3.136 \(\int (a+a \sin (e+f x))^m \tan ^2(e+f x) \, dx\)
Optimal. Leaf size=157 \[ \frac {2^{m-\frac {1}{2}} \left (-m^2-m+1\right ) \sec (e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^m \, _2F_1\left (-\frac {1}{2},\frac {3}{2}-m;\frac {1}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{f (1-m) m}+\frac {\sec (e+f x) (a \sin (e+f x)+a)^m}{f (1-m) m}-\frac {\sec (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f m} \]
[Out]
sec(f*x+e)*(a+a*sin(f*x+e))^m/f/(1-m)/m+2^(-1/2+m)*(-m^2-m+1)*hypergeom([-1/2, 3/2-m],[1/2],1/2-1/2*sin(f*x+e)
)*sec(f*x+e)*(1+sin(f*x+e))^(1/2-m)*(a+a*sin(f*x+e))^m/f/(1-m)/m-sec(f*x+e)*(a+a*sin(f*x+e))^(1+m)/a/f/m
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Rubi [A] time = 0.25, antiderivative size = 157, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used =
{2713, 2860, 2689, 70, 69} \[ \frac {2^{m-\frac {1}{2}} \left (-m^2-m+1\right ) \sec (e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^m \, _2F_1\left (-\frac {1}{2},\frac {3}{2}-m;\frac {1}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{f (1-m) m}+\frac {\sec (e+f x) (a \sin (e+f x)+a)^m}{f (1-m) m}-\frac {\sec (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f m} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])^m*Tan[e + f*x]^2,x]
[Out]
(Sec[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1 - m)*m) + (2^(-1/2 + m)*(1 - m - m^2)*Hypergeometric2F1[-1/2, 3/2
- m, 1/2, (1 - Sin[e + f*x])/2]*Sec[e + f*x]*(1 + Sin[e + f*x])^(1/2 - m)*(a + a*Sin[e + f*x])^m)/(f*(1 - m)*m
) - (Sec[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(a*f*m)
Rule 69
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))
Rule 70
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[m] && !IntegerQ[n] &&
(RationalQ[m] || !SimplerQ[n + 1, m + 1])
Rule 2689
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[(a^2*
(g*Cos[e + f*x])^(p + 1))/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] && !IntegerQ[m]
Rule 2713
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> -Simp[(a + b*Sin[e + f
*x])^(m + 1)/(b*f*m*Cos[e + f*x]), x] + Dist[1/(b*m), Int[((a + b*Sin[e + f*x])^m*(b*(m + 1) + a*Sin[e + f*x])
)/Cos[e + f*x]^2, x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m] && !LtQ[m, 0]
Rule 2860
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 1, 0]
Rubi steps
\begin {align*} \int (a+a \sin (e+f x))^m \tan ^2(e+f x) \, dx &=-\frac {\sec (e+f x) (a+a \sin (e+f x))^{1+m}}{a f m}+\frac {\int \sec ^2(e+f x) (a+a \sin (e+f x))^m (a (1+m)+a \sin (e+f x)) \, dx}{a m}\\ &=\frac {\sec (e+f x) (a+a \sin (e+f x))^m}{f (1-m) m}-\frac {\sec (e+f x) (a+a \sin (e+f x))^{1+m}}{a f m}+\frac {\left (1-m-m^2\right ) \int \sec ^2(e+f x) (a+a \sin (e+f x))^m \, dx}{(1-m) m}\\ &=\frac {\sec (e+f x) (a+a \sin (e+f x))^m}{f (1-m) m}-\frac {\sec (e+f x) (a+a \sin (e+f x))^{1+m}}{a f m}+\frac {\left (a^2 \left (1-m-m^2\right ) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{-\frac {3}{2}+m}}{(a-a x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{f (1-m) m}\\ &=\frac {\sec (e+f x) (a+a \sin (e+f x))^m}{f (1-m) m}-\frac {\sec (e+f x) (a+a \sin (e+f x))^{1+m}}{a f m}+\frac {\left (2^{-\frac {3}{2}+m} a \left (1-m-m^2\right ) \sec (e+f x) \sqrt {a-a \sin (e+f x)} (a+a \sin (e+f x))^m \left (\frac {a+a \sin (e+f x)}{a}\right )^{\frac {1}{2}-m}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {x}{2}\right )^{-\frac {3}{2}+m}}{(a-a x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{f (1-m) m}\\ &=\frac {\sec (e+f x) (a+a \sin (e+f x))^m}{f (1-m) m}+\frac {2^{-\frac {1}{2}+m} \left (1-m-m^2\right ) \, _2F_1\left (-\frac {1}{2},\frac {3}{2}-m;\frac {1}{2};\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f (1-m) m}-\frac {\sec (e+f x) (a+a \sin (e+f x))^{1+m}}{a f m}\\ \end {align*}
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Mathematica [C] time = 6.32, size = 4043, normalized size = 25.75 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + a*Sin[e + f*x])^m*Tan[e + f*x]^2,x]
[Out]
(-2*Cos[(-e + Pi/2 - f*x)/2]*Hypergeometric2F1[1/2, (1 + 2*m)/2, (3 + 2*m)/2, Cos[(-e + Pi/2 - f*x)/2]^2]*Sin[
(-e + Pi/2 - f*x)/2]*(a + a*Sin[e + f*x])^m)/(f*(1 + 2*m)*Sqrt[Sin[(-e + Pi/2 - f*x)/2]^2]) - ((Cos[(-e + Pi/2
- f*x)/4]^2)^(2*m)*Cot[(-e + Pi/2 - f*x)/4]*(a + a*Sin[e + f*x])^m*(-(AppellF1[-1/2, -2*m, 2*m, 1/2, Tan[(-e
+ Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*(Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m)) + (3*AppellF1[1/2, -2*m,
2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Tan[(-e + Pi/2 - f*x)/4]^2*(1 - Tan[(-e + P
i/2 - f*x)/4]^2)^(2*m))/(3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]
^2] - 4*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + AppellF
1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*Tan[(-e + Pi/2 - f*x)/4]^
2)))/(4*f*(Cos[Pi/4 + (e - Pi/2 + f*x)/2] - Sin[Pi/4 + (e - Pi/2 + f*x)/2])^2*(-1/2*(m*(Cos[(-e + Pi/2 - f*x)/
4]^2)^(2*m)*(-(AppellF1[-1/2, -2*m, 2*m, 1/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*(Sec[(-
e + Pi/2 - f*x)/4]^2)^(2*m)) + (3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 -
f*x)/4]^2]*Tan[(-e + Pi/2 - f*x)/4]^2*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))/(3*AppellF1[1/2, -2*m, 2*m, 3/2,
Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 4*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-e + Pi
/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -
Tan[(-e + Pi/2 - f*x)/4]^2])*Tan[(-e + Pi/2 - f*x)/4]^2))) - ((Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Csc[(-e + Pi/
2 - f*x)/4]^2*(-(AppellF1[-1/2, -2*m, 2*m, 1/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*(Sec[
(-e + Pi/2 - f*x)/4]^2)^(2*m)) + (3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2
- f*x)/4]^2]*Tan[(-e + Pi/2 - f*x)/4]^2*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))/(3*AppellF1[1/2, -2*m, 2*m, 3/
2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 4*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-e +
Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2,
-Tan[(-e + Pi/2 - f*x)/4]^2])*Tan[(-e + Pi/2 - f*x)/4]^2)))/8 + ((Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Cot[(-e +
Pi/2 - f*x)/4]*(-(m*AppellF1[-1/2, -2*m, 2*m, 1/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*(
Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Tan[(-e + Pi/2 - f*x)/4]) - (Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m)*(m*AppellF1[1
/2, 1 - 2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan
[(-e + Pi/2 - f*x)/4] + m*AppellF1[1/2, -2*m, 1 + 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)
/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4]) + (3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2
- f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4]*(1 - Tan[(-e + P
i/2 - f*x)/4]^2)^(2*m))/(2*(3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)
/4]^2] - 4*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + Appe
llF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*Tan[(-e + Pi/2 - f*x)/
4]^2)) + (3*Tan[(-e + Pi/2 - f*x)/4]^2*(-1/3*(m*AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -
Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4]) - (m*AppellF1[3/2, -2*m, 1 +
2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 -
f*x)/4])/3)*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))/(3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]
^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 4*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e
+ Pi/2 - f*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^
2])*Tan[(-e + Pi/2 - f*x)/4]^2) - (3*m*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi
/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4]^3*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(-1 + 2
*m))/(3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 4*m*(AppellF1
[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*
m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*Tan[(-e + Pi/2 - f*x)/4]^2) - (3*AppellF1[1/
2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Tan[(-e + Pi/2 - f*x)/4]^2*(1 - Ta
n[(-e + Pi/2 - f*x)/4]^2)^(2*m)*(-2*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e +
Pi/2 - f*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2
])*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4] + 3*(-1/3*(m*AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-e +
Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4]) - (m*Appe
llF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4
]^2*Tan[(-e + Pi/2 - f*x)/4])/3) - 4*m*Tan[(-e + Pi/2 - f*x)/4]^2*((-6*m*AppellF1[5/2, 1 - 2*m, 1 + 2*m, 7/2,
Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/
5 + (3*(1 - 2*m)*AppellF1[5/2, 2 - 2*m, 2*m, 7/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec
[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/10 - (3*(1 + 2*m)*AppellF1[5/2, -2*m, 2 + 2*m, 7/2, Tan[(-e
+ Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/10)))/(3
*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 4*m*(AppellF1[3/2, 1
- 2*m, 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*m, 5/2,
Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*Tan[(-e + Pi/2 - f*x)/4]^2)^2))/2)) + (Hypergeometr
ic2F1[1/2, (-1 + 2*m)/2, (1 + 2*m)/2, Cos[(-e + Pi/2 - f*x)/2]^2]*(a + a*Sin[e + f*x])^m*Tan[(-e + Pi/2 - f*x)
/2])/(2*f*(-1 + 2*m)*Sqrt[Sin[(-e + Pi/2 - f*x)/2]^2])
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{m} \tan \left (f x + e\right )^{2}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^m*tan(f*x+e)^2,x, algorithm="fricas")
[Out]
integral((a*sin(f*x + e) + a)^m*tan(f*x + e)^2, x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \tan \left (f x + e\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^m*tan(f*x+e)^2,x, algorithm="giac")
[Out]
integrate((a*sin(f*x + e) + a)^m*tan(f*x + e)^2, x)
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maple [F] time = 0.22, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (\tan ^{2}\left (f x +e \right )\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(f*x+e))^m*tan(f*x+e)^2,x)
[Out]
int((a+a*sin(f*x+e))^m*tan(f*x+e)^2,x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \tan \left (f x + e\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^m*tan(f*x+e)^2,x, algorithm="maxima")
[Out]
integrate((a*sin(f*x + e) + a)^m*tan(f*x + e)^2, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(e + f*x)^2*(a + a*sin(e + f*x))^m,x)
[Out]
int(tan(e + f*x)^2*(a + a*sin(e + f*x))^m, x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \tan ^{2}{\left (e + f x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))**m*tan(f*x+e)**2,x)
[Out]
Integral((a*(sin(e + f*x) + 1))**m*tan(e + f*x)**2, x)
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