3.124 \(\int (a+a \sin (e+f x))^2 (g \tan (e+f x))^p \, dx\)
Optimal. Leaf size=187 \[ \frac{a^2 (g \tan (e+f x))^{p+3} \, _2F_1\left (2,\frac{p+3}{2};\frac{p+5}{2};-\tan ^2(e+f x)\right )}{f g^3 (p+3)}+\frac{a^2 (g \tan (e+f x))^{p+1} \, _2F_1\left (1,\frac{p+1}{2};\frac{p+3}{2};-\tan ^2(e+f x)\right )}{f g (p+1)}+\frac{2 a^2 \sin (e+f x) \cos ^2(e+f x)^{\frac{p+1}{2}} (g \tan (e+f x))^{p+1} \, _2F_1\left (\frac{p+1}{2},\frac{p+2}{2};\frac{p+4}{2};\sin ^2(e+f x)\right )}{f g (p+2)} \]
[Out]
(a^2*Hypergeometric2F1[1, (1 + p)/2, (3 + p)/2, -Tan[e + f*x]^2]*(g*Tan[e + f*x])^(1 + p))/(f*g*(1 + p)) + (2*
a^2*(Cos[e + f*x]^2)^((1 + p)/2)*Hypergeometric2F1[(1 + p)/2, (2 + p)/2, (4 + p)/2, Sin[e + f*x]^2]*Sin[e + f*
x]*(g*Tan[e + f*x])^(1 + p))/(f*g*(2 + p)) + (a^2*Hypergeometric2F1[2, (3 + p)/2, (5 + p)/2, -Tan[e + f*x]^2]*
(g*Tan[e + f*x])^(3 + p))/(f*g^3*(3 + p))
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Rubi [A] time = 0.226524, antiderivative size = 187, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used =
{2710, 3476, 364, 2602, 2577, 2591} \[ \frac{a^2 (g \tan (e+f x))^{p+3} \, _2F_1\left (2,\frac{p+3}{2};\frac{p+5}{2};-\tan ^2(e+f x)\right )}{f g^3 (p+3)}+\frac{a^2 (g \tan (e+f x))^{p+1} \, _2F_1\left (1,\frac{p+1}{2};\frac{p+3}{2};-\tan ^2(e+f x)\right )}{f g (p+1)}+\frac{2 a^2 \sin (e+f x) \cos ^2(e+f x)^{\frac{p+1}{2}} (g \tan (e+f x))^{p+1} \, _2F_1\left (\frac{p+1}{2},\frac{p+2}{2};\frac{p+4}{2};\sin ^2(e+f x)\right )}{f g (p+2)} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])^2*(g*Tan[e + f*x])^p,x]
[Out]
(a^2*Hypergeometric2F1[1, (1 + p)/2, (3 + p)/2, -Tan[e + f*x]^2]*(g*Tan[e + f*x])^(1 + p))/(f*g*(1 + p)) + (2*
a^2*(Cos[e + f*x]^2)^((1 + p)/2)*Hypergeometric2F1[(1 + p)/2, (2 + p)/2, (4 + p)/2, Sin[e + f*x]^2]*Sin[e + f*
x]*(g*Tan[e + f*x])^(1 + p))/(f*g*(2 + p)) + (a^2*Hypergeometric2F1[2, (3 + p)/2, (5 + p)/2, -Tan[e + f*x]^2]*
(g*Tan[e + f*x])^(3 + p))/(f*g^3*(3 + p))
Rule 2710
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Int[Expan
dIntegrand[(g*Tan[e + f*x])^p, (a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2
, 0] && IGtQ[m, 0]
Rule 3476
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
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 2602
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[e + f
*x]^(n + 1)*(b*Tan[e + f*x])^(n + 1))/(b*(a*Sin[e + f*x])^(n + 1)), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^
n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[n]
Rule 2577
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b^(2*IntPart
[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]
Rule 2591
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]
Rubi steps
\begin{align*} \int (a+a \sin (e+f x))^2 (g \tan (e+f x))^p \, dx &=\int \left (a^2 (g \tan (e+f x))^p+2 a^2 \sin (e+f x) (g \tan (e+f x))^p+a^2 \sin ^2(e+f x) (g \tan (e+f x))^p\right ) \, dx\\ &=a^2 \int (g \tan (e+f x))^p \, dx+a^2 \int \sin ^2(e+f x) (g \tan (e+f x))^p \, dx+\left (2 a^2\right ) \int \sin (e+f x) (g \tan (e+f x))^p \, dx\\ &=\frac{\left (a^2 g\right ) \operatorname{Subst}\left (\int \frac{x^{2+p}}{\left (g^2+x^2\right )^2} \, dx,x,g \tan (e+f x)\right )}{f}+\frac{\left (a^2 g\right ) \operatorname{Subst}\left (\int \frac{x^p}{g^2+x^2} \, dx,x,g \tan (e+f x)\right )}{f}+\frac{\left (2 a^2 \cos ^{1+p}(e+f x) \sin ^{-1-p}(e+f x) (g \tan (e+f x))^{1+p}\right ) \int \cos ^{-p}(e+f x) \sin ^{1+p}(e+f x) \, dx}{g}\\ &=\frac{a^2 \, _2F_1\left (1,\frac{1+p}{2};\frac{3+p}{2};-\tan ^2(e+f x)\right ) (g \tan (e+f x))^{1+p}}{f g (1+p)}+\frac{2 a^2 \cos ^2(e+f x)^{\frac{1+p}{2}} \, _2F_1\left (\frac{1+p}{2},\frac{2+p}{2};\frac{4+p}{2};\sin ^2(e+f x)\right ) \sin (e+f x) (g \tan (e+f x))^{1+p}}{f g (2+p)}+\frac{a^2 \, _2F_1\left (2,\frac{3+p}{2};\frac{5+p}{2};-\tan ^2(e+f x)\right ) (g \tan (e+f x))^{3+p}}{f g^3 (3+p)}\\ \end{align*}
Mathematica [C] time = 18.053, size = 2054, normalized size = 10.98 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + a*Sin[e + f*x])^2*(g*Tan[e + f*x])^p,x]
[Out]
(2*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^p*(a + a*Sin[e + f*x])^2*Tan[(e + f*x)/2]*((2 + p)*AppellF1[(1 + p)/2, p,
1, (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 4*(2 + p)*AppellF1[(1 + p)/2, p, 2, (3 + p)/2, Tan[(
e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 4*(2 + p)*AppellF1[(1 + p)/2, p, 3, (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[
(e + f*x)/2]^2] + 4*(1 + p)*AppellF1[1 + p/2, p, 2, 2 + p/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Tan[(e +
f*x)/2])*(g*Tan[e + f*x])^p*(Cos[(e + f*x)/2]^4*Tan[e + f*x]^p + 4*Cos[(e + f*x)/2]^3*Sin[(e + f*x)/2]*Tan[e
+ f*x]^p + 6*Cos[(e + f*x)/2]^2*Sin[(e + f*x)/2]^2*Tan[e + f*x]^p + 4*Cos[(e + f*x)/2]*Sin[(e + f*x)/2]^3*Tan[
e + f*x]^p + Sin[(e + f*x)/2]^4*Tan[e + f*x]^p))/(f*(1 + p)*(2 + p)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*((
2*p*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^p*Sec[e + f*x]^2*Tan[(e + f*x)/2]*((2 + p)*AppellF1[(1 + p)/2, p, 1, (3
+ p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 4*(2 + p)*AppellF1[(1 + p)/2, p, 2, (3 + p)/2, Tan[(e + f*x
)/2]^2, -Tan[(e + f*x)/2]^2] - 4*(2 + p)*AppellF1[(1 + p)/2, p, 3, (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*
x)/2]^2] + 4*(1 + p)*AppellF1[1 + p/2, p, 2, 2 + p/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2
])*Tan[e + f*x]^(-1 + p))/((1 + p)*(2 + p)) + (Sec[(e + f*x)/2]^2*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^p*((2 + p)
*AppellF1[(1 + p)/2, p, 1, (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 4*(2 + p)*AppellF1[(1 + p)/2,
p, 2, (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 4*(2 + p)*AppellF1[(1 + p)/2, p, 3, (3 + p)/2, Ta
n[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 4*(1 + p)*AppellF1[1 + p/2, p, 2, 2 + p/2, Tan[(e + f*x)/2]^2, -Tan[(
e + f*x)/2]^2]*Tan[(e + f*x)/2])*Tan[e + f*x]^p)/((1 + p)*(2 + p)) + (2*p*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(-
1 + p)*Tan[(e + f*x)/2]*((2 + p)*AppellF1[(1 + p)/2, p, 1, (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]
+ 4*(2 + p)*AppellF1[(1 + p)/2, p, 2, (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 4*(2 + p)*AppellF
1[(1 + p)/2, p, 3, (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 4*(1 + p)*AppellF1[1 + p/2, p, 2, 2 +
p/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2])*(-(Sec[(e + f*x)/2]^2*Sin[e + f*x]) + Cos[e +
f*x]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])*Tan[e + f*x]^p)/((1 + p)*(2 + p)) + (2*(Cos[e + f*x]*Sec[(e + f*x)/
2]^2)^p*Tan[(e + f*x)/2]*(2*(1 + p)*AppellF1[1 + p/2, p, 2, 2 + p/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*
Sec[(e + f*x)/2]^2 + 4*(1 + p)*Tan[(e + f*x)/2]*((-2*(1 + p/2)*AppellF1[2 + p/2, p, 3, 3 + p/2, Tan[(e + f*x)/
2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/(2 + p/2) + ((1 + p/2)*p*AppellF1[2 + p/2, 1 +
p, 2, 3 + p/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/(2 + p/2)) + (2
+ p)*(-(((1 + p)*AppellF1[1 + (1 + p)/2, p, 2, 1 + (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e
+ f*x)/2]^2*Tan[(e + f*x)/2])/(3 + p)) + (p*(1 + p)*AppellF1[1 + (1 + p)/2, 1 + p, 1, 1 + (3 + p)/2, Tan[(e +
f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/(3 + p)) + 4*(2 + p)*((-2*(1 + p)*AppellF
1[1 + (1 + p)/2, p, 3, 1 + (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x
)/2])/(3 + p) + (p*(1 + p)*AppellF1[1 + (1 + p)/2, 1 + p, 2, 1 + (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)
/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/(3 + p)) - 4*(2 + p)*((-3*(1 + p)*AppellF1[1 + (1 + p)/2, p, 4, 1
+ (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/(3 + p) + (p*(1 + p
)*AppellF1[1 + (1 + p)/2, 1 + p, 3, 1 + (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2
*Tan[(e + f*x)/2])/(3 + p)))*Tan[e + f*x]^p)/((1 + p)*(2 + p))))
________________________________________________________________________________________
Maple [F] time = 1.421, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{2} \left ( g\tan \left ( fx+e \right ) \right ) ^{p}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(f*x+e))^2*(g*tan(f*x+e))^p,x)
[Out]
int((a+a*sin(f*x+e))^2*(g*tan(f*x+e))^p,x)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{2} \left (g \tan \left (f x + e\right )\right )^{p}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^2*(g*tan(f*x+e))^p,x, algorithm="maxima")
[Out]
integrate((a*sin(f*x + e) + a)^2*(g*tan(f*x + e))^p, x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}\right )} \left (g \tan \left (f x + e\right )\right )^{p}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^2*(g*tan(f*x+e))^p,x, algorithm="fricas")
[Out]
integral(-(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2)*(g*tan(f*x + e))^p, x)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int \left (g \tan{\left (e + f x \right )}\right )^{p}\, dx + \int 2 \left (g \tan{\left (e + f x \right )}\right )^{p} \sin{\left (e + f x \right )}\, dx + \int \left (g \tan{\left (e + f x \right )}\right )^{p} \sin ^{2}{\left (e + f x \right )}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))**2*(g*tan(f*x+e))**p,x)
[Out]
a**2*(Integral((g*tan(e + f*x))**p, x) + Integral(2*(g*tan(e + f*x))**p*sin(e + f*x), x) + Integral((g*tan(e +
f*x))**p*sin(e + f*x)**2, x))
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{2} \left (g \tan \left (f x + e\right )\right )^{p}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^2*(g*tan(f*x+e))^p,x, algorithm="giac")
[Out]
integrate((a*sin(f*x + e) + a)^2*(g*tan(f*x + e))^p, x)