3.187 \(\int \frac{\tan ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=333 \[ \frac{2 a^5 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac{8 a^3 b^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac{a^4 b \cos (c+d x)}{d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac{(3 a+b) \cos (c+d x)}{4 d (a+b)^3 (1-\sin (c+d x))}+\frac{\cos (c+d x)}{12 d (a+b)^2 (1-\sin (c+d x))}+\frac{(3 a-b) \cos (c+d x)}{4 d (a-b)^3 (\sin (c+d x)+1)}-\frac{\cos (c+d x)}{12 d (a-b)^2 (\sin (c+d x)+1)}+\frac{\cos (c+d x)}{12 d (a+b)^2 (1-\sin (c+d x))^2}-\frac{\cos (c+d x)}{12 d (a-b)^2 (\sin (c+d x)+1)^2} \]

[Out]

(2*a^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2)*d) + (8*a^3*b^2*ArcTan[(b + a*Tan[
(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2)*d) + Cos[c + d*x]/(12*(a + b)^2*d*(1 - Sin[c + d*x])^2) + C
os[c + d*x]/(12*(a + b)^2*d*(1 - Sin[c + d*x])) - ((3*a + b)*Cos[c + d*x])/(4*(a + b)^3*d*(1 - Sin[c + d*x]))
- Cos[c + d*x]/(12*(a - b)^2*d*(1 + Sin[c + d*x])^2) - Cos[c + d*x]/(12*(a - b)^2*d*(1 + Sin[c + d*x])) + ((3*
a - b)*Cos[c + d*x])/(4*(a - b)^3*d*(1 + Sin[c + d*x])) + (a^4*b*Cos[c + d*x])/((a^2 - b^2)^3*d*(a + b*Sin[c +
d*x]))

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Rubi [A]  time = 0.626607, antiderivative size = 333, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {2731, 2650, 2648, 2664, 12, 2660, 618, 204} \[ \frac{2 a^5 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac{8 a^3 b^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac{a^4 b \cos (c+d x)}{d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac{(3 a+b) \cos (c+d x)}{4 d (a+b)^3 (1-\sin (c+d x))}+\frac{\cos (c+d x)}{12 d (a+b)^2 (1-\sin (c+d x))}+\frac{(3 a-b) \cos (c+d x)}{4 d (a-b)^3 (\sin (c+d x)+1)}-\frac{\cos (c+d x)}{12 d (a-b)^2 (\sin (c+d x)+1)}+\frac{\cos (c+d x)}{12 d (a+b)^2 (1-\sin (c+d x))^2}-\frac{\cos (c+d x)}{12 d (a-b)^2 (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/(a + b*Sin[c + d*x])^2,x]

[Out]

(2*a^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2)*d) + (8*a^3*b^2*ArcTan[(b + a*Tan[
(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2)*d) + Cos[c + d*x]/(12*(a + b)^2*d*(1 - Sin[c + d*x])^2) + C
os[c + d*x]/(12*(a + b)^2*d*(1 - Sin[c + d*x])) - ((3*a + b)*Cos[c + d*x])/(4*(a + b)^3*d*(1 - Sin[c + d*x]))
- Cos[c + d*x]/(12*(a - b)^2*d*(1 + Sin[c + d*x])^2) - Cos[c + d*x]/(12*(a - b)^2*d*(1 + Sin[c + d*x])) + ((3*
a - b)*Cos[c + d*x])/(4*(a - b)^3*d*(1 + Sin[c + d*x])) + (a^4*b*Cos[c + d*x])/((a^2 - b^2)^3*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 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))^2} \, dx &=\int \left (\frac{1}{4 (a+b)^2 (-1+\sin (c+d x))^2}+\frac{3 a+b}{4 (a+b)^3 (-1+\sin (c+d x))}+\frac{1}{4 (a-b)^2 (1+\sin (c+d x))^2}+\frac{-3 a+b}{4 (a-b)^3 (1+\sin (c+d x))}+\frac{a^4}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac{4 a^3 b^2}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))}\right ) \, dx\\ &=\frac{\int \frac{1}{(1+\sin (c+d x))^2} \, dx}{4 (a-b)^2}-\frac{(3 a-b) \int \frac{1}{1+\sin (c+d x)} \, dx}{4 (a-b)^3}+\frac{\int \frac{1}{(-1+\sin (c+d x))^2} \, dx}{4 (a+b)^2}+\frac{(3 a+b) \int \frac{1}{-1+\sin (c+d x)} \, dx}{4 (a+b)^3}+\frac{\left (4 a^3 b^2\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^3}+\frac{a^4 \int \frac{1}{(a+b \sin (c+d x))^2} \, dx}{\left (a^2-b^2\right )^2}\\ &=\frac{\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))^2}-\frac{(3 a+b) \cos (c+d x)}{4 (a+b)^3 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))^2}+\frac{(3 a-b) \cos (c+d x)}{4 (a-b)^3 d (1+\sin (c+d x))}+\frac{a^4 b \cos (c+d x)}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac{\int \frac{1}{1+\sin (c+d x)} \, dx}{12 (a-b)^2}-\frac{\int \frac{1}{-1+\sin (c+d x)} \, dx}{12 (a+b)^2}+\frac{a^4 \int \frac{a}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^3}+\frac{\left (8 a^3 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 )^3 d}\\ &=\frac{\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))^2}+\frac{\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))}-\frac{(3 a+b) \cos (c+d x)}{4 (a+b)^3 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))^2}-\frac{\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))}+\frac{(3 a-b) \cos (c+d x)}{4 (a-b)^3 d (1+\sin (c+d x))}+\frac{a^4 b \cos (c+d x)}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac{a^5 \int \frac{1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^3}-\frac{\left (16 a^3 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 )^3 d}\\ &=\frac{8 a^3 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 )^{7/2} d}+\frac{\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))^2}+\frac{\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))}-\frac{(3 a+b) \cos (c+d x)}{4 (a+b)^3 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))^2}-\frac{\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))}+\frac{(3 a-b) \cos (c+d x)}{4 (a-b)^3 d (1+\sin (c+d x))}+\frac{a^4 b \cos (c+d x)}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac{\left (2 a^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 )}{\left (a^2-b^2\right )^3 d}\\ &=\frac{8 a^3 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 )^{7/2} d}+\frac{\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))^2}+\frac{\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))}-\frac{(3 a+b) \cos (c+d x)}{4 (a+b)^3 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))^2}-\frac{\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))}+\frac{(3 a-b) \cos (c+d x)}{4 (a-b)^3 d (1+\sin (c+d x))}+\frac{a^4 b \cos (c+d x)}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac{\left (4 a^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 )}{\left (a^2-b^2\right )^3 d}\\ &=\frac{2 a^5 \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 )^{7/2} d}+\frac{8 a^3 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 )^{7/2} d}+\frac{\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))^2}+\frac{\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))}-\frac{(3 a+b) \cos (c+d x)}{4 (a+b)^3 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))^2}-\frac{\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))}+\frac{(3 a-b) \cos (c+d x)}{4 (a-b)^3 d (1+\sin (c+d x))}+\frac{a^4 b \cos (c+d x)}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.96747, size = 341, normalized size = 1.02 \[ \frac{\frac{24 a^3 \left (a^2+4 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac{12 a^4 b \cos (c+d x)}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))}-\frac{4 (4 a+b) \sin \left (\frac{1}{2} (c+d x)\right )}{(a+b)^3 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{2 \sin \left (\frac{1}{2} (c+d x)\right )}{(a+b)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{4 (b-4 a) \sin \left (\frac{1}{2} (c+d x)\right )}{(a-b)^3 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{1}{(a+b)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{1}{(a-b)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 \sin \left (\frac{1}{2} (c+d x)\right )}{(a-b)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4/(a + b*Sin[c + d*x])^2,x]

[Out]

((24*a^3*(a^2 + 4*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(7/2) + 1/((a + b)^2*(Cos
[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (2*Sin[(c + d*x)/2])/((a + b)^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3
) - (4*(4*a + b)*Sin[(c + d*x)/2])/((a + b)^3*(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])^3) - 1/((a - b)^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (4
*(-4*a + b)*Sin[(c + d*x)/2])/((a - b)^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (12*a^4*b*Cos[c + d*x])/((a
- b)^3*(a + b)^3*(a + b*Sin[c + d*x])))/(12*d)

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Maple [A]  time = 0.098, size = 382, normalized size = 1.2 \begin{align*} -{\frac{1}{3\,d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{2\,d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{a}{d \left ( a+b \right ) ^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+2\,{\frac{{a}^{3}{b}^{2}\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( a-b \right ) ^{3} \left ( a+b \right ) ^{3} \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{{a}^{4}b}{d \left ( a-b \right ) ^{3} \left ( a+b \right ) ^{3} \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{{a}^{5}}{d \left ( a-b \right ) ^{3} \left ( a+b \right ) ^{3}\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 ) }+8\,{\frac{{a}^{3}{b}^{2}}{d \left ( a-b \right ) ^{3} \left ( a+b \right ) ^{3}\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}{3\,d \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{1}{2\,d \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{a}{d \left ( a-b \right ) ^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}} \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))^2,x)

[Out]

-1/3/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)^3-1/2/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)^2+1/d*a/(a+b)^3/(tan(1/2*d*x+1/2*
c)-1)+2/d*a^3/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*b^2*tan(1/2*d*x+1/2*c)+2/d*a^4
/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*b+2/d*a^5/(a-b)^3/(a+b)^3/(a^2-b^2)^(1/2)*a
rctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+8/d*a^3/(a-b)^3/(a+b)^3/(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^2-1/3/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)^3+1/2/d/(a-b)^2/(tan(1/2*d
*x+1/2*c)+1)^2+1/d*a/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)

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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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.50435, size = 1790, normalized size = 5.38 \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))^2,x, algorithm="fricas")

[Out]

[-1/6*(2*a^6*b - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 - 2*(7*a^6*b + 2*a^4*b^3 - 10*a^2*b^5 + b^7)*cos(d*x + c)^4 - 2
*(7*a^6*b - 16*a^4*b^3 + 11*a^2*b^5 - 2*b^7)*cos(d*x + c)^2 - 3*((a^5*b + 4*a^3*b^3)*cos(d*x + c)^3*sin(d*x +
c) + (a^6 + 4*a^4*b^2)*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*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 - (4*a^7 - 7*a^5*b^2 + 2*a^3*b^4 + a*b^6)
*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7 + b^9)*d*cos(d*x + c)^3*sin(d*x + c
) + (a^9 - 4*a^7*b^2 + 6*a^5*b^4 - 4*a^3*b^6 + a*b^8)*d*cos(d*x + c)^3), -1/3*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 -
b^7 - (7*a^6*b + 2*a^4*b^3 - 10*a^2*b^5 + b^7)*cos(d*x + c)^4 - (7*a^6*b - 16*a^4*b^3 + 11*a^2*b^5 - 2*b^7)*c
os(d*x + c)^2 + 3*((a^5*b + 4*a^3*b^3)*cos(d*x + c)^3*sin(d*x + c) + (a^6 + 4*a^4*b^2)*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))) - (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 -
(4*a^7 - 7*a^5*b^2 + 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2
*b^7 + b^9)*d*cos(d*x + c)^3*sin(d*x + c) + (a^9 - 4*a^7*b^2 + 6*a^5*b^4 - 4*a^3*b^6 + a*b^8)*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 )^{2}}\, 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))**2,x)

[Out]

Integral(tan(c + d*x)**4/(a + b*sin(c + d*x))**2, x)

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Giac [A]  time = 2.7838, size = 548, normalized size = 1.65 \begin{align*} \frac{2 \,{\left (\frac{3 \,{\left (a^{5} + 4 \, a^{3} b^{2}\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^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt{a^{2} - b^{2}}} + \frac{3 \,{\left (a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{4} b\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a\right )}} + \frac{3 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 6 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 10 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 18 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 10 \, a^{3} b - 2 \, a b^{3}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )}}{3 \, d} \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))^2,x, algorithm="giac")

[Out]

2/3*(3*(a^5 + 4*a^3*b^2)*(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^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(a^2 - b^2)) + 3*(a^3*b^2*tan(1/2*d*x + 1/2*c) + a^4*b)/(
(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)) + (3*a^4*tan(1/
2*d*x + 1/2*c)^5 + 9*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*a^3*b*tan(1/2*d*x + 1/2*c)^4 - 6*a*b^3*tan(1/2*d*x + 1
/2*c)^4 - 10*a^4*tan(1/2*d*x + 1/2*c)^3 - 18*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 4*b^4*tan(1/2*d*x + 1/2*c)^3 + 2
4*a^3*b*tan(1/2*d*x + 1/2*c)^2 + 3*a^4*tan(1/2*d*x + 1/2*c) + 9*a^2*b^2*tan(1/2*d*x + 1/2*c) - 10*a^3*b - 2*a*
b^3)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(tan(1/2*d*x + 1/2*c)^2 - 1)^3))/d