3.1353 \(\int \frac{\sec (c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=142 \[ -\frac{2 a^3 b \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}-\frac{b \tan ^3(c+d x)}{3 d \left (a^2-b^2\right )}+\frac{a \sec ^3(c+d x)}{3 d \left (a^2-b^2\right )}-\frac{a^2 \sec (c+d x) (a-b \sin (c+d x))}{d \left (a^2-b^2\right )^2} \]

[Out]

(-2*a^3*b*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(5/2)*d) + (a*Sec[c + d*x]^3)/(3*(a^2
- b^2)*d) - (a^2*Sec[c + d*x]*(a - b*Sin[c + d*x]))/((a^2 - b^2)^2*d) - (b*Tan[c + d*x]^3)/(3*(a^2 - b^2)*d)

________________________________________________________________________________________

Rubi [A]  time = 0.238903, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2902, 2606, 30, 2607, 2866, 12, 2660, 618, 204} \[ -\frac{2 a^3 b \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}-\frac{b \tan ^3(c+d x)}{3 d \left (a^2-b^2\right )}+\frac{a \sec ^3(c+d x)}{3 d \left (a^2-b^2\right )}-\frac{a^2 \sec (c+d x) (a-b \sin (c+d x))}{d \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^3*b*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(5/2)*d) + (a*Sec[c + d*x]^3)/(3*(a^2
- b^2)*d) - (a^2*Sec[c + d*x]*(a - b*Sin[c + d*x]))/((a^2 - b^2)^2*d) - (b*Tan[c + d*x]^3)/(3*(a^2 - b^2)*d)

Rule 2902

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[(a*d^2)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-D
ist[(b*d)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[(a^2*d^2)/(g^2*(a^2 - b^
2)), Int[((g*Cos[e + f*x])^(p + 2)*(d*Sin[e + f*x])^(n - 2))/(a + b*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, d,
e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1]

Rule 2606

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2607

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

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

Mathematica [A]  time = 1.41831, size = 184, normalized size = 1.3 \[ \frac{\frac{\sec ^3(c+d x) \left (3 a \left (5 a^2+b^2\right ) \cos (c+d x)+8 a^2 b \sin (3 (c+d x))-12 a^3 \cos (2 (c+d x))+5 a^3 \cos (3 (c+d x))-4 a^3+a b^2 \cos (3 (c+d x))-8 a b^2+6 b^3 \sin (c+d x)-2 b^3 \sin (3 (c+d x))\right )}{(a-b)^2 (a+b)^2}-\frac{48 a^3 b \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-48*a^3*b*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (Sec[c + d*x]^3*(-4*a^3 - 8*
a*b^2 + 3*a*(5*a^2 + b^2)*Cos[c + d*x] - 12*a^3*Cos[2*(c + d*x)] + 5*a^3*Cos[3*(c + d*x)] + a*b^2*Cos[3*(c + d
*x)] + 6*b^3*Sin[c + d*x] + 8*a^2*b*Sin[3*(c + d*x)] - 2*b^3*Sin[3*(c + d*x)]))/((a - b)^2*(a + b)^2))/(24*d)

________________________________________________________________________________________

Maple [A]  time = 0.105, size = 222, normalized size = 1.6 \begin{align*} -{\frac{16}{3\,d \left ( 16\,a+16\,b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-8\,{\frac{1}{d \left ( 16\,a+16\,b \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{2}}}+{\frac{a}{2\,d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-2\,{\frac{{a}^{3}b}{d \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 ) }-8\,{\frac{1}{d \left ( 16\,a-16\,b \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}+{\frac{16}{3\,d \left ( 16\,a-16\,b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{a}{2\,d \left ( a-b \right ) ^{2}} \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(sec(d*x+c)^4*sin(d*x+c)^3/(a+b*sin(d*x+c)),x)

[Out]

-16/3/d/(tan(1/2*d*x+1/2*c)-1)^3/(16*a+16*b)-8/d/(16*a+16*b)/(tan(1/2*d*x+1/2*c)-1)^2+1/2/d/(a+b)^2/(tan(1/2*d
*x+1/2*c)-1)*a-2/d*a^3*b/(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))-8/d/(16*a-16*b)/(tan(1/2*d*x+1/2*c)+1)^2+16/3/d/(tan(1/2*d*x+1/2*c)+1)^3/(16*a-16*b)-1/2/d/(a-b)^2/(tan(1/
2*d*x+1/2*c)+1)*a

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.63637, size = 1027, normalized size = 7.23 \begin{align*} \left [-\frac{3 \, \sqrt{-a^{2} + b^{2}} a^{3} b \cos \left (d x + c\right )^{3} \log \left (-\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \,{\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, a^{5} + 4 \, a^{3} b^{2} - 2 \, a b^{4} + 6 \,{\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5} -{\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}, \frac{3 \, \sqrt{a^{2} - b^{2}} a^{3} b \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} + a^{5} - 2 \, a^{3} b^{2} + a b^{4} - 3 \,{\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} -{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5} -{\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(3*sqrt(-a^2 + b^2)*a^3*b*cos(d*x + c)^3*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^5 + 4*a^3*b^2 - 2*a*b^4 + 6*(a^5 - a^3*b^2)*cos(d*x + c)^2 + 2*(a^4*b - 2*a^2*b^3 + b
^5 - (4*a^4*b - 5*a^2*b^3 + b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x
+ c)^3), 1/3*(3*sqrt(a^2 - b^2)*a^3*b*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*cos(d*x + c
)^3 + a^5 - 2*a^3*b^2 + a*b^4 - 3*(a^5 - a^3*b^2)*cos(d*x + c)^2 - (a^4*b - 2*a^2*b^3 + b^5 - (4*a^4*b - 5*a^2
*b^3 + b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x + c)^3)]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.22595, size = 306, normalized size = 2.15 \begin{align*} -\frac{2 \,{\left (\frac{3 \,{\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 )} a^{3} b}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} - b^{2}}} + \frac{3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 10 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a^{3} - a b^{2}}{{\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}}\right )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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