3.198 \(\int \frac{\tan ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=474 \[ \frac{a^4 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac{8 a^4 b^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac{12 a^2 b^2 \left (a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac{3 a^5 b \cos (c+d x)}{2 d \left (a^2-b^2\right )^4 (a+b \sin (c+d x))}+\frac{a^4 b \cos (c+d x)}{2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))^2}+\frac{4 a^3 b^3 \cos (c+d x)}{d \left (a^2-b^2\right )^4 (a+b \sin (c+d x))}-\frac{3 a \cos (c+d x)}{4 d (a+b)^4 (1-\sin (c+d x))}+\frac{3 a \cos (c+d x)}{4 d (a-b)^4 (\sin (c+d x)+1)}+\frac{\cos (c+d x)}{12 d (a+b)^3 (1-\sin (c+d x))}-\frac{\cos (c+d x)}{12 d (a-b)^3 (\sin (c+d x)+1)}+\frac{\cos (c+d x)}{12 d (a+b)^3 (1-\sin (c+d x))^2}-\frac{\cos (c+d x)}{12 d (a-b)^3 (\sin (c+d x)+1)^2} \]

[Out]

(8*a^4*b^2*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(9/2)*d) + (12*a^2*b^2*(a^2 + b^2)*A
rcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(9/2)*d) + (a^4*(2*a^2 + b^2)*ArcTan[(b + a*Tan[
(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(9/2)*d) + Cos[c + d*x]/(12*(a + b)^3*d*(1 - Sin[c + d*x])^2) - (
3*a*Cos[c + d*x])/(4*(a + b)^4*d*(1 - Sin[c + d*x])) + Cos[c + d*x]/(12*(a + b)^3*d*(1 - Sin[c + d*x])) - Cos[
c + d*x]/(12*(a - b)^3*d*(1 + Sin[c + d*x])^2) + (3*a*Cos[c + d*x])/(4*(a - b)^4*d*(1 + Sin[c + d*x])) - Cos[c
+ d*x]/(12*(a - b)^3*d*(1 + Sin[c + d*x])) + (a^4*b*Cos[c + d*x])/(2*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x])^2)
+ (3*a^5*b*Cos[c + d*x])/(2*(a^2 - b^2)^4*d*(a + b*Sin[c + d*x])) + (4*a^3*b^3*Cos[c + d*x])/((a^2 - b^2)^4*d*
(a + b*Sin[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.871767, antiderivative size = 474, normalized size of antiderivative = 1., number of steps used = 22, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {2731, 2650, 2648, 2664, 2754, 12, 2660, 618, 204} \[ \frac{a^4 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac{8 a^4 b^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac{12 a^2 b^2 \left (a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac{3 a^5 b \cos (c+d x)}{2 d \left (a^2-b^2\right )^4 (a+b \sin (c+d x))}+\frac{a^4 b \cos (c+d x)}{2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))^2}+\frac{4 a^3 b^3 \cos (c+d x)}{d \left (a^2-b^2\right )^4 (a+b \sin (c+d x))}-\frac{3 a \cos (c+d x)}{4 d (a+b)^4 (1-\sin (c+d x))}+\frac{3 a \cos (c+d x)}{4 d (a-b)^4 (\sin (c+d x)+1)}+\frac{\cos (c+d x)}{12 d (a+b)^3 (1-\sin (c+d x))}-\frac{\cos (c+d x)}{12 d (a-b)^3 (\sin (c+d x)+1)}+\frac{\cos (c+d x)}{12 d (a+b)^3 (1-\sin (c+d x))^2}-\frac{\cos (c+d x)}{12 d (a-b)^3 (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^4*b^2*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(9/2)*d) + (12*a^2*b^2*(a^2 + b^2)*A
rcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(9/2)*d) + (a^4*(2*a^2 + b^2)*ArcTan[(b + a*Tan[
(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(9/2)*d) + Cos[c + d*x]/(12*(a + b)^3*d*(1 - Sin[c + d*x])^2) - (
3*a*Cos[c + d*x])/(4*(a + b)^4*d*(1 - Sin[c + d*x])) + Cos[c + d*x]/(12*(a + b)^3*d*(1 - Sin[c + d*x])) - Cos[
c + d*x]/(12*(a - b)^3*d*(1 + Sin[c + d*x])^2) + (3*a*Cos[c + d*x])/(4*(a - b)^4*d*(1 + Sin[c + d*x])) - Cos[c
+ d*x]/(12*(a - b)^3*d*(1 + Sin[c + d*x])) + (a^4*b*Cos[c + d*x])/(2*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x])^2)
+ (3*a^5*b*Cos[c + d*x])/(2*(a^2 - b^2)^4*d*(a + b*Sin[c + d*x])) + (4*a^3*b^3*Cos[c + d*x])/((a^2 - b^2)^4*d*
(a + b*Sin[c + d*x]))

Rule 2731

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Int[ExpandIntegrand
[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^m)/(1 - Sin[e + f*x]^2)^(p/2), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a
^2 - b^2, 0] && IntegersQ[m, p/2]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[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 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -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{\tan ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\int \left (\frac{1}{4 (a+b)^3 (-1+\sin (c+d x))^2}+\frac{3 a}{4 (a+b)^4 (-1+\sin (c+d x))}+\frac{1}{4 (a-b)^3 (1+\sin (c+d x))^2}-\frac{3 a}{4 (a-b)^4 (1+\sin (c+d x))}+\frac{a^4}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^3}+\frac{4 a^3 b^2}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))^2}+\frac{6 a^2 b^2 \left (a^2+b^2\right )}{\left (a^2-b^2\right )^4 (a+b \sin (c+d x))}\right ) \, dx\\ &=-\frac{(3 a) \int \frac{1}{1+\sin (c+d x)} \, dx}{4 (a-b)^4}+\frac{\int \frac{1}{(1+\sin (c+d x))^2} \, dx}{4 (a-b)^3}+\frac{(3 a) \int \frac{1}{-1+\sin (c+d x)} \, dx}{4 (a+b)^4}+\frac{\int \frac{1}{(-1+\sin (c+d x))^2} \, dx}{4 (a+b)^3}+\frac{\left (4 a^3 b^2\right ) \int \frac{1}{(a+b \sin (c+d x))^2} \, dx}{\left (a^2-b^2\right )^3}+\frac{a^4 \int \frac{1}{(a+b \sin (c+d x))^3} \, dx}{\left (a^2-b^2\right )^2}+\frac{\left (6 a^2 b^2 \left (a^2+b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^4}\\ &=\frac{\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac{3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac{3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}+\frac{a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac{4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac{\int \frac{1}{1+\sin (c+d x)} \, dx}{12 (a-b)^3}-\frac{\int \frac{1}{-1+\sin (c+d x)} \, dx}{12 (a+b)^3}+\frac{\left (4 a^3 b^2\right ) \int \frac{a}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^4}-\frac{a^4 \int \frac{-2 a+b \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )^3}+\frac{\left (12 a^2 b^2 \left (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 )}{\left (a^2-b^2\right )^4 d}\\ &=\frac{\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac{3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}+\frac{\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac{3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}-\frac{\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))}+\frac{a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac{3 a^5 b \cos (c+d x)}{2 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac{4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac{a^4 \int \frac{2 a^2+b^2}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^4}+\frac{\left (4 a^4 b^2\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^4}-\frac{\left (24 a^2 b^2 \left (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 )}{\left (a^2-b^2\right )^4 d}\\ &=\frac{12 a^2 b^2 \left (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 )}{\left (a^2-b^2\right )^{9/2} d}+\frac{\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac{3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}+\frac{\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac{3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}-\frac{\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))}+\frac{a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac{3 a^5 b \cos (c+d x)}{2 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac{4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac{\left (a^4 \left (2 a^2+b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^4}+\frac{\left (8 a^4 b^2\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 )}{\left (a^2-b^2\right )^4 d}\\ &=\frac{12 a^2 b^2 \left (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 )}{\left (a^2-b^2\right )^{9/2} d}+\frac{\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac{3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}+\frac{\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac{3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}-\frac{\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))}+\frac{a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac{3 a^5 b \cos (c+d x)}{2 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac{4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}-\frac{\left (16 a^4 b^2\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 )}{\left (a^2-b^2\right )^4 d}+\frac{\left (a^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 )}{\left (a^2-b^2\right )^4 d}\\ &=\frac{8 a^4 b^2 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2} d}+\frac{12 a^2 b^2 \left (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 )}{\left (a^2-b^2\right )^{9/2} d}+\frac{\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac{3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}+\frac{\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac{3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}-\frac{\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))}+\frac{a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac{3 a^5 b \cos (c+d x)}{2 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac{4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}-\frac{\left (2 a^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 )}{\left (a^2-b^2\right )^4 d}\\ &=\frac{8 a^4 b^2 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2} d}+\frac{12 a^2 b^2 \left (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 )}{\left (a^2-b^2\right )^{9/2} d}+\frac{a^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 )}{\left (a^2-b^2\right )^{9/2} d}+\frac{\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))^2}-\frac{3 a \cos (c+d x)}{4 (a+b)^4 d (1-\sin (c+d x))}+\frac{\cos (c+d x)}{12 (a+b)^3 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))^2}+\frac{3 a \cos (c+d x)}{4 (a-b)^4 d (1+\sin (c+d x))}-\frac{\cos (c+d x)}{12 (a-b)^3 d (1+\sin (c+d x))}+\frac{a^4 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}+\frac{3 a^5 b \cos (c+d x)}{2 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac{4 a^3 b^3 \cos (c+d x)}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.043, size = 351, normalized size = 0.74 \[ \frac{\frac{96 a^2 \left (21 a^2 b^2+2 a^4+12 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2}}-\frac{\sec ^3(c+d x) \left (22 a^5 b^2 \sin (c+d x)-91 a^5 b^2 \sin (3 (c+d x))-17 a^5 b^2 \sin (5 (c+d x))-264 a^3 b^4 \sin (c+d x)-244 a^3 b^4 \sin (3 (c+d x))-76 a^3 b^4 \sin (5 (c+d x))-8 \left (8 a^2 b^5+55 a^4 b^3+44 a^6 b-2 b^7\right ) \cos (2 (c+d x))-2 \left (89 a^4 b^3-12 a^2 b^5+28 a^6 b\right ) \cos (4 (c+d x))-358 a^4 b^3+8 a^2 b^5-264 a^6 b+32 a^7 \sin (3 (c+d x))+32 a b^6 \sin (c+d x)-12 a b^6 \sin (3 (c+d x))-12 a b^6 \sin (5 (c+d x))-16 b^7\right )}{\left (a^2-b^2\right )^4 (a+b \sin (c+d x))^2}}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((96*a^2*(2*a^4 + 21*a^2*b^2 + 12*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(9/2) - (
Sec[c + d*x]^3*(-264*a^6*b - 358*a^4*b^3 + 8*a^2*b^5 - 16*b^7 - 8*(44*a^6*b + 55*a^4*b^3 + 8*a^2*b^5 - 2*b^7)*
Cos[2*(c + d*x)] - 2*(28*a^6*b + 89*a^4*b^3 - 12*a^2*b^5)*Cos[4*(c + d*x)] + 22*a^5*b^2*Sin[c + d*x] - 264*a^3
*b^4*Sin[c + d*x] + 32*a*b^6*Sin[c + d*x] + 32*a^7*Sin[3*(c + d*x)] - 91*a^5*b^2*Sin[3*(c + d*x)] - 244*a^3*b^
4*Sin[3*(c + d*x)] - 12*a*b^6*Sin[3*(c + d*x)] - 17*a^5*b^2*Sin[5*(c + d*x)] - 76*a^3*b^4*Sin[5*(c + d*x)] - 1
2*a*b^6*Sin[5*(c + d*x)]))/((a^2 - b^2)^4*(a + b*Sin[c + d*x])^2))/(96*d)

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Maple [B]  time = 0.131, size = 922, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4/(a+b*sin(d*x+c))^3,x)

[Out]

-1/3/d/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)^3-1/2/d/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)^2+1/d/(a+b)^4/(tan(1/2*d*x+1/2*c)
-1)*a-1/2/d/(a+b)^4/(tan(1/2*d*x+1/2*c)-1)*b+5/d*a^5/(a-b)^4/(a+b)^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2
*c)*b+a)^2*tan(1/2*d*x+1/2*c)^3*b^2+6/d*a^3/(a-b)^4/(a+b)^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^
2*tan(1/2*d*x+1/2*c)^3*b^4+4/d*a^6/(a-b)^4/(a+b)^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2
*d*x+1/2*c)^2*b+15/d*a^4/(a-b)^4/(a+b)^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c
)^2*b^3+14/d*b^5/(a-b)^4/(a+b)^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*a^2*tan(1/2*d*x+1/2*c)^2+
11/d*a^5/(a-b)^4/(a+b)^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)*b^2+22/d*a^3/(
a-b)^4/(a+b)^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)*b^4+4/d*a^6/(a-b)^4/(a+b
)^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*b+7/d*a^4/(a-b)^4/(a+b)^4/(tan(1/2*d*x+1/2*c)^2*a+2*ta
n(1/2*d*x+1/2*c)*b+a)^2*b^3+2/d*a^6/(a-b)^4/(a+b)^4/(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))+21/d*a^4/(a-b)^4/(a+b)^4/(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
))*b^2+12/d*b^4/(a-b)^4/(a+b)^4/(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))*a^2-1
/3/d/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)^3+1/2/d/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)^2+1/d/(a-b)^4/(tan(1/2*d*x+1/2*c)+1
)*a+1/2/d/(a-b)^4/(tan(1/2*d*x+1/2*c)+1)*b

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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(tan(d*x+c)^4/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.85301, size = 2765, normalized size = 5.83 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/12*(4*a^8*b - 16*a^6*b^3 + 24*a^4*b^5 - 16*a^2*b^7 + 4*b^9 - 2*(28*a^8*b + 61*a^6*b^3 - 101*a^4*b^5 + 12*a^
2*b^7)*cos(d*x + c)^4 - 4*(8*a^8*b - 25*a^6*b^3 + 27*a^4*b^5 - 11*a^2*b^7 + b^9)*cos(d*x + c)^2 - 3*((2*a^6*b^
2 + 21*a^4*b^4 + 12*a^2*b^6)*cos(d*x + c)^5 - 2*(2*a^7*b + 21*a^5*b^3 + 12*a^3*b^5)*cos(d*x + c)^3*sin(d*x + c
) - (2*a^8 + 23*a^6*b^2 + 33*a^4*b^4 + 12*a^2*b^6)*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)) - 2*(2*a^9 - 8*a^7*b^2 + 12*a^5*b^4 - 8*a^3*b^6 + 2*a*b
^8 + (17*a^7*b^2 + 59*a^5*b^4 - 64*a^3*b^6 - 12*a*b^8)*cos(d*x + c)^4 - 2*(4*a^9 - 9*a^7*b^2 + 3*a^5*b^4 + 5*a
^3*b^6 - 3*a*b^8)*cos(d*x + c)^2)*sin(d*x + c))/((a^10*b^2 - 5*a^8*b^4 + 10*a^6*b^6 - 10*a^4*b^8 + 5*a^2*b^10
- b^12)*d*cos(d*x + c)^5 - 2*(a^11*b - 5*a^9*b^3 + 10*a^7*b^5 - 10*a^5*b^7 + 5*a^3*b^9 - a*b^11)*d*cos(d*x + c
)^3*sin(d*x + c) - (a^12 - 4*a^10*b^2 + 5*a^8*b^4 - 5*a^4*b^8 + 4*a^2*b^10 - b^12)*d*cos(d*x + c)^3), 1/6*(2*a
^8*b - 8*a^6*b^3 + 12*a^4*b^5 - 8*a^2*b^7 + 2*b^9 - (28*a^8*b + 61*a^6*b^3 - 101*a^4*b^5 + 12*a^2*b^7)*cos(d*x
+ c)^4 - 2*(8*a^8*b - 25*a^6*b^3 + 27*a^4*b^5 - 11*a^2*b^7 + b^9)*cos(d*x + c)^2 - 3*((2*a^6*b^2 + 21*a^4*b^4
+ 12*a^2*b^6)*cos(d*x + c)^5 - 2*(2*a^7*b + 21*a^5*b^3 + 12*a^3*b^5)*cos(d*x + c)^3*sin(d*x + c) - (2*a^8 + 2
3*a^6*b^2 + 33*a^4*b^4 + 12*a^2*b^6)*cos(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))) - (2*a^9 - 8*a^7*b^2 + 12*a^5*b^4 - 8*a^3*b^6 + 2*a*b^8 + (17*a^7*b^2 + 59*a^5*b^4 - 64*a^
3*b^6 - 12*a*b^8)*cos(d*x + c)^4 - 2*(4*a^9 - 9*a^7*b^2 + 3*a^5*b^4 + 5*a^3*b^6 - 3*a*b^8)*cos(d*x + c)^2)*sin
(d*x + c))/((a^10*b^2 - 5*a^8*b^4 + 10*a^6*b^6 - 10*a^4*b^8 + 5*a^2*b^10 - b^12)*d*cos(d*x + c)^5 - 2*(a^11*b
- 5*a^9*b^3 + 10*a^7*b^5 - 10*a^5*b^7 + 5*a^3*b^9 - a*b^11)*d*cos(d*x + c)^3*sin(d*x + c) - (a^12 - 4*a^10*b^2
+ 5*a^8*b^4 - 5*a^4*b^8 + 4*a^2*b^10 - b^12)*d*cos(d*x + c)^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{4}{\left (c + d x \right )}}{\left (a + b \sin{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**4/(a+b*sin(d*x+c))**3,x)

[Out]

Integral(tan(c + d*x)**4/(a + b*sin(c + d*x))**3, x)

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Giac [A]  time = 3.39856, size = 853, normalized size = 1.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/3*(3*(2*a^6 + 21*a^4*b^2 + 12*a^2*b^4)*(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 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*sqrt(a^2 - b^2)) + 3*(5*a^5*b^2*t
an(1/2*d*x + 1/2*c)^3 + 6*a^3*b^4*tan(1/2*d*x + 1/2*c)^3 + 4*a^6*b*tan(1/2*d*x + 1/2*c)^2 + 15*a^4*b^3*tan(1/2
*d*x + 1/2*c)^2 + 14*a^2*b^5*tan(1/2*d*x + 1/2*c)^2 + 11*a^5*b^2*tan(1/2*d*x + 1/2*c) + 22*a^3*b^4*tan(1/2*d*x
+ 1/2*c) + 4*a^6*b + 7*a^4*b^3)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(a*tan(1/2*d*x + 1/2*c)^2 +
2*b*tan(1/2*d*x + 1/2*c) + a)^2) + 2*(3*a^5*tan(1/2*d*x + 1/2*c)^5 + 24*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 9*a*b
^4*tan(1/2*d*x + 1/2*c)^5 - 9*a^4*b*tan(1/2*d*x + 1/2*c)^4 - 24*a^2*b^3*tan(1/2*d*x + 1/2*c)^4 - 3*b^5*tan(1/2
*d*x + 1/2*c)^4 - 10*a^5*tan(1/2*d*x + 1/2*c)^3 - 56*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 6*a*b^4*tan(1/2*d*x + 1/
2*c)^3 + 36*a^4*b*tan(1/2*d*x + 1/2*c)^2 + 36*a^2*b^3*tan(1/2*d*x + 1/2*c)^2 + 3*a^5*tan(1/2*d*x + 1/2*c) + 24
*a^3*b^2*tan(1/2*d*x + 1/2*c) + 9*a*b^4*tan(1/2*d*x + 1/2*c) - 15*a^4*b - 20*a^2*b^3 - b^5)/((a^8 - 4*a^6*b^2
+ 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(tan(1/2*d*x + 1/2*c)^2 - 1)^3))/d