3.1032 \(\int \sec ^6(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\)

Optimal. Leaf size=135 \[ \frac {2^{m-\frac {5}{2}} (A (5-m)-B m) \sec ^5(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m+2} \, _2F_1\left (-\frac {5}{2},\frac {7}{2}-m;-\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{5 a^2 f (5-m)}+\frac {B \sec ^5(e+f x) (a \sin (e+f x)+a)^m}{f (5-m)} \]

[Out]

B*sec(f*x+e)^5*(a+a*sin(f*x+e))^m/f/(5-m)+1/5*2^(-5/2+m)*(A*(5-m)-B*m)*hypergeom([-5/2, 7/2-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/f/(5-m)

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Rubi [A]  time = 0.19, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2860, 2689, 70, 69} \[ \frac {2^{m-\frac {5}{2}} (A (5-m)-B m) \sec ^5(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m+2} \, _2F_1\left (-\frac {5}{2},\frac {7}{2}-m;-\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{5 a^2 f (5-m)}+\frac {B \sec ^5(e+f x) (a \sin (e+f x)+a)^m}{f (5-m)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^6*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

(B*Sec[e + f*x]^5*(a + a*Sin[e + f*x])^m)/(f*(5 - m)) + (2^(-5/2 + m)*(A*(5 - m) - B*m)*Hypergeometric2F1[-5/2
, 7/2 - 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*f*(5 - 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]

Rubi steps

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

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Mathematica [F]  time = 3.91, size = 0, normalized size = 0.00 \[ \int \sec ^6(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sec[e + f*x]^6*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

Integrate[Sec[e + f*x]^6*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]), x]

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \sec \left (f x + e\right )^{6} \sin \left (f x + e\right ) + A \sec \left (f x + e\right )^{6}\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="fricas")

[Out]

integral((B*sec(f*x + e)^6*sin(f*x + e) + A*sec(f*x + e)^6)*(a*sin(f*x + e) + a)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(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)^6, x)

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maple [F]  time = 0.94, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{6}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^6*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)

[Out]

int(sec(f*x+e)^6*(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 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(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)^6, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\cos \left (e+f\,x\right )}^6} \,d x \]

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)^6,x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/cos(e + f*x)^6, 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)**6*(a+a*sin(f*x+e))**m*(A+B*sin(f*x+e)),x)

[Out]

Timed out

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