3.753 \(\int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx\)

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

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Rubi [A]  time = 0.10, 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 = {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 12

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

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]

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 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 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]

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 = 0.04, 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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fricas [B]  time = 0.44, size = 80, normalized size = 2.05 \[ -\frac {a d x - a \cos \left (d x + c\right )^{2} + {\left (a d x - 2 \, a\right )} \cos \left (d x + c\right ) - {\left (a d x - a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) - a}{d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c)),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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giac [B]  time = 0.17, size = 81, normalized size = 2.08 \[ -\frac {{\left (d x + c\right )} a + \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-((d*x + c)*a + 2*(a*tan(1/2*d*x + 1/2*c)^2 - a*tan(1/2*d*x + 1/2*c) + 2*a)/(tan(1/2*d*x + 1/2*c)^3 - tan(1/2*
d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c) - 1))/d

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maple [A]  time = 0.30, size = 59, normalized size = 1.51 \[ \frac {a \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+a \left (\tan \left (d x +c \right )-d x -c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c)),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 = 0.42, size = 39, normalized size = 1.00 \[ -\frac {{\left (d x + c - \tan \left (d x + c\right )\right )} a - a {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c)),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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mupad [B]  time = 9.21, size = 99, normalized size = 2.54 \[ \frac {\left (a\,\left (d\,x-2\right )-a\,d\,x\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (a\,d\,x-a\,\left (d\,x-2\right )\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\,\left (d\,x-4\right )-a\,d\,x}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-a\,x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*sin(d*x+c)**2*(a+a*sin(d*x+c)),x)

[Out]

a*(Integral(sin(c + d*x)**2*sec(c + d*x)**2, x) + Integral(sin(c + d*x)**3*sec(c + d*x)**2, x))

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