3.481 \(\int \frac{A+B \sin (x)}{(1-\sin (x))^4} \, dx\)

Optimal. Leaf size=81 \[ \frac{2 (3 A-4 B) \cos (x)}{105 (1-\sin (x))}+\frac{2 (3 A-4 B) \cos (x)}{105 (1-\sin (x))^2}+\frac{(3 A-4 B) \cos (x)}{35 (1-\sin (x))^3}+\frac{(A+B) \cos (x)}{7 (1-\sin (x))^4} \]

[Out]

((A + B)*Cos[x])/(7*(1 - Sin[x])^4) + ((3*A - 4*B)*Cos[x])/(35*(1 - Sin[x])^3) + (2*(3*A - 4*B)*Cos[x])/(105*(
1 - Sin[x])^2) + (2*(3*A - 4*B)*Cos[x])/(105*(1 - Sin[x]))

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Rubi [A]  time = 0.0663277, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2750, 2650, 2648} \[ \frac{2 (3 A-4 B) \cos (x)}{105 (1-\sin (x))}+\frac{2 (3 A-4 B) \cos (x)}{105 (1-\sin (x))^2}+\frac{(3 A-4 B) \cos (x)}{35 (1-\sin (x))^3}+\frac{(A+B) \cos (x)}{7 (1-\sin (x))^4} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[x])/(1 - Sin[x])^4,x]

[Out]

((A + B)*Cos[x])/(7*(1 - Sin[x])^4) + ((3*A - 4*B)*Cos[x])/(35*(1 - Sin[x])^3) + (2*(3*A - 4*B)*Cos[x])/(105*(
1 - Sin[x])^2) + (2*(3*A - 4*B)*Cos[x])/(105*(1 - Sin[x]))

Rule 2750

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)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

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]

Rubi steps

\begin{align*} \int \frac{A+B \sin (x)}{(1-\sin (x))^4} \, dx &=\frac{(A+B) \cos (x)}{7 (1-\sin (x))^4}+\frac{1}{7} (3 A-4 B) \int \frac{1}{(1-\sin (x))^3} \, dx\\ &=\frac{(A+B) \cos (x)}{7 (1-\sin (x))^4}+\frac{(3 A-4 B) \cos (x)}{35 (1-\sin (x))^3}+\frac{1}{35} (2 (3 A-4 B)) \int \frac{1}{(1-\sin (x))^2} \, dx\\ &=\frac{(A+B) \cos (x)}{7 (1-\sin (x))^4}+\frac{(3 A-4 B) \cos (x)}{35 (1-\sin (x))^3}+\frac{2 (3 A-4 B) \cos (x)}{105 (1-\sin (x))^2}+\frac{1}{105} (2 (3 A-4 B)) \int \frac{1}{1-\sin (x)} \, dx\\ &=\frac{(A+B) \cos (x)}{7 (1-\sin (x))^4}+\frac{(3 A-4 B) \cos (x)}{35 (1-\sin (x))^3}+\frac{2 (3 A-4 B) \cos (x)}{105 (1-\sin (x))^2}+\frac{2 (3 A-4 B) \cos (x)}{105 (1-\sin (x))}\\ \end{align*}

Mathematica [A]  time = 0.0650276, size = 54, normalized size = 0.67 \[ \frac{\cos (x) \left ((8 B-6 A) \sin ^3(x)+8 (3 A-4 B) \sin ^2(x)+(52 B-39 A) \sin (x)+36 A-13 B\right )}{105 (\sin (x)-1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[x])/(1 - Sin[x])^4,x]

[Out]

(Cos[x]*(36*A - 13*B + (-39*A + 52*B)*Sin[x] + 8*(3*A - 4*B)*Sin[x]^2 + (-6*A + 8*B)*Sin[x]^3))/(105*(-1 + Sin
[x])^4)

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Maple [A]  time = 0.036, size = 115, normalized size = 1.4 \begin{align*} -{(6\,A+2\,B) \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}-{\frac{24\,A+24\,B}{3} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-6}}-{\frac{72\,A+64\,B}{5} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-5}}-{\frac{32\,A+24\,B}{2} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-4}}-{\frac{16\,A+16\,B}{7} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-7}}-{\frac{36\,A+20\,B}{3} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-3}}-2\,{\frac{A}{\tan \left ( x/2 \right ) -1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(x))/(1-sin(x))^4,x)

[Out]

-(6*A+2*B)/(tan(1/2*x)-1)^2-1/3*(24*A+24*B)/(tan(1/2*x)-1)^6-2/5*(36*A+32*B)/(tan(1/2*x)-1)^5-1/2*(32*A+24*B)/
(tan(1/2*x)-1)^4-2/7*(8*A+8*B)/(tan(1/2*x)-1)^7-2/3*(18*A+10*B)/(tan(1/2*x)-1)^3-2*A/(tan(1/2*x)-1)

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Maxima [B]  time = 1.53041, size = 417, normalized size = 5.15 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(x))/(1-sin(x))^4,x, algorithm="maxima")

[Out]

-2/105*B*(91*sin(x)/(cos(x) + 1) - 168*sin(x)^2/(cos(x) + 1)^2 + 280*sin(x)^3/(cos(x) + 1)^3 - 175*sin(x)^4/(c
os(x) + 1)^4 + 105*sin(x)^5/(cos(x) + 1)^5 - 13)/(7*sin(x)/(cos(x) + 1) - 21*sin(x)^2/(cos(x) + 1)^2 + 35*sin(
x)^3/(cos(x) + 1)^3 - 35*sin(x)^4/(cos(x) + 1)^4 + 21*sin(x)^5/(cos(x) + 1)^5 - 7*sin(x)^6/(cos(x) + 1)^6 + si
n(x)^7/(cos(x) + 1)^7 - 1) + 2/35*A*(49*sin(x)/(cos(x) + 1) - 147*sin(x)^2/(cos(x) + 1)^2 + 210*sin(x)^3/(cos(
x) + 1)^3 - 210*sin(x)^4/(cos(x) + 1)^4 + 105*sin(x)^5/(cos(x) + 1)^5 - 35*sin(x)^6/(cos(x) + 1)^6 - 12)/(7*si
n(x)/(cos(x) + 1) - 21*sin(x)^2/(cos(x) + 1)^2 + 35*sin(x)^3/(cos(x) + 1)^3 - 35*sin(x)^4/(cos(x) + 1)^4 + 21*
sin(x)^5/(cos(x) + 1)^5 - 7*sin(x)^6/(cos(x) + 1)^6 + sin(x)^7/(cos(x) + 1)^7 - 1)

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Fricas [B]  time = 1.53891, size = 429, normalized size = 5.3 \begin{align*} -\frac{2 \,{\left (3 \, A - 4 \, B\right )} \cos \left (x\right )^{4} + 8 \,{\left (3 \, A - 4 \, B\right )} \cos \left (x\right )^{3} - 9 \,{\left (3 \, A - 4 \, B\right )} \cos \left (x\right )^{2} - 15 \,{\left (4 \, A - 3 \, B\right )} \cos \left (x\right ) -{\left (2 \,{\left (3 \, A - 4 \, B\right )} \cos \left (x\right )^{3} - 6 \,{\left (3 \, A - 4 \, B\right )} \cos \left (x\right )^{2} - 15 \,{\left (3 \, A - 4 \, B\right )} \cos \left (x\right ) + 15 \, A + 15 \, B\right )} \sin \left (x\right ) - 15 \, A - 15 \, B}{105 \,{\left (\cos \left (x\right )^{4} - 3 \, \cos \left (x\right )^{3} - 8 \, \cos \left (x\right )^{2} +{\left (\cos \left (x\right )^{3} + 4 \, \cos \left (x\right )^{2} - 4 \, \cos \left (x\right ) - 8\right )} \sin \left (x\right ) + 4 \, \cos \left (x\right ) + 8\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(x))/(1-sin(x))^4,x, algorithm="fricas")

[Out]

-1/105*(2*(3*A - 4*B)*cos(x)^4 + 8*(3*A - 4*B)*cos(x)^3 - 9*(3*A - 4*B)*cos(x)^2 - 15*(4*A - 3*B)*cos(x) - (2*
(3*A - 4*B)*cos(x)^3 - 6*(3*A - 4*B)*cos(x)^2 - 15*(3*A - 4*B)*cos(x) + 15*A + 15*B)*sin(x) - 15*A - 15*B)/(co
s(x)^4 - 3*cos(x)^3 - 8*cos(x)^2 + (cos(x)^3 + 4*cos(x)^2 - 4*cos(x) - 8)*sin(x) + 4*cos(x) + 8)

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Sympy [B]  time = 19.0602, size = 818, normalized size = 10.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(x))/(1-sin(x))**4,x)

[Out]

-30*A*tan(x/2)**7/(105*tan(x/2)**7 - 735*tan(x/2)**6 + 2205*tan(x/2)**5 - 3675*tan(x/2)**4 + 3675*tan(x/2)**3
- 2205*tan(x/2)**2 + 735*tan(x/2) - 105) - 210*A*tan(x/2)**4/(105*tan(x/2)**7 - 735*tan(x/2)**6 + 2205*tan(x/2
)**5 - 3675*tan(x/2)**4 + 3675*tan(x/2)**3 - 2205*tan(x/2)**2 + 735*tan(x/2) - 105) + 210*A*tan(x/2)**3/(105*t
an(x/2)**7 - 735*tan(x/2)**6 + 2205*tan(x/2)**5 - 3675*tan(x/2)**4 + 3675*tan(x/2)**3 - 2205*tan(x/2)**2 + 735
*tan(x/2) - 105) - 252*A*tan(x/2)**2/(105*tan(x/2)**7 - 735*tan(x/2)**6 + 2205*tan(x/2)**5 - 3675*tan(x/2)**4
+ 3675*tan(x/2)**3 - 2205*tan(x/2)**2 + 735*tan(x/2) - 105) + 84*A*tan(x/2)/(105*tan(x/2)**7 - 735*tan(x/2)**6
+ 2205*tan(x/2)**5 - 3675*tan(x/2)**4 + 3675*tan(x/2)**3 - 2205*tan(x/2)**2 + 735*tan(x/2) - 105) - 42*A/(105
*tan(x/2)**7 - 735*tan(x/2)**6 + 2205*tan(x/2)**5 - 3675*tan(x/2)**4 + 3675*tan(x/2)**3 - 2205*tan(x/2)**2 + 7
35*tan(x/2) - 105) - 210*B*tan(x/2)**5/(105*tan(x/2)**7 - 735*tan(x/2)**6 + 2205*tan(x/2)**5 - 3675*tan(x/2)**
4 + 3675*tan(x/2)**3 - 2205*tan(x/2)**2 + 735*tan(x/2) - 105) + 350*B*tan(x/2)**4/(105*tan(x/2)**7 - 735*tan(x
/2)**6 + 2205*tan(x/2)**5 - 3675*tan(x/2)**4 + 3675*tan(x/2)**3 - 2205*tan(x/2)**2 + 735*tan(x/2) - 105) - 560
*B*tan(x/2)**3/(105*tan(x/2)**7 - 735*tan(x/2)**6 + 2205*tan(x/2)**5 - 3675*tan(x/2)**4 + 3675*tan(x/2)**3 - 2
205*tan(x/2)**2 + 735*tan(x/2) - 105) + 336*B*tan(x/2)**2/(105*tan(x/2)**7 - 735*tan(x/2)**6 + 2205*tan(x/2)**
5 - 3675*tan(x/2)**4 + 3675*tan(x/2)**3 - 2205*tan(x/2)**2 + 735*tan(x/2) - 105) - 182*B*tan(x/2)/(105*tan(x/2
)**7 - 735*tan(x/2)**6 + 2205*tan(x/2)**5 - 3675*tan(x/2)**4 + 3675*tan(x/2)**3 - 2205*tan(x/2)**2 + 735*tan(x
/2) - 105) + 26*B/(105*tan(x/2)**7 - 735*tan(x/2)**6 + 2205*tan(x/2)**5 - 3675*tan(x/2)**4 + 3675*tan(x/2)**3
- 2205*tan(x/2)**2 + 735*tan(x/2) - 105)

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Giac [A]  time = 1.32475, size = 151, normalized size = 1.86 \begin{align*} -\frac{2 \,{\left (105 \, A \tan \left (\frac{1}{2} \, x\right )^{6} - 315 \, A \tan \left (\frac{1}{2} \, x\right )^{5} + 105 \, B \tan \left (\frac{1}{2} \, x\right )^{5} + 630 \, A \tan \left (\frac{1}{2} \, x\right )^{4} - 175 \, B \tan \left (\frac{1}{2} \, x\right )^{4} - 630 \, A \tan \left (\frac{1}{2} \, x\right )^{3} + 280 \, B \tan \left (\frac{1}{2} \, x\right )^{3} + 441 \, A \tan \left (\frac{1}{2} \, x\right )^{2} - 168 \, B \tan \left (\frac{1}{2} \, x\right )^{2} - 147 \, A \tan \left (\frac{1}{2} \, x\right ) + 91 \, B \tan \left (\frac{1}{2} \, x\right ) + 36 \, A - 13 \, B\right )}}{105 \,{\left (\tan \left (\frac{1}{2} \, x\right ) - 1\right )}^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(x))/(1-sin(x))^4,x, algorithm="giac")

[Out]

-2/105*(105*A*tan(1/2*x)^6 - 315*A*tan(1/2*x)^5 + 105*B*tan(1/2*x)^5 + 630*A*tan(1/2*x)^4 - 175*B*tan(1/2*x)^4
- 630*A*tan(1/2*x)^3 + 280*B*tan(1/2*x)^3 + 441*A*tan(1/2*x)^2 - 168*B*tan(1/2*x)^2 - 147*A*tan(1/2*x) + 91*B
*tan(1/2*x) + 36*A - 13*B)/(tan(1/2*x) - 1)^7