3.201 \(\int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.33, 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 = {2967, 2860, 2689, 70, 69} \[ \frac {B \sec ^3(e+f x) (a \sin (e+f x)+a)^{m+2}}{a^2 c^2 f (1-m)}+\frac {2^{m+\frac {1}{2}} (A (1-m)-B (m+2)) \sec ^3(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (-\frac {3}{2},\frac {1}{2}-m;-\frac {1}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{3 a c^2 f (1-m)} \]

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])^2,x]

[Out]

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

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), 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 70

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 2689

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 2860

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] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 1, 0]

Rule 2967

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))^2} \, dx &=\frac {\int \sec ^4(e+f x) (a+a \sin (e+f x))^{2+m} (A+B \sin (e+f x)) \, dx}{a^2 c^2}\\ &=\frac {B \sec ^3(e+f x) (a+a \sin (e+f x))^{2+m}}{a^2 c^2 f (1-m)}+\frac {\left (A-\frac {B (2+m)}{1-m}\right ) \int \sec ^4(e+f x) (a+a \sin (e+f x))^{2+m} \, dx}{a^2 c^2}\\ &=\frac {B \sec ^3(e+f x) (a+a \sin (e+f x))^{2+m}}{a^2 c^2 f (1-m)}+\frac {\left (\left (A-\frac {B (2+m)}{1-m}\right ) \sec ^3(e+f x) (a-a \sin (e+f x))^{3/2} (a+a \sin (e+f x))^{3/2}\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m}}{(a-a x)^{5/2}} \, dx,x,\sin (e+f x)\right )}{c^2 f}\\ &=\frac {B \sec ^3(e+f x) (a+a \sin (e+f x))^{2+m}}{a^2 c^2 f (1-m)}+\frac {\left (2^{-\frac {1}{2}+m} \left (A-\frac {B (2+m)}{1-m}\right ) \sec ^3(e+f x) (a-a \sin (e+f x))^{3/2} (a+a \sin (e+f x))^{1+m} \left (\frac {a+a \sin (e+f x)}{a}\right )^{\frac {1}{2}-m}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {x}{2}\right )^{-\frac {1}{2}+m}}{(a-a x)^{5/2}} \, dx,x,\sin (e+f x)\right )}{c^2 f}\\ &=\frac {2^{\frac {1}{2}+m} \left (A-\frac {B (2+m)}{1-m}\right ) \, _2F_1\left (-\frac {3}{2},\frac {1}{2}-m;-\frac {1}{2};\frac {1}{2} (1-\sin (e+f x))\right ) \sec ^3(e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^{1+m}}{3 a c^2 f}+\frac {B \sec ^3(e+f x) (a+a \sin (e+f x))^{2+m}}{a^2 c^2 f (1-m)}\\ \end {align*}

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Mathematica [C]  time = 6.93, size = 9240, normalized size = 62.43 \[ \text {Result too large to show} \]

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])^2,x]

[Out]

Result too large to show

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c^{2} \sin \left (f x + e\right ) - 2 \, c^{2}}, x\right ) \]

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))^2,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}}\,{d x} \]

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))^2,x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m/(c*sin(f*x + e) - c)^2, x)

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maple [F]  time = 8.47, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )}{\left (c -c \sin \left (f x +e \right )\right )^{2}}\, dx \]

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))^2,x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}}\,{d x} \]

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))^2,x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m/(c*sin(f*x + e) - c)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]

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))^2,x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \left (a \sin {\left (e + f x \right )} + a\right )^{m}}{\sin ^{2}{\left (e + f x \right )} - 2 \sin {\left (e + f x \right )} + 1}\, dx + \int \frac {B \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}}{\sin ^{2}{\left (e + f x \right )} - 2 \sin {\left (e + f x \right )} + 1}\, dx}{c^{2}} \]

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))**2,x)

[Out]

(Integral(A*(a*sin(e + f*x) + a)**m/(sin(e + f*x)**2 - 2*sin(e + f*x) + 1), x) + Integral(B*(a*sin(e + f*x) +
a)**m*sin(e + f*x)/(sin(e + f*x)**2 - 2*sin(e + f*x) + 1), x))/c**2

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