2)*d*(b + a*Csc[c + d*x])) + (b*Cot[(c + d*x)/2]*Csc[c + d*x]*(a + b*Sin[c + d*
x]))/(32*a^2*d*(b + a*Csc[c + d*x])) - (Csc[(c + d*x)/2]^2*Csc[c + d*x]*(a + b*Sin[c + d*x]))/(128*a*d*(b + a*
Csc[c + d*x])) + ((-5*a^2 - 2*b^2)*Csc[c + d*x]*Log[Cos[(c + d*x)/2]]*(a + b*Sin[c + d*x]))/(32*a^3*d*(b + a*C
sc[c + d*x])) + ((5*a^2 + 2*b^2)*Csc[c + d*x]*Log[Sin[(c + d*x)/2]]*(a + b*Sin[c + d*x]))/(32*a^3*d*(b + a*Csc
[c + d*x])) + (Csc[c + d*x]*Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x]))/(128*a*d*(b + a*Csc[c + d*x])) + (Csc[c +
d*x]*(a + b*Sin[c + d*x]))/(192*(a + b)*d*(b + a*Csc[c + d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (Cs
c[c + d*x]*Sin[(c + d*x)/2]*(a + b*Sin[c + d*x]))/(96*(a + b)*d*(b + a*Csc[c + d*x])*(Cos[(c + d*x)/2] - Sin[(
c + d*x)/2])^3) - (Csc[c + d*x]*Sin[(c + d*x)/2]*(a + b*Sin[c + d*x]))/(96*(a - b)*d*(b + a*Csc[c + d*x])*(Cos
[(c + d*x)/2] + Sin[(c + d*x)/2])^3) + (Csc[c + d*x]*(a + b*Sin[c + d*x]))/(192*(a - b)*d*(b + a*Csc[c + d*x])
*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (Csc[c + d*x]*(-13*a*Sin[(c + d*x)/2] + 16*b*Sin[(c + d*x)/2])*(a
+ b*Sin[c + d*x]))/(96*(a - b)^2*d*(b + a*Csc[c + d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (Csc[c + d*x]
*(13*a*Sin[(c + d*x)/2] + 16*b*Sin[(c + d*x)/2])*(a + b*Sin[c + d*x]))/(96*(a + b)^2*d*(b + a*Csc[c + d*x])*(C
os[(c + d*x)/2] - Sin[(c + d*x)/2])) - (b*Csc[c + d*x]*(a + b*Sin[c + d*x])*Tan[(c + d*x)/2])/(32*a^2*d*(b + a
*Csc[c + d*x])))

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Maple [A]  time = 0.138, size = 376, normalized size = 1.1 \begin{align*}{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{b}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{5\,a}{2\,d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-3\,{\frac{b}{d \left ( a+b \right ) ^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }}-{\frac{1}{3\,d \left ( a+b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{2\,d \left ( a+b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-2\,{\frac{{b}^{7}}{d{a}^{3} \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{5\,a}{2\,d \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-3\,{\frac{b}{d \left ( a-b \right ) ^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}+{\frac{1}{3\,d \left ( a-b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{1}{2\,d \left ( a-b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{5}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{{b}^{2}}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{b}{2\,d{a}^{2}} \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)^4/(a+b*sin(d*x+c)),x)

[Out]

1/8/d/a*tan(1/2*d*x+1/2*c)^2-1/2/d/a^2*tan(1/2*d*x+1/2*c)*b-5/2/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)*a-3/d/(a+b)^2
/(tan(1/2*d*x+1/2*c)-1)*b-1/3/d/(a+b)/(tan(1/2*d*x+1/2*c)-1)^3-1/2/d/(a+b)/(tan(1/2*d*x+1/2*c)-1)^2-2/d/a^3/(a
+b)^2/(a-b)^2*b^7/(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))+5/2/d/(a-b)^2/(tan(
1/2*d*x+1/2*c)+1)*a-3/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)*b+1/3/d/(a-b)/(tan(1/2*d*x+1/2*c)+1)^3-1/2/d/(a-b)/(tan
(1/2*d*x+1/2*c)+1)^2-1/8/d/a/tan(1/2*d*x+1/2*c)^2+5/2/d/a*ln(tan(1/2*d*x+1/2*c))+1/d/a^3*ln(tan(1/2*d*x+1/2*c)
)*b^2+1/2/d/a^2*b/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)^4/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 10.2822, size = 2596, normalized size = 7.82 \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+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

[-1/12*(4*a^8 - 8*a^6*b^2 + 4*a^4*b^4 - 6*(5*a^8 - 13*a^6*b^2 + 9*a^4*b^4 - a^2*b^6)*cos(d*x + c)^4 + 4*(5*a^8
- 13*a^6*b^2 + 8*a^4*b^4)*cos(d*x + c)^2 + 6*(b^7*cos(d*x + c)^5 - b^7*cos(d*x + c)^3)*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*((5*a^8 - 13*a^6*b^2 + 9*a^4*
b^4 + a^2*b^6 - 2*b^8)*cos(d*x + c)^5 - (5*a^8 - 13*a^6*b^2 + 9*a^4*b^4 + a^2*b^6 - 2*b^8)*cos(d*x + c)^3)*log
(1/2*cos(d*x + c) + 1/2) - 3*((5*a^8 - 13*a^6*b^2 + 9*a^4*b^4 + a^2*b^6 - 2*b^8)*cos(d*x + c)^5 - (5*a^8 - 13*
a^6*b^2 + 9*a^4*b^4 + a^2*b^6 - 2*b^8)*cos(d*x + c)^3)*log(-1/2*cos(d*x + c) + 1/2) - 4*(a^7*b - 2*a^5*b^3 + a
^3*b^5 - (8*a^7*b - 22*a^5*b^3 + 17*a^3*b^5 - 3*a*b^7)*cos(d*x + c)^4 + (4*a^7*b - 11*a^5*b^3 + 7*a^3*b^5)*cos
(d*x + c)^2)*sin(d*x + c))/((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*d*cos(d*x + c)^5 - (a^9 - 3*a^7*b^2 + 3*a^
5*b^4 - a^3*b^6)*d*cos(d*x + c)^3), -1/12*(4*a^8 - 8*a^6*b^2 + 4*a^4*b^4 - 6*(5*a^8 - 13*a^6*b^2 + 9*a^4*b^4 -
a^2*b^6)*cos(d*x + c)^4 + 4*(5*a^8 - 13*a^6*b^2 + 8*a^4*b^4)*cos(d*x + c)^2 - 12*(b^7*cos(d*x + c)^5 - b^7*co
s(d*x + c)^3)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + 3*((5*a^8 - 13*a^
6*b^2 + 9*a^4*b^4 + a^2*b^6 - 2*b^8)*cos(d*x + c)^5 - (5*a^8 - 13*a^6*b^2 + 9*a^4*b^4 + a^2*b^6 - 2*b^8)*cos(d
*x + c)^3)*log(1/2*cos(d*x + c) + 1/2) - 3*((5*a^8 - 13*a^6*b^2 + 9*a^4*b^4 + a^2*b^6 - 2*b^8)*cos(d*x + c)^5
- (5*a^8 - 13*a^6*b^2 + 9*a^4*b^4 + a^2*b^6 - 2*b^8)*cos(d*x + c)^3)*log(-1/2*cos(d*x + c) + 1/2) - 4*(a^7*b -
2*a^5*b^3 + a^3*b^5 - (8*a^7*b - 22*a^5*b^3 + 17*a^3*b^5 - 3*a*b^7)*cos(d*x + c)^4 + (4*a^7*b - 11*a^5*b^3 +
7*a^3*b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*d*cos(d*x + c)^5 - (a^9 - 3*
a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*d*cos(d*x + c)^3)]

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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+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.30251, size = 563, normalized size = 1.7 \begin{align*} -\frac{\frac{48 \,{\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 )} b^{7}}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \sqrt{a^{2} - b^{2}}} - \frac{3 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a^{2}} - \frac{12 \,{\left (5 \, a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{16 \,{\left (6 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 12 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 8 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 14 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 18 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 6 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 9 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 7 \, a^{3} + 10 \, a b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{3}} + \frac{3 \,{\left (30 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2}\right )}}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}}{24 \, 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+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/24*(48*(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)))*b^7
/((a^7 - 2*a^5*b^2 + a^3*b^4)*sqrt(a^2 - b^2)) - 3*(a*tan(1/2*d*x + 1/2*c)^2 - 4*b*tan(1/2*d*x + 1/2*c))/a^2 -
12*(5*a^2 + 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 16*(6*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 9*b^3*tan(1/2*d*
x + 1/2*c)^5 - 9*a^3*tan(1/2*d*x + 1/2*c)^4 + 12*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 8*a^2*b*tan(1/2*d*x + 1/2*c)^3
+ 14*b^3*tan(1/2*d*x + 1/2*c)^3 + 12*a^3*tan(1/2*d*x + 1/2*c)^2 - 18*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 6*a^2*b*t
an(1/2*d*x + 1/2*c) - 9*b^3*tan(1/2*d*x + 1/2*c) - 7*a^3 + 10*a*b^2)/((a^4 - 2*a^2*b^2 + b^4)*(tan(1/2*d*x + 1
/2*c)^2 - 1)^3) + 3*(30*a^2*tan(1/2*d*x + 1/2*c)^2 + 12*b^2*tan(1/2*d*x + 1/2*c)^2 - 4*a*b*tan(1/2*d*x + 1/2*c
) + a^2)/(a^3*tan(1/2*d*x + 1/2*c)^2))/d