3.1025 \(\int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\)

Optimal. Leaf size=100 \[ \frac {a^2 (A+B) (a \sin (e+f x)+a)^{m-1}}{2 f (a-a \sin (e+f x))}-\frac {a (A (2-m)-B m) (a \sin (e+f x)+a)^{m-1} \, _2F_1\left (1,m-1;m;\frac {1}{2} (\sin (e+f x)+1)\right )}{4 f (1-m)} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2836, 78, 68} \[ \frac {a^2 (A+B) (a \sin (e+f x)+a)^{m-1}}{2 f (a-a \sin (e+f x))}-\frac {a (A (2-m)-B m) (a \sin (e+f x)+a)^{m-1} \, _2F_1\left (1,m-1;m;\frac {1}{2} (\sin (e+f x)+1)\right )}{4 f (1-m)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 78

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)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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

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Mathematica [A]  time = 0.17, size = 82, normalized size = 0.82 \[ -\frac {a (a (\sin (e+f x)+1))^{m-1} \left ((A (m-2)+B m) (\sin (e+f x)-1) \, _2F_1\left (1,m-1;m;\frac {1}{2} (\sin (e+f x)+1)\right )+2 (m-1) (A+B)\right )}{4 f (m-1) (\sin (e+f x)-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \sec \left (f x + e\right )^{3} \sin \left (f x + e\right ) + A \sec \left (f x + e\right )^{3}\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*sec(f*x + e)^3*sin(f*x + e) + A*sec(f*x + e)^3)*(a*sin(f*x + e) + a)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3*(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)^3, x)

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maple [F]  time = 0.75, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{3}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(sec(f*x+e)^3*(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 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3*(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)^3, x)

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

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)^3,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**3*(a+a*sin(f*x+e))**m*(A+B*sin(f*x+e)),x)

[Out]

Timed out

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