3.1024 \(\int \sec (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\)

Optimal. Leaf size=80 \[ \frac {(A+B) (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {1}{2} (\sin (e+f x)+1)\right )}{4 a f (m+1)}+\frac {(A-B) (a \sin (e+f x)+a)^m}{2 f m} \]

[Out]

1/2*(A-B)*(a+a*sin(f*x+e))^m/f/m+1/4*(A+B)*hypergeom([1, 1+m],[2+m],1/2+1/2*sin(f*x+e))*(a+a*sin(f*x+e))^(1+m)
/a/f/(1+m)

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Rubi [A]  time = 0.11, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2836, 79, 68} \[ \frac {(A+B) (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {1}{2} (\sin (e+f x)+1)\right )}{4 a f (m+1)}+\frac {(A-B) (a \sin (e+f x)+a)^m}{2 f m} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

((A - B)*(a + a*Sin[e + f*x])^m)/(2*f*m) + ((A + B)*Hypergeometric2F1[1, 1 + m, 2 + m, (1 + Sin[e + f*x])/2]*(
a + a*Sin[e + f*x])^(1 + m))/(4*a*f*(1 + m))

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

Rule 2836

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 71, normalized size = 0.89 \[ \frac {(a (\sin (e+f x)+1))^m \left (m (A+B) (\sin (e+f x)+1) \, _2F_1\left (1,m+1;m+2;\frac {1}{2} (\sin (e+f x)+1)\right )+2 (m+1) (A-B)\right )}{4 f m (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

((a*(1 + Sin[e + f*x]))^m*(2*(A - B)*(1 + m) + (A + B)*m*Hypergeometric2F1[1, 1 + m, 2 + m, (1 + Sin[e + f*x])
/2]*(1 + Sin[e + f*x])))/(4*f*m*(1 + m))

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \sec \left (f x + e\right ) \sin \left (f x + e\right ) + A \sec \left (f x + e\right )\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="fricas")

[Out]

integral((B*sec(f*x + e)*sin(f*x + e) + A*sec(f*x + e))*(a*sin(f*x + e) + a)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*sec(f*x + e), x)

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maple [F]  time = 2.12, size = 0, normalized size = 0.00 \[ \int \sec \left (f x +e \right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)

[Out]

int(sec(f*x+e)*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*sec(f*x + e), x)

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

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)/cos(e + f*x),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sin(f*x+e))**m*(A+B*sin(f*x+e)),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**m*(A + B*sin(e + f*x))*sec(e + f*x), x)

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