3.3.3 \(\int (a+b \sin (e+f x))^3 (g \tan (e+f x))^p \, dx\) [203]
Optimal. Leaf size=271 \[ \frac {a^3 \, _2F_1\left (1,\frac {1+p}{2};\frac {3+p}{2};-\tan ^2(e+f x)\right ) (g \tan (e+f x))^{1+p}}{f g (1+p)}+\frac {3 a^2 b \cos ^2(e+f x)^{\frac {1+p}{2}} \, _2F_1\left (\frac {1+p}{2},\frac {2+p}{2};\frac {4+p}{2};\sin ^2(e+f x)\right ) \sin (e+f x) (g \tan (e+f x))^{1+p}}{f g (2+p)}+\frac {b^3 \cos ^2(e+f x)^{\frac {1+p}{2}} \, _2F_1\left (\frac {1+p}{2},\frac {4+p}{2};\frac {6+p}{2};\sin ^2(e+f x)\right ) \sin ^3(e+f x) (g \tan (e+f x))^{1+p}}{f g (4+p)}+\frac {3 a b^2 \, _2F_1\left (2,\frac {3+p}{2};\frac {5+p}{2};-\tan ^2(e+f x)\right ) (g \tan (e+f x))^{3+p}}{f g^3 (3+p)} \]
[Out]
a^3*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)+3*a^2*b*(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)+b
^3*(cos(f*x+e)^2)^(1/2+1/2*p)*hypergeom([2+1/2*p, 1/2+1/2*p],[3+1/2*p],sin(f*x+e)^2)*sin(f*x+e)^3*(g*tan(f*x+e
))^(1+p)/f/g/(4+p)+3*a*b^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 = 0.27, antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2801, 3557,
371, 2682, 2657, 2671} \begin {gather*} \frac {a^3 (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 {3 a^2 b \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)}+\frac {3 a b^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 {b^3 \sin ^3(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+4}{2};\frac {p+6}{2};\sin ^2(e+f x)\right )}{f g (p+4)} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + b*Sin[e + f*x])^3*(g*Tan[e + f*x])^p,x]
[Out]
(a^3*Hypergeometric2F1[1, (1 + p)/2, (3 + p)/2, -Tan[e + f*x]^2]*(g*Tan[e + f*x])^(1 + p))/(f*g*(1 + p)) + (3*
a^2*b*(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)) + (b^3*(Cos[e + f*x]^2)^((1 + p)/2)*Hypergeometric2F1[(1 + p)/2,
(4 + p)/2, (6 + p)/2, Sin[e + f*x]^2]*Sin[e + f*x]^3*(g*Tan[e + f*x])^(1 + p))/(f*g*(4 + p)) + (3*a*b^2*Hyperg
eometric2F1[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 2801
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] && NeQ[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+b \sin (e+f x))^3 (g \tan (e+f x))^p \, dx &=\int \left (a^3 (g \tan (e+f x))^p+3 a^2 b \sin (e+f x) (g \tan (e+f x))^p+3 a b^2 \sin ^2(e+f x) (g \tan (e+f x))^p+b^3 \sin ^3(e+f x) (g \tan (e+f x))^p\right ) \, dx\\ &=a^3 \int (g \tan (e+f x))^p \, dx+\left (3 a^2 b\right ) \int \sin (e+f x) (g \tan (e+f x))^p \, dx+\left (3 a b^2\right ) \int \sin ^2(e+f x) (g \tan (e+f x))^p \, dx+b^3 \int \sin ^3(e+f x) (g \tan (e+f x))^p \, dx\\ &=\frac {\left (a^3 g\right ) \text {Subst}\left (\int \frac {x^p}{g^2+x^2} \, dx,x,g \tan (e+f x)\right )}{f}+\frac {\left (3 a b^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 (3 a^2 b \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 {\left (b^3 \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 ^{3+p}(e+f x) \, dx}{g}\\ &=\frac {a^3 \, _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 {3 a^2 b \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 {b^3 \cos ^2(e+f x)^{\frac {1+p}{2}} \, _2F_1\left (\frac {1+p}{2},\frac {4+p}{2};\frac {6+p}{2};\sin ^2(e+f x)\right ) \sin ^3(e+f x) (g \tan (e+f x))^{1+p}}{f g (4+p)}+\frac {3 a b^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 = 14.62, size = 4791, normalized size = 17.68 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + b*Sin[e + f*x])^3*(g*Tan[e + f*x])^p,x]
[Out]
(2*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^p*Tan[(e + f*x)/2]*(a^3*(2 + p)*AppellF1[(1 + p)/2, p, 1, (3 + p)/2, Tan[
(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*b*(6*a*b*(2 + p)*AppellF1[(1 + p)/2, p, 2, (3 + p)/2, Tan[(e + f*x)/2
]^2, -Tan[(e + f*x)/2]^2] - 6*a*b*(2 + p)*AppellF1[(1 + p)/2, p, 3, (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f
*x)/2]^2] + (1 + p)*(3*a^2*AppellF1[1 + p/2, p, 2, 2 + p/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 4*b^2*(
AppellF1[1 + p/2, p, 3, 2 + p/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - AppellF1[1 + p/2, p, 4, 2 + p/2, T
an[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]))*Tan[(e + f*x)/2]))*(g*Tan[e + f*x])^p*(-1/8*(b^3*Sin[3*(e + f*x)]*Ta
n[e + f*x]^p) - a^3*Sin[e + f*x]^3*Sin[3*(e + f*x)]*Tan[e + f*x]^p + ((3*I)/8)*b^3*Sin[2*(e + f*x)]*Sin[3*(e +
f*x)]*Tan[e + f*x]^p + (3*b^3*Sin[2*(e + f*x)]^2*Sin[3*(e + f*x)]*Tan[e + f*x]^p)/8 - (I/8)*b^3*Sin[2*(e + f*
x)]^3*Sin[3*(e + f*x)]*Tan[e + f*x]^p + Cos[e + f*x]^3*(a^3*Cos[3*(e + f*x)]*Tan[e + f*x]^p - I*a^3*Sin[3*(e +
f*x)]*Tan[e + f*x]^p) + Cos[2*(e + f*x)]^3*((I/8)*b^3*Cos[3*(e + f*x)]*Tan[e + f*x]^p + (b^3*Sin[3*(e + f*x)]
*Tan[e + f*x]^p)/8) + Sin[e + f*x]^2*((-3*a^2*b*Sin[3*(e + f*x)]*Tan[e + f*x]^p)/2 + ((3*I)/2)*a^2*b*Sin[2*(e
+ f*x)]*Sin[3*(e + f*x)]*Tan[e + f*x]^p) + Sin[e + f*x]*((-3*a*b^2*Sin[3*(e + f*x)]*Tan[e + f*x]^p)/4 + ((3*I)
/2)*a*b^2*Sin[2*(e + f*x)]*Sin[3*(e + f*x)]*Tan[e + f*x]^p + (3*a*b^2*Sin[2*(e + f*x)]^2*Sin[3*(e + f*x)]*Tan[
e + f*x]^p)/4) + Cos[2*(e + f*x)]^2*((-3*b^3*Sin[3*(e + f*x)]*Tan[e + f*x]^p)/8 - (3*a*b^2*Sin[e + f*x]*Sin[3*
(e + f*x)]*Tan[e + f*x]^p)/4 + ((3*I)/8)*b^3*Sin[2*(e + f*x)]*Sin[3*(e + f*x)]*Tan[e + f*x]^p + Cos[3*(e + f*x
)]*(((-3*I)/8)*b^3*Tan[e + f*x]^p - ((3*I)/4)*a*b^2*Sin[e + f*x]*Tan[e + f*x]^p - (3*b^3*Sin[2*(e + f*x)]*Tan[
e + f*x]^p)/8)) + Cos[3*(e + f*x)]*((-1/8*I)*b^3*Tan[e + f*x]^p - I*a^3*Sin[e + f*x]^3*Tan[e + f*x]^p - (3*b^3
*Sin[2*(e + f*x)]*Tan[e + f*x]^p)/8 + ((3*I)/8)*b^3*Sin[2*(e + f*x)]^2*Tan[e + f*x]^p + (b^3*Sin[2*(e + f*x)]^
3*Tan[e + f*x]^p)/8 + Sin[e + f*x]^2*(((-3*I)/2)*a^2*b*Tan[e + f*x]^p - (3*a^2*b*Sin[2*(e + f*x)]*Tan[e + f*x]
^p)/2) + Sin[e + f*x]*(((-3*I)/4)*a*b^2*Tan[e + f*x]^p - (3*a*b^2*Sin[2*(e + f*x)]*Tan[e + f*x]^p)/2 + ((3*I)/
4)*a*b^2*Sin[2*(e + f*x)]^2*Tan[e + f*x]^p)) + Cos[e + f*x]^2*((3*a^2*b*Sin[3*(e + f*x)]*Tan[e + f*x]^p)/2 + 3
*a^3*Sin[e + f*x]*Sin[3*(e + f*x)]*Tan[e + f*x]^p - ((3*I)/2)*a^2*b*Sin[2*(e + f*x)]*Sin[3*(e + f*x)]*Tan[e +
f*x]^p + Cos[3*(e + f*x)]*(((3*I)/2)*a^2*b*Tan[e + f*x]^p + (3*I)*a^3*Sin[e + f*x]*Tan[e + f*x]^p + (3*a^2*b*S
in[2*(e + f*x)]*Tan[e + f*x]^p)/2) + Cos[2*(e + f*x)]*(((-3*I)/2)*a^2*b*Cos[3*(e + f*x)]*Tan[e + f*x]^p - (3*a
^2*b*Sin[3*(e + f*x)]*Tan[e + f*x]^p)/2)) + Cos[e + f*x]*(((3*I)/4)*a*b^2*Sin[3*(e + f*x)]*Tan[e + f*x]^p + (3
*I)*a^3*Sin[e + f*x]^2*Sin[3*(e + f*x)]*Tan[e + f*x]^p + (3*a*b^2*Sin[2*(e + f*x)]*Sin[3*(e + f*x)]*Tan[e + f*
x]^p)/2 - ((3*I)/4)*a*b^2*Sin[2*(e + f*x)]^2*Sin[3*(e + f*x)]*Tan[e + f*x]^p + Cos[2*(e + f*x)]^2*((-3*a*b^2*C
os[3*(e + f*x)]*Tan[e + f*x]^p)/4 + ((3*I)/4)*a*b^2*Sin[3*(e + f*x)]*Tan[e + f*x]^p) + Sin[e + f*x]*((3*I)*a^2
*b*Sin[3*(e + f*x)]*Tan[e + f*x]^p + 3*a^2*b*Sin[2*(e + f*x)]*Sin[3*(e + f*x)]*Tan[e + f*x]^p) + Cos[3*(e + f*
x)]*((-3*a*b^2*Tan[e + f*x]^p)/4 - 3*a^3*Sin[e + f*x]^2*Tan[e + f*x]^p + ((3*I)/2)*a*b^2*Sin[2*(e + f*x)]*Tan[
e + f*x]^p + (3*a*b^2*Sin[2*(e + f*x)]^2*Tan[e + f*x]^p)/4 + Sin[e + f*x]*(-3*a^2*b*Tan[e + f*x]^p + (3*I)*a^2
*b*Sin[2*(e + f*x)]*Tan[e + f*x]^p)) + Cos[2*(e + f*x)]*(((-3*I)/2)*a*b^2*Sin[3*(e + f*x)]*Tan[e + f*x]^p - (3
*I)*a^2*b*Sin[e + f*x]*Sin[3*(e + f*x)]*Tan[e + f*x]^p - (3*a*b^2*Sin[2*(e + f*x)]*Sin[3*(e + f*x)]*Tan[e + f*
x]^p)/2 + Cos[3*(e + f*x)]*((3*a*b^2*Tan[e + f*x]^p)/2 + 3*a^2*b*Sin[e + f*x]*Tan[e + f*x]^p - ((3*I)/2)*a*b^2
*Sin[2*(e + f*x)]*Tan[e + f*x]^p))) + Cos[2*(e + f*x)]*((3*b^3*Sin[3*(e + f*x)]*Tan[e + f*x]^p)/8 + (3*a^2*b*S
in[e + f*x]^2*Sin[3*(e + f*x)]*Tan[e + f*x]^p)/2 - ((3*I)/4)*b^3*Sin[2*(e + f*x)]*Sin[3*(e + f*x)]*Tan[e + f*x
]^p - (3*b^3*Sin[2*(e + f*x)]^2*Sin[3*(e + f*x)]*Tan[e + f*x]^p)/8 + Sin[e + f*x]*((3*a*b^2*Sin[3*(e + f*x)]*T
an[e + f*x]^p)/2 - ((3*I)/2)*a*b^2*Sin[2*(e + f*x)]*Sin[3*(e + f*x)]*Tan[e + f*x]^p) + Cos[3*(e + f*x)]*(((3*I
)/8)*b^3*Tan[e + f*x]^p + ((3*I)/2)*a^2*b*Sin[e + f*x]^2*Tan[e + f*x]^p + (3*b^3*Sin[2*(e + f*x)]*Tan[e + f*x]
^p)/4 - ((3*I)/8)*b^3*Sin[2*(e + f*x)]^2*Tan[e + f*x]^p + Sin[e + f*x]*(((3*I)/2)*a*b^2*Tan[e + f*x]^p + (3*a*
b^2*Sin[2*(e + f*x)]*Tan[e + f*x]^p)/2)))))/(f*(1 + p)*(2 + p)*((2*p*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^p*Sec[e
+ f*x]^2*Tan[(e + f*x)/2]*(a^3*(2 + p)*AppellF1[(1 + p)/2, p, 1, (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x
)/2]^2] + 2*b*(6*a*b*(2 + p)*AppellF1[(1 + p)/2, p, 2, (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 6
*a*b*(2 + p)*AppellF1[(1 + p)/2, p, 3, (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (1 + p)*(3*a^2*Ap
pellF1[1 + p/2, p, 2, 2 + p/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 4*b^2*(AppellF1[1 + p/2, p, 3, 2 + p
/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - ...
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Maple [F]
time = 1.12, size = 0, normalized size = 0.00 \[\int \left (a +b \sin \left (f x +e \right )\right )^{3} \left (g \tan \left (f x +e \right )\right )^{p}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*sin(f*x+e))^3*(g*tan(f*x+e))^p,x)
[Out]
int((a+b*sin(f*x+e))^3*(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+b*sin(f*x+e))^3*(g*tan(f*x+e))^p,x, algorithm="maxima")
[Out]
integrate((b*sin(f*x + e) + a)^3*(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+b*sin(f*x+e))^3*(g*tan(f*x+e))^p,x, algorithm="fricas")
[Out]
integral(-(3*a*b^2*cos(f*x + e)^2 - a^3 - 3*a*b^2 + (b^3*cos(f*x + e)^2 - 3*a^2*b - b^3)*sin(f*x + e))*(g*tan(
f*x + e))^p, x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (g \tan {\left (e + f x \right )}\right )^{p} \left (a + b \sin {\left (e + f x \right )}\right )^{3}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sin(f*x+e))**3*(g*tan(f*x+e))**p,x)
[Out]
Integral((g*tan(e + f*x))**p*(a + b*sin(e + f*x))**3, 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+b*sin(f*x+e))^3*(g*tan(f*x+e))^p,x, algorithm="giac")
[Out]
integrate((b*sin(f*x + e) + a)^3*(g*tan(f*x + e))^p, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (g\,\mathrm {tan}\left (e+f\,x\right )\right )}^p\,{\left (a+b\,\sin \left (e+f\,x\right )\right )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((g*tan(e + f*x))^p*(a + b*sin(e + f*x))^3,x)
[Out]
int((g*tan(e + f*x))^p*(a + b*sin(e + f*x))^3, x)
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