3.3.2 \(\int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx\) [202]

Optimal. Leaf size=148 \[ \frac {2^{\frac {1}{2}+m} (A (2-m)-B (3+m)) \, _2F_1\left (-\frac {5}{2},\frac {1}{2}-m;-\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right ) \sec ^5(e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^{2+m}}{5 a^2 c^3 f (2-m)}+\frac {B \sec ^5(e+f x) (a+a \sin (e+f x))^{3+m}}{a^3 c^3 f (2-m)} \]

[Out]

1/5*2^(1/2+m)*(A*(2-m)-B*(3+m))*hypergeom([-5/2, 1/2-m],[-3/2],1/2-1/2*sin(f*x+e))*sec(f*x+e)^5*(1+sin(f*x+e))
^(1/2-m)*(a+a*sin(f*x+e))^(2+m)/a^2/c^3/f/(2-m)+B*sec(f*x+e)^5*(a+a*sin(f*x+e))^(3+m)/a^3/c^3/f/(2-m)

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Rubi [A]
time = 0.22, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3046, 2939, 2768, 72, 71} \begin {gather*} \frac {B \sec ^5(e+f x) (a \sin (e+f x)+a)^{m+3}}{a^3 c^3 f (2-m)}+\frac {2^{m+\frac {1}{2}} (A (2-m)-B (m+3)) \sec ^5(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m+2} \, _2F_1\left (-\frac {5}{2},\frac {1}{2}-m;-\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{5 a^2 c^3 f (2-m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^3,x]

[Out]

(2^(1/2 + m)*(A*(2 - m) - B*(3 + m))*Hypergeometric2F1[-5/2, 1/2 - m, -3/2, (1 - Sin[e + f*x])/2]*Sec[e + f*x]
^5*(1 + Sin[e + f*x])^(1/2 - m)*(a + a*Sin[e + f*x])^(2 + m))/(5*a^2*c^3*f*(2 - m)) + (B*Sec[e + f*x]^5*(a + a
*Sin[e + f*x])^(3 + m))/(a^3*c^3*f*(2 - m))

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 2768

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[a^2*(
(g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2))), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rule 2939

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x
] + Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; F
reeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 1, 0]

Rule 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx &=\frac {\int \sec ^6(e+f x) (a+a \sin (e+f x))^{3+m} (A+B \sin (e+f x)) \, dx}{a^3 c^3}\\ &=\frac {B \sec ^5(e+f x) (a+a \sin (e+f x))^{3+m}}{a^3 c^3 f (2-m)}+\frac {\left (A-\frac {B (3+m)}{2-m}\right ) \int \sec ^6(e+f x) (a+a \sin (e+f x))^{3+m} \, dx}{a^3 c^3}\\ &=\frac {B \sec ^5(e+f x) (a+a \sin (e+f x))^{3+m}}{a^3 c^3 f (2-m)}+\frac {\left (\left (A-\frac {B (3+m)}{2-m}\right ) \sec ^5(e+f x) (a-a \sin (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}\right ) \text {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m}}{(a-a x)^{7/2}} \, dx,x,\sin (e+f x)\right )}{a c^3 f}\\ &=\frac {B \sec ^5(e+f x) (a+a \sin (e+f x))^{3+m}}{a^3 c^3 f (2-m)}+\frac {\left (2^{-\frac {1}{2}+m} \left (A-\frac {B (3+m)}{2-m}\right ) \sec ^5(e+f x) (a-a \sin (e+f x))^{5/2} (a+a \sin (e+f x))^{2+m} \left (\frac {a+a \sin (e+f x)}{a}\right )^{\frac {1}{2}-m}\right ) \text {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {x}{2}\right )^{-\frac {1}{2}+m}}{(a-a x)^{7/2}} \, dx,x,\sin (e+f x)\right )}{a c^3 f}\\ &=\frac {2^{\frac {1}{2}+m} \left (A-\frac {B (3+m)}{2-m}\right ) \, _2F_1\left (-\frac {5}{2},\frac {1}{2}-m;-\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right ) \sec ^5(e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^{2+m}}{5 a^2 c^3 f}+\frac {B \sec ^5(e+f x) (a+a \sin (e+f x))^{3+m}}{a^3 c^3 f (2-m)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 6.82, size = 12302, normalized size = 83.12 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^3,x]

[Out]

Result too large to show

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Maple [F]
time = 2.00, size = 0, normalized size = 0.00 \[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )}{\left (c -c \sin \left (f x +e \right )\right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

-integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m/(c*sin(f*x + e) - c)^3, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

integral(-(B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m/(3*c^3*cos(f*x + e)^2 - 4*c^3 - (c^3*cos(f*x + e)^2 - 4*
c^3)*sin(f*x + e)), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algorithm="giac")

[Out]

integrate(-(B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m/(c*sin(f*x + e) - c)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^3,x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^3, x)

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