3.10 \(\int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx\)
Optimal. Leaf size=39 \[ \frac{a \cos (c+d x)}{d}+\frac{a \cos (c+d x)}{d (1-\sin (c+d x))}-a x \]
[Out]
-(a*x) + (a*Cos[c + d*x])/d + (a*Cos[c + d*x])/(d*(1 - Sin[c + d*x]))
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Rubi [A] time = 0.104128, antiderivative size = 39, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used =
{2708, 2746, 12, 2735, 2648} \[ \frac{a \cos (c+d x)}{d}+\frac{a \cos (c+d x)}{d (1-\sin (c+d x))}-a x \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[c + d*x])*Tan[c + d*x]^2,x]
[Out]
-(a*x) + (a*Cos[c + d*x])/d + (a*Cos[c + d*x])/(d*(1 - Sin[c + d*x]))
Rule 2708
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Sin[
e + f*x]^p/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] &&
EqQ[p, 2*m]
Rule 2746
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b^2
*Cos[e + f*x])/(d*f), x] + Dist[1/d, Int[Simp[a^2*d - b*(b*c - 2*a*d)*Sin[e + f*x], x]/(c + d*Sin[e + f*x]), x
], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 2735
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]
Rule 2648
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Rubi steps
\begin{align*} \int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx &=a^2 \int \frac{\sin ^2(c+d x)}{a-a \sin (c+d x)} \, dx\\ &=\frac{a \cos (c+d x)}{d}+a \int \frac{a \sin (c+d x)}{a-a \sin (c+d x)} \, dx\\ &=\frac{a \cos (c+d x)}{d}+a^2 \int \frac{\sin (c+d x)}{a-a \sin (c+d x)} \, dx\\ &=-a x+\frac{a \cos (c+d x)}{d}+a^2 \int \frac{1}{a-a \sin (c+d x)} \, dx\\ &=-a x+\frac{a \cos (c+d x)}{d}+\frac{a^2 \cos (c+d x)}{d (a-a \sin (c+d x))}\\ \end{align*}
Mathematica [A] time = 0.0452538, size = 47, normalized size = 1.21 \[ \frac{a \cos (c+d x)}{d}-\frac{a \tan ^{-1}(\tan (c+d x))}{d}+\frac{a \tan (c+d x)}{d}+\frac{a \sec (c+d x)}{d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sin[c + d*x])*Tan[c + d*x]^2,x]
[Out]
-((a*ArcTan[Tan[c + d*x]])/d) + (a*Cos[c + d*x])/d + (a*Sec[c + d*x])/d + (a*Tan[c + d*x])/d
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Maple [A] time = 0.039, size = 59, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ( a \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{\cos \left ( dx+c \right ) }}+ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +a \left ( \tan \left ( dx+c \right ) -dx-c \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(d*x+c))*tan(d*x+c)^2,x)
[Out]
1/d*(a*(sin(d*x+c)^4/cos(d*x+c)+(2+sin(d*x+c)^2)*cos(d*x+c))+a*(tan(d*x+c)-d*x-c))
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Maxima [A] time = 1.68792, size = 53, normalized size = 1.36 \begin{align*} -\frac{{\left (d x + c - \tan \left (d x + c\right )\right )} a - a{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(d*x+c))*tan(d*x+c)^2,x, algorithm="maxima")
[Out]
-((d*x + c - tan(d*x + c))*a - a*(1/cos(d*x + c) + cos(d*x + c)))/d
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Fricas [B] time = 1.42136, size = 194, normalized size = 4.97 \begin{align*} -\frac{a d x - a \cos \left (d x + c\right )^{2} +{\left (a d x - 2 \, a\right )} \cos \left (d x + c\right ) -{\left (a d x - a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) - a}{d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(d*x+c))*tan(d*x+c)^2,x, algorithm="fricas")
[Out]
-(a*d*x - a*cos(d*x + c)^2 + (a*d*x - 2*a)*cos(d*x + c) - (a*d*x - a*cos(d*x + c) + a)*sin(d*x + c) - a)/(d*co
s(d*x + c) - d*sin(d*x + c) + d)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int \sin{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}\, dx + \int \tan ^{2}{\left (c + d x \right )}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(d*x+c))*tan(d*x+c)**2,x)
[Out]
a*(Integral(sin(c + d*x)*tan(c + d*x)**2, x) + Integral(tan(c + d*x)**2, x))
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Giac [B] time = 3.47197, size = 1361, normalized size = 34.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(d*x+c))*tan(d*x+c)^2,x, algorithm="giac")
[Out]
-(a*d*x*tan(d*x)*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c) - a*d*x*tan(1/2*d*x)^4*tan(1/2*c)^4 - 4*a*d*x*tan(d*x)*tan
(1/2*d*x)^3*tan(1/2*c)^3*tan(c) - 2*a*tan(d*x)*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c) + a*tan(d*x)*tan(1/2*d*x)^4*
tan(1/2*c)^4 + a*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c) + 4*a*d*x*tan(1/2*d*x)^3*tan(1/2*c)^3 + 2*a*tan(1/2*d*x)^4
*tan(1/2*c)^4 - a*d*x*tan(d*x)*tan(1/2*d*x)^4*tan(c) - 4*a*d*x*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c)*tan(c) - 4*a
*d*x*tan(d*x)*tan(1/2*d*x)*tan(1/2*c)^3*tan(c) + 8*a*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c) - a*d*x*tan(d
*x)*tan(1/2*c)^4*tan(c) - 4*a*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c)^3 - 4*a*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c) +
a*d*x*tan(1/2*d*x)^4 + 4*a*d*x*tan(1/2*d*x)^3*tan(1/2*c) + 4*a*d*x*tan(1/2*d*x)*tan(1/2*c)^3 - 8*a*tan(1/2*d*x
)^3*tan(1/2*c)^3 + a*d*x*tan(1/2*c)^4 - 2*a*tan(d*x)*tan(1/2*d*x)^4*tan(c) - 4*a*d*x*tan(d*x)*tan(1/2*d*x)*tan
(1/2*c)*tan(c) - 8*a*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c)*tan(c) - 24*a*tan(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2*tan
(c) - 8*a*tan(d*x)*tan(1/2*d*x)*tan(1/2*c)^3*tan(c) - 2*a*tan(d*x)*tan(1/2*c)^4*tan(c) - a*tan(d*x)*tan(1/2*d*
x)^4 - 4*a*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c) - 4*a*tan(d*x)*tan(1/2*d*x)*tan(1/2*c)^3 - a*tan(d*x)*tan(1/2*c)
^4 - a*tan(1/2*d*x)^4*tan(c) - 4*a*tan(1/2*d*x)^3*tan(1/2*c)*tan(c) - 4*a*tan(1/2*d*x)*tan(1/2*c)^3*tan(c) - a
*tan(1/2*c)^4*tan(c) + 2*a*tan(1/2*d*x)^4 + 4*a*d*x*tan(1/2*d*x)*tan(1/2*c) + 8*a*tan(1/2*d*x)^3*tan(1/2*c) +
24*a*tan(1/2*d*x)^2*tan(1/2*c)^2 + 8*a*tan(1/2*d*x)*tan(1/2*c)^3 + 2*a*tan(1/2*c)^4 + a*d*x*tan(d*x)*tan(c) +
8*a*tan(d*x)*tan(1/2*d*x)*tan(1/2*c)*tan(c) - 4*a*tan(d*x)*tan(1/2*d*x)*tan(1/2*c) - 4*a*tan(1/2*d*x)*tan(1/2*
c)*tan(c) - a*d*x - 8*a*tan(1/2*d*x)*tan(1/2*c) - 2*a*tan(d*x)*tan(c) + a*tan(d*x) + a*tan(c) + 2*a)/(d*tan(d*
x)*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c) - d*tan(1/2*d*x)^4*tan(1/2*c)^4 - 4*d*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c)
^3*tan(c) + 4*d*tan(1/2*d*x)^3*tan(1/2*c)^3 - d*tan(d*x)*tan(1/2*d*x)^4*tan(c) - 4*d*tan(d*x)*tan(1/2*d*x)^3*t
an(1/2*c)*tan(c) - 4*d*tan(d*x)*tan(1/2*d*x)*tan(1/2*c)^3*tan(c) - d*tan(d*x)*tan(1/2*c)^4*tan(c) + d*tan(1/2*
d*x)^4 + 4*d*tan(1/2*d*x)^3*tan(1/2*c) + 4*d*tan(1/2*d*x)*tan(1/2*c)^3 + d*tan(1/2*c)^4 - 4*d*tan(d*x)*tan(1/2
*d*x)*tan(1/2*c)*tan(c) + 4*d*tan(1/2*d*x)*tan(1/2*c) + d*tan(d*x)*tan(c) - d)