3.135 \(\int (a+b \sec ^2(e+f x))^p \sin (e+f x) \, dx\)

Optimal. Leaf size=68 \[ -\frac {\cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \sec ^2(e+f x)}{a}\right )}{f} \]

[Out]

-cos(f*x+e)*hypergeom([-1/2, -p],[1/2],-b*sec(f*x+e)^2/a)*(a+b*sec(f*x+e)^2)^p/f/((1+b*sec(f*x+e)^2/a)^p)

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Rubi [A]  time = 0.05, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4134, 365, 364} \[ -\frac {\cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \sec ^2(e+f x)}{a}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[e + f*x]^2)^p*Sin[e + f*x],x]

[Out]

-((Cos[e + f*x]*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Sec[e + f*x]^2)/a)]*(a + b*Sec[e + f*x]^2)^p)/(f*(1 + (b
*Sec[e + f*x]^2)/a)^p))

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 4134

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

Rubi steps

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

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Mathematica [A]  time = 1.65, size = 68, normalized size = 1.00 \[ -\frac {\cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \sec ^2(e+f x)}{a}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[e + f*x]^2)^p*Sin[e + f*x],x]

[Out]

-((Cos[e + f*x]*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Sec[e + f*x]^2)/a)]*(a + b*Sec[e + f*x]^2)^p)/(f*(1 + (b
*Sec[e + f*x]^2)/a)^p))

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^p*sin(f*x+e),x, algorithm="fricas")

[Out]

integral((b*sec(f*x + e)^2 + a)^p*sin(f*x + e), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^p*sin(f*x+e),x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e)^2 + a)^p*sin(f*x + e), x)

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maple [F]  time = 1.37, size = 0, normalized size = 0.00 \[ \int \left (a +b \left (\sec ^{2}\left (f x +e \right )\right )\right )^{p} \sin \left (f x +e \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(f*x+e)^2)^p*sin(f*x+e),x)

[Out]

int((a+b*sec(f*x+e)^2)^p*sin(f*x+e),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^p*sin(f*x+e),x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e)^2 + a)^p*sin(f*x + e), x)

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mupad [B]  time = 4.92, size = 79, normalized size = 1.16 \[ \frac {\cos \left (e+f\,x\right )\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2}-p,-p;\ \frac {3}{2}-p;\ -\frac {a\,{\cos \left (e+f\,x\right )}^2}{b}\right )}{f\,\left (2\,p-1\right )\,{\left (\frac {a\,{\cos \left (e+f\,x\right )}^2}{b}+1\right )}^p} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)*(a + b/cos(e + f*x)^2)^p,x)

[Out]

(cos(e + f*x)*(a + b/cos(e + f*x)^2)^p*hypergeom([1/2 - p, -p], 3/2 - p, -(a*cos(e + f*x)^2)/b))/(f*(2*p - 1)*
((a*cos(e + f*x)^2)/b + 1)^p)

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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((a+b*sec(f*x+e)**2)**p*sin(f*x+e),x)

[Out]

Timed out

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