3.1.10 \(\int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx\) [10]

Optimal. Leaf size=39 \[ -a x+\frac {a \cos (c+d x)}{d}+\frac {a \cos (c+d x)}{d (1-\sin (c+d x))} \]

[Out]

-a*x+a*cos(d*x+c)/d+a*cos(d*x+c)/d/(1-sin(d*x+c))

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Rubi [A]
time = 0.08, antiderivative size = 39, normalized size of antiderivative = 1.00, 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 = {2787, 2825, 12, 2814, 2727} \begin {gather*} \frac {a \cos (c+d x)}{d}+\frac {a \cos (c+d x)}{d (1-\sin (c+d x))}-a x \end {gather*}

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2727

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]

Rule 2787

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 2814

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 2825

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]

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*}

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Mathematica [A]
time = 0.03, size = 47, normalized size = 1.21 \begin {gather*} -\frac {a \tan ^{-1}(\tan (c+d x))}{d}+\frac {a \cos (c+d x)}{d}+\frac {a \sec (c+d x)}{d}+\frac {a \tan (c+d x)}{d} \end {gather*}

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 = 0.13, size = 59, normalized size = 1.51

method	result	size
risch	\(-a x +\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {2 a}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\)	\(56\)
derivativedivides	\(\frac {a \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+a \left (\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(59\)
default	\(\frac {a \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+a \left (\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))*tan(d*x+c)^2,x,method=_RETURNVERBOSE)

[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 = 0.51, size = 39, normalized size = 1.00 \begin {gather*} -\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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (38) = 76\).
time = 0.34, size = 80, normalized size = 2.05 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 1008 vs. \(2 (38) = 76\).
time = 7.40, size = 1008, normalized size = 25.85 \begin {gather*} \text {Too large to display} \end {gather*}

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)

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Mupad [B]
time = 6.77, size = 111, normalized size = 2.85 \begin {gather*} \frac {\left (a\,\left (c+d\,x-2\right )-a\,\left (c+d\,x\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (a\,\left (c+d\,x\right )-a\,\left (c+d\,x-2\right )\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\left (c+d\,x\right )+a\,\left (c+d\,x-4\right )}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-a\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^2*(a + a*sin(c + d*x)),x)

[Out]

(tan(c/2 + (d*x)/2)*(a*(c + d*x) - a*(c + d*x - 2)) - tan(c/2 + (d*x)/2)^2*(a*(c + d*x) - a*(c + d*x - 2)) - a
*(c + d*x) + a*(c + d*x - 4))/(d*(tan(c/2 + (d*x)/2) - 1)*(tan(c/2 + (d*x)/2)^2 + 1)) - a*x

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