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]

1/7*(A+B)*cos(x)/(1-sin(x))^4+1/35*(3*A-4*B)*cos(x)/(1-sin(x))^3+2/105*(3*A-4*B)*cos(x)/(1-sin(x))^2+2/105*(3*
A-4*B)*cos(x)/(1-sin(x))

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Rubi [A]  time = 0.07, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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 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 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)]

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*}

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Mathematica [A]  time = 0.07, 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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fricas [B]  time = 0.43, size = 150, normalized size = 1.85 \[ -\frac {2 \, {\left (3 \, A - 4 \, B\right )} \cos \relax (x)^{4} + 8 \, {\left (3 \, A - 4 \, B\right )} \cos \relax (x)^{3} - 9 \, {\left (3 \, A - 4 \, B\right )} \cos \relax (x)^{2} - 15 \, {\left (4 \, A - 3 \, B\right )} \cos \relax (x) - {\left (2 \, {\left (3 \, A - 4 \, B\right )} \cos \relax (x)^{3} - 6 \, {\left (3 \, A - 4 \, B\right )} \cos \relax (x)^{2} - 15 \, {\left (3 \, A - 4 \, B\right )} \cos \relax (x) + 15 \, A + 15 \, B\right )} \sin \relax (x) - 15 \, A - 15 \, B}{105 \, {\left (\cos \relax (x)^{4} - 3 \, \cos \relax (x)^{3} - 8 \, \cos \relax (x)^{2} + {\left (\cos \relax (x)^{3} + 4 \, \cos \relax (x)^{2} - 4 \, \cos \relax (x) - 8\right )} \sin \relax (x) + 4 \, \cos \relax (x) + 8\right )}} \]

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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giac [A]  time = 0.27, size = 112, normalized size = 1.38 \[ -\frac {2 \, {\left (105 \, A \tan \left (\frac {1}{2} \, x\right )^{6} - 315 \, A \tan \left (\frac {1}{2} \, x\right )^{5} + 105 \, B \tan \left (\frac {1}{2} \, x\right )^{5} + 630 \, A \tan \left (\frac {1}{2} \, x\right )^{4} - 175 \, B \tan \left (\frac {1}{2} \, x\right )^{4} - 630 \, A \tan \left (\frac {1}{2} \, x\right )^{3} + 280 \, B \tan \left (\frac {1}{2} \, x\right )^{3} + 441 \, A \tan \left (\frac {1}{2} \, x\right )^{2} - 168 \, B \tan \left (\frac {1}{2} \, x\right )^{2} - 147 \, A \tan \left (\frac {1}{2} \, x\right ) + 91 \, B \tan \left (\frac {1}{2} \, x\right ) + 36 \, A - 13 \, B\right )}}{105 \, {\left (\tan \left (\frac {1}{2} \, x\right ) - 1\right )}^{7}} \]

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

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maple [A]  time = 0.10, size = 115, normalized size = 1.42 \[ -\frac {2 \left (18 A +10 B \right )}{3 \left (\tan \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {24 A +24 B}{3 \left (\tan \left (\frac {x}{2}\right )-1\right )^{6}}-\frac {2 \left (8 A +8 B \right )}{7 \left (\tan \left (\frac {x}{2}\right )-1\right )^{7}}-\frac {2 A}{\tan \left (\frac {x}{2}\right )-1}-\frac {2 \left (36 A +32 B \right )}{5 \left (\tan \left (\frac {x}{2}\right )-1\right )^{5}}-\frac {6 A +2 B}{\left (\tan \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {32 A +24 B}{2 \left (\tan \left (\frac {x}{2}\right )-1\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.33, size = 309, normalized size = 3.81 \[ -\frac {2 \, B {\left (\frac {91 \, \sin \relax (x)}{\cos \relax (x) + 1} - \frac {168 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {280 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} - \frac {175 \, \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {105 \, \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} - 13\right )}}{105 \, {\left (\frac {7 \, \sin \relax (x)}{\cos \relax (x) + 1} - \frac {21 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {35 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} - \frac {35 \, \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {21 \, \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} - \frac {7 \, \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} + \frac {\sin \relax (x)^{7}}{{\left (\cos \relax (x) + 1\right )}^{7}} - 1\right )}} + \frac {2 \, A {\left (\frac {49 \, \sin \relax (x)}{\cos \relax (x) + 1} - \frac {147 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {210 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} - \frac {210 \, \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {105 \, \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} - \frac {35 \, \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} - 12\right )}}{35 \, {\left (\frac {7 \, \sin \relax (x)}{\cos \relax (x) + 1} - \frac {21 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {35 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} - \frac {35 \, \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {21 \, \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} - \frac {7 \, \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} + \frac {\sin \relax (x)^{7}}{{\left (\cos \relax (x) + 1\right )}^{7}} - 1\right )}} \]

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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mupad [B]  time = 7.06, size = 97, normalized size = 1.20 \[ -\frac {2\,A\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6+\left (2\,B-6\,A\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5+\left (12\,A-\frac {10\,B}{3}\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+\left (\frac {16\,B}{3}-12\,A\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+\left (\frac {42\,A}{5}-\frac {16\,B}{5}\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+\left (\frac {26\,B}{15}-\frac {14\,A}{5}\right )\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {24\,A}{35}-\frac {26\,B}{105}}{{\left (\mathrm {tan}\left (\frac {x}{2}\right )-1\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sin(x))/(sin(x) - 1)^4,x)

[Out]

-((24*A)/35 - (26*B)/105 + 2*A*tan(x/2)^6 - tan(x/2)*((14*A)/5 - (26*B)/15) - tan(x/2)^5*(6*A - 2*B) + tan(x/2
)^4*(12*A - (10*B)/3) - tan(x/2)^3*(12*A - (16*B)/3) + tan(x/2)^2*((42*A)/5 - (16*B)/5))/(tan(x/2) - 1)^7

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sympy [B]  time = 7.87, size = 887, normalized size = 10.95 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-210*A*tan(x/2)**6/(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) + 630*A*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) - 1260*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 + 7
35*tan(x/2) - 105) + 1260*A*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 - 2205*tan(x/2)**2 + 735*tan(x/2) - 105) - 882*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) + 29
4*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 - 220
5*tan(x/2)**2 + 735*tan(x/2) - 105) - 72*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 + 735*tan(x/2) - 105) - 210*B*tan(x/2)**5/(105*tan(x/2)**7 - 735*t
an(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 - 2205*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 + 73
5*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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