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

Optimal. Leaf size=49 \[ -\frac {\sin (c+d x) \cos (c+d x) (a+b \tan (c+d x))}{2 d}+\frac {a x}{2}-\frac {b \log (\cos (c+d x))}{d} \]

[Out]

1/2*a*x-b*ln(cos(d*x+c))/d-1/2*cos(d*x+c)*sin(d*x+c)*(a+b*tan(d*x+c))/d

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Rubi [A]  time = 0.08, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {819, 635, 203, 260} \[ -\frac {\sin (c+d x) \cos (c+d x) (a+b \tan (c+d x))}{2 d}+\frac {a x}{2}-\frac {b \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^2*(a + b*Tan[c + d*x]),x]

[Out]

(a*x)/2 - (b*Log[Cos[c + d*x]])/d - (Cos[c + d*x]*Sin[c + d*x]*(a + b*Tan[c + d*x]))/(2*d)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rubi steps

\begin {align*} \int \sin ^2(c+d x) (a+b \tan (c+d x)) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2 (a+b x)}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))}{2 d}+\frac {\operatorname {Subst}\left (\int \frac {a+2 b x}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac {\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))}{2 d}+\frac {a \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {b \operatorname {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {a x}{2}-\frac {b \log (\cos (c+d x))}{d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 56, normalized size = 1.14 \[ \frac {a (c+d x)}{2 d}-\frac {a \sin (2 (c+d x))}{4 d}-\frac {b \left (\log (\cos (c+d x))-\frac {1}{2} \cos ^2(c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^2*(a + b*Tan[c + d*x]),x]

[Out]

(a*(c + d*x))/(2*d) - (b*(-1/2*Cos[c + d*x]^2 + Log[Cos[c + d*x]]))/d - (a*Sin[2*(c + d*x)])/(4*d)

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fricas [A]  time = 0.54, size = 47, normalized size = 0.96 \[ \frac {a d x + b \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, b \log \left (-\cos \left (d x + c\right )\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2*(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(a*d*x + b*cos(d*x + c)^2 - a*cos(d*x + c)*sin(d*x + c) - 2*b*log(-cos(d*x + c)))/d

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giac [B]  time = 0.54, size = 413, normalized size = 8.43 \[ \frac {2 \, a d x \tan \left (d x\right )^{2} \tan \relax (c)^{2} - 2 \, b \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{4} \tan \relax (c)^{2} - 2 \, \tan \left (d x\right )^{3} \tan \relax (c) + \tan \left (d x\right )^{2} \tan \relax (c)^{2} + \tan \left (d x\right )^{2} - 2 \, \tan \left (d x\right ) \tan \relax (c) + 1\right )}}{\tan \relax (c)^{2} + 1}\right ) \tan \left (d x\right )^{2} \tan \relax (c)^{2} + 2 \, a d x \tan \left (d x\right )^{2} + 2 \, a d x \tan \relax (c)^{2} + b \tan \left (d x\right )^{2} \tan \relax (c)^{2} - 2 \, b \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{4} \tan \relax (c)^{2} - 2 \, \tan \left (d x\right )^{3} \tan \relax (c) + \tan \left (d x\right )^{2} \tan \relax (c)^{2} + \tan \left (d x\right )^{2} - 2 \, \tan \left (d x\right ) \tan \relax (c) + 1\right )}}{\tan \relax (c)^{2} + 1}\right ) \tan \left (d x\right )^{2} + 2 \, a \tan \left (d x\right )^{2} \tan \relax (c) - 2 \, b \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{4} \tan \relax (c)^{2} - 2 \, \tan \left (d x\right )^{3} \tan \relax (c) + \tan \left (d x\right )^{2} \tan \relax (c)^{2} + \tan \left (d x\right )^{2} - 2 \, \tan \left (d x\right ) \tan \relax (c) + 1\right )}}{\tan \relax (c)^{2} + 1}\right ) \tan \relax (c)^{2} + 2 \, a \tan \left (d x\right ) \tan \relax (c)^{2} + 2 \, a d x - b \tan \left (d x\right )^{2} - 4 \, b \tan \left (d x\right ) \tan \relax (c) - b \tan \relax (c)^{2} - 2 \, b \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{4} \tan \relax (c)^{2} - 2 \, \tan \left (d x\right )^{3} \tan \relax (c) + \tan \left (d x\right )^{2} \tan \relax (c)^{2} + \tan \left (d x\right )^{2} - 2 \, \tan \left (d x\right ) \tan \relax (c) + 1\right )}}{\tan \relax (c)^{2} + 1}\right ) - 2 \, a \tan \left (d x\right ) - 2 \, a \tan \relax (c) + b}{4 \, {\left (d \tan \left (d x\right )^{2} \tan \relax (c)^{2} + d \tan \left (d x\right )^{2} + d \tan \relax (c)^{2} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*d*x*tan(d*x)^2*tan(c)^2 - 2*b*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2
+ tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 + 2*a*d*x*tan(d*x)^2 + 2*a*d*x*tan(c
)^2 + b*tan(d*x)^2*tan(c)^2 - 2*b*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan
(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^2 + 2*a*tan(d*x)^2*tan(c) - 2*b*log(4*(tan(d*x)^4*ta
n(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(c
)^2 + 2*a*tan(d*x)*tan(c)^2 + 2*a*d*x - b*tan(d*x)^2 - 4*b*tan(d*x)*tan(c) - b*tan(c)^2 - 2*b*log(4*(tan(d*x)^
4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1)) -
2*a*tan(d*x) - 2*a*tan(c) + b)/(d*tan(d*x)^2*tan(c)^2 + d*tan(d*x)^2 + d*tan(c)^2 + d)

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maple [A]  time = 0.18, size = 58, normalized size = 1.18 \[ -\frac {a \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {a x}{2}+\frac {c a}{2 d}-\frac {b \left (\sin ^{2}\left (d x +c \right )\right )}{2 d}-\frac {b \ln \left (\cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*cos(d*x+c)*sin(d*x+c)/d+1/2*a*x+1/2/d*c*a-1/2/d*b*sin(d*x+c)^2-b*ln(cos(d*x+c))/d

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maxima [A]  time = 0.57, size = 52, normalized size = 1.06 \[ \frac {{\left (d x + c\right )} a + b \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - \frac {a \tan \left (d x + c\right ) - b}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2*(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.82, size = 50, normalized size = 1.02 \[ \frac {\frac {b\,{\cos \left (c+d\,x\right )}^2}{2}-\frac {a\,\sin \left (c+d\,x\right )\,\cos \left (c+d\,x\right )}{2}+\frac {b\,\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}{2}+\frac {a\,d\,x}{2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*log(tan(c + d*x)^2 + 1))/2 + (b*cos(c + d*x)^2)/2 - (a*cos(c + d*x)*sin(c + d*x))/2 + (a*d*x)/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*tan(c + d*x))*sin(c + d*x)**2, x)

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