3.2.24 \(\int (a+a \sin (e+f x))^2 (g \tan (e+f x))^p \, dx\) [124]

Optimal. Leaf size=187 \[ \frac {a^2 \, _2F_1\left (1,\frac {1+p}{2};\frac {3+p}{2};-\tan ^2(e+f x)\right ) (g \tan (e+f x))^{1+p}}{f g (1+p)}+\frac {2 a^2 \cos ^2(e+f x)^{\frac {1+p}{2}} \, _2F_1\left (\frac {1+p}{2},\frac {2+p}{2};\frac {4+p}{2};\sin ^2(e+f x)\right ) \sin (e+f x) (g \tan (e+f x))^{1+p}}{f g (2+p)}+\frac {a^2 \, _2F_1\left (2,\frac {3+p}{2};\frac {5+p}{2};-\tan ^2(e+f x)\right ) (g \tan (e+f x))^{3+p}}{f g^3 (3+p)} \]

[Out]

a^2*hypergeom([1, 1/2+1/2*p],[3/2+1/2*p],-tan(f*x+e)^2)*(g*tan(f*x+e))^(1+p)/f/g/(1+p)+2*a^2*(cos(f*x+e)^2)^(1
/2+1/2*p)*hypergeom([1+1/2*p, 1/2+1/2*p],[2+1/2*p],sin(f*x+e)^2)*sin(f*x+e)*(g*tan(f*x+e))^(1+p)/f/g/(2+p)+a^2
*hypergeom([2, 3/2+1/2*p],[5/2+1/2*p],-tan(f*x+e)^2)*(g*tan(f*x+e))^(3+p)/f/g^3/(3+p)

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Rubi [A]
time = 0.17, antiderivative size = 187, normalized size of antiderivative = 1.00, 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 = {2789, 3557, 371, 2682, 2657, 2671} \begin {gather*} \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)} \end {gather*}

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2657

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)/(a*f*(m + 1)*(Cos[e + f*x]^
2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 2671

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]/ff
)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 2682

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 2789

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 3557

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]

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 ) \text {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 ) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 13.24, size = 1452, normalized size = 7.76 \begin {gather*} \frac {2 a^2 (1+\sin (e+f x))^2 \tan \left (\frac {1}{2} (e+f x)\right ) \left ((2+p) F_1\left (\frac {1+p}{2};p,1;\frac {3+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+4 (2+p) F_1\left (\frac {1+p}{2};p,2;\frac {3+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-4 (2+p) F_1\left (\frac {1+p}{2};p,3;\frac {3+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+4 (1+p) F_1\left (\frac {2+p}{2};p,2;\frac {4+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )\right ) (g \tan (e+f x))^p}{f \left (\sec ^2\left (\frac {1}{2} (e+f x)\right ) \left ((2+p) F_1\left (\frac {1+p}{2};p,1;\frac {3+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+4 (2+p) F_1\left (\frac {1+p}{2};p,2;\frac {3+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-4 (2+p) F_1\left (\frac {1+p}{2};p,3;\frac {3+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+4 (1+p) F_1\left (\frac {2+p}{2};p,2;\frac {4+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )\right )-16 p \cos \left (\frac {1}{2} (e+f x)\right ) \csc ^3(e+f x) \sec (e+f x) \sin ^5\left (\frac {1}{2} (e+f x)\right ) \left ((2+p) F_1\left (\frac {1+p}{2};p,1;\frac {3+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+4 (2+p) F_1\left (\frac {1+p}{2};p,2;\frac {3+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-4 (2+p) F_1\left (\frac {1+p}{2};p,3;\frac {3+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+4 (1+p) F_1\left (\frac {2+p}{2};p,2;\frac {4+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )\right )+2 p \csc (e+f x) \sec (e+f x) \tan \left (\frac {1}{2} (e+f x)\right ) \left ((2+p) F_1\left (\frac {1+p}{2};p,1;\frac {3+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+4 (2+p) F_1\left (\frac {1+p}{2};p,2;\frac {3+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-4 (2+p) F_1\left (\frac {1+p}{2};p,3;\frac {3+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+4 (1+p) F_1\left (\frac {2+p}{2};p,2;\frac {4+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )\right )+2 (1+p) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \tan \left (\frac {1}{2} (e+f x)\right ) \left (2 F_1\left (\frac {2+p}{2};p,2;\frac {4+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+\frac {(2+p) \left (-F_1\left (\frac {3+p}{2};p,2;\frac {5+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+p F_1\left (\frac {3+p}{2};1+p,1;\frac {5+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )}{3+p}+\frac {4 (2+p) \left (-2 F_1\left (\frac {3+p}{2};p,3;\frac {5+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+p F_1\left (\frac {3+p}{2};1+p,2;\frac {5+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )}{3+p}-\frac {4 (2+p) \left (-3 F_1\left (\frac {3+p}{2};p,4;\frac {5+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+p F_1\left (\frac {3+p}{2};1+p,3;\frac {5+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )}{3+p}+\frac {4 (2+p) \left (-2 F_1\left (\frac {4+p}{2};p,3;\frac {6+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+p F_1\left (\frac {4+p}{2};1+p,2;\frac {6+p}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{4+p}\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*a^2*(1 + 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)*Appell
F1[(2 + p)/2, p, 2, (4 + 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)/
(f*(Sec[(e + f*x)/2]^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[(2 + p)/2, p, 2, (4
+ p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]) - 16*p*Cos[(e + f*x)/2]*Csc[e + f*x]^3*Sec
[e + f*x]*Sin[(e + f*x)/2]^5*((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)*Ap
pellF1[(1 + p)/2, p, 3, (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 4*(1 + p)*AppellF1[(2 + p)/2, p,
2, (4 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]) + 2*p*Csc[e + f*x]*Sec[e + f*x]*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[(2 + p)/2, p, 2, (4 + p)/2, T
an[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]) + 2*(1 + p)*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]*(2*A
ppellF1[(2 + p)/2, p, 2, (4 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + ((2 + p)*(-AppellF1[(3 + p)/2,
p, 2, (5 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + p*AppellF1[(3 + p)/2, 1 + p, 1, (5 + p)/2, Tan[(e
+ f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2])/(3 + p) + (4*(2 + p)*(-2*AppellF1[(3 + p)/2, p, 3, (5 + p
)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + p*AppellF1[(3 + p)/2, 1 + p, 2, (5 + p)/2, Tan[(e + f*x)/2]^2,
-Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2])/(3 + p) - (4*(2 + p)*(-3*AppellF1[(3 + p)/2, p, 4, (5 + p)/2, Tan[(e
+ f*x)/2]^2, -Tan[(e + f*x)/2]^2] + p*AppellF1[(3 + p)/2, 1 + p, 3, (5 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f
*x)/2]^2])*Tan[(e + f*x)/2])/(3 + p) + (4*(2 + p)*(-2*AppellF1[(4 + p)/2, p, 3, (6 + p)/2, Tan[(e + f*x)/2]^2,
-Tan[(e + f*x)/2]^2] + p*AppellF1[(4 + p)/2, 1 + p, 2, (6 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*T
an[(e + f*x)/2]^2)/(4 + p))))

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Maple [F]
time = 0.97, size = 0, normalized size = 0.00 \[\int \left (a +a \sin \left (f x +e \right )\right )^{2} \left (g \tan \left (f x +e \right )\right )^{p}\, dx\]

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)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (g\,\mathrm {tan}\left (e+f\,x\right )\right )}^p\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*tan(e + f*x))^p*(a + a*sin(e + f*x))^2,x)

[Out]

int((g*tan(e + f*x))^p*(a + a*sin(e + f*x))^2, x)

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