3.133 \(\int (e x)^{-1+n} (a+b \sin (c+d x^n))^p \, dx\)

Optimal. Leaf size=132 \[ -\frac {\sqrt {2} x^{-n} (e x)^n \cos \left (c+d x^n\right ) \left (a+b \sin \left (c+d x^n\right )\right )^p \left (\frac {a+b \sin \left (c+d x^n\right )}{a+b}\right )^{-p} F_1\left (\frac {1}{2};\frac {1}{2},-p;\frac {3}{2};\frac {1}{2} \left (1-\sin \left (d x^n+c\right )\right ),\frac {b \left (1-\sin \left (d x^n+c\right )\right )}{a+b}\right )}{d e n \sqrt {\sin \left (c+d x^n\right )+1}} \]

[Out]

-(e*x)^n*AppellF1(1/2,-p,1/2,3/2,b*(1-sin(c+d*x^n))/(a+b),1/2-1/2*sin(c+d*x^n))*cos(c+d*x^n)*(a+b*sin(c+d*x^n)
)^p*2^(1/2)/d/e/n/(x^n)/(((a+b*sin(c+d*x^n))/(a+b))^p)/(1+sin(c+d*x^n))^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {3381, 3379, 2665, 139, 138} \[ -\frac {\sqrt {2} x^{-n} (e x)^n \cos \left (c+d x^n\right ) \left (a+b \sin \left (c+d x^n\right )\right )^p \left (\frac {a+b \sin \left (c+d x^n\right )}{a+b}\right )^{-p} F_1\left (\frac {1}{2};\frac {1}{2},-p;\frac {3}{2};\frac {1}{2} \left (1-\sin \left (d x^n+c\right )\right ),\frac {b \left (1-\sin \left (d x^n+c\right )\right )}{a+b}\right )}{d e n \sqrt {\sin \left (c+d x^n\right )+1}} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^(-1 + n)*(a + b*Sin[c + d*x^n])^p,x]

[Out]

-((Sqrt[2]*(e*x)^n*AppellF1[1/2, 1/2, -p, 3/2, (1 - Sin[c + d*x^n])/2, (b*(1 - Sin[c + d*x^n]))/(a + b)]*Cos[c
+ d*x^n]*(a + b*Sin[c + d*x^n])^p)/(d*e*n*x^n*Sqrt[1 + Sin[c + d*x^n]]*((a + b*Sin[c + d*x^n])/(a + b))^p))

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 2665

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[c + d*x]/(d*Sqrt[1 + Sin[c + d*x]]*Sqrt
[1 - Sin[c + d*x]]), Subst[Int[(a + b*x)^n/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Sin[c + d*x]], x] /; FreeQ[{a, b,
c, d, n}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*n]

Rule 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3381

Int[((e_)*(x_))^(m_)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[(e^IntPart[m]*(e*x)
^FracPart[m])/x^FracPart[m], Int[x^m*(a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] &&
IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int (e x)^{-1+n} \left (a+b \sin \left (c+d x^n\right )\right )^p \, dx &=\frac {\left (x^{-n} (e x)^n\right ) \int x^{-1+n} \left (a+b \sin \left (c+d x^n\right )\right )^p \, dx}{e}\\ &=\frac {\left (x^{-n} (e x)^n\right ) \operatorname {Subst}\left (\int (a+b \sin (c+d x))^p \, dx,x,x^n\right )}{e n}\\ &=\frac {\left (x^{-n} (e x)^n \cos \left (c+d x^n\right )\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^p}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin \left (c+d x^n\right )\right )}{d e n \sqrt {1-\sin \left (c+d x^n\right )} \sqrt {1+\sin \left (c+d x^n\right )}}\\ &=\frac {\left (x^{-n} (e x)^n \cos \left (c+d x^n\right ) \left (a+b \sin \left (c+d x^n\right )\right )^p \left (-\frac {a+b \sin \left (c+d x^n\right )}{-a-b}\right )^{-p}\right ) \operatorname {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^p}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin \left (c+d x^n\right )\right )}{d e n \sqrt {1-\sin \left (c+d x^n\right )} \sqrt {1+\sin \left (c+d x^n\right )}}\\ &=-\frac {\sqrt {2} x^{-n} (e x)^n F_1\left (\frac {1}{2};\frac {1}{2},-p;\frac {3}{2};\frac {1}{2} \left (1-\sin \left (c+d x^n\right )\right ),\frac {b \left (1-\sin \left (c+d x^n\right )\right )}{a+b}\right ) \cos \left (c+d x^n\right ) \left (a+b \sin \left (c+d x^n\right )\right )^p \left (\frac {a+b \sin \left (c+d x^n\right )}{a+b}\right )^{-p}}{d e n \sqrt {1+\sin \left (c+d x^n\right )}}\\ \end {align*}

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Mathematica [A]  time = 0.46, size = 148, normalized size = 1.12 \[ \frac {x^{-n} (e x)^n \sec \left (c+d x^n\right ) \sqrt {-\frac {b \left (\sin \left (c+d x^n\right )-1\right )}{a+b}} \sqrt {\frac {b \left (\sin \left (c+d x^n\right )+1\right )}{b-a}} \left (a+b \sin \left (c+d x^n\right )\right )^{p+1} F_1\left (p+1;\frac {1}{2},\frac {1}{2};p+2;\frac {a+b \sin \left (d x^n+c\right )}{a-b},\frac {a+b \sin \left (d x^n+c\right )}{a+b}\right )}{b d e n (p+1)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(e*x)^(-1 + n)*(a + b*Sin[c + d*x^n])^p,x]

[Out]

((e*x)^n*AppellF1[1 + p, 1/2, 1/2, 2 + p, (a + b*Sin[c + d*x^n])/(a - b), (a + b*Sin[c + d*x^n])/(a + b)]*Sec[
c + d*x^n]*Sqrt[-((b*(-1 + Sin[c + d*x^n]))/(a + b))]*Sqrt[(b*(1 + Sin[c + d*x^n]))/(-a + b)]*(a + b*Sin[c + d
*x^n])^(1 + p))/(b*d*e*n*(1 + p)*x^n)

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (e x\right )^{n - 1} {\left (b \sin \left (d x^{n} + c\right ) + a\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(-1+n)*(a+b*sin(c+d*x^n))^p,x, algorithm="fricas")

[Out]

integral((e*x)^(n - 1)*(b*sin(d*x^n + c) + a)^p, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e x\right )^{n - 1} {\left (b \sin \left (d x^{n} + c\right ) + a\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(-1+n)*(a+b*sin(c+d*x^n))^p,x, algorithm="giac")

[Out]

integrate((e*x)^(n - 1)*(b*sin(d*x^n + c) + a)^p, x)

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maple [F]  time = 1.38, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{-1+n} \left (a +b \sin \left (c +d \,x^{n}\right )\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(-1+n)*(a+b*sin(c+d*x^n))^p,x)

[Out]

int((e*x)^(-1+n)*(a+b*sin(c+d*x^n))^p,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e x\right )^{n - 1} {\left (b \sin \left (d x^{n} + c\right ) + a\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(-1+n)*(a+b*sin(c+d*x^n))^p,x, algorithm="maxima")

[Out]

integrate((e*x)^(n - 1)*(b*sin(d*x^n + c) + a)^p, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (e\,x\right )}^{n-1}\,{\left (a+b\,\sin \left (c+d\,x^n\right )\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(n - 1)*(a + b*sin(c + d*x^n))^p,x)

[Out]

int((e*x)^(n - 1)*(a + b*sin(c + d*x^n))^p, 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((e*x)**(-1+n)*(a+b*sin(c+d*x**n))**p,x)

[Out]

Timed out

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