3.513 \(\int \cos (c+d x) (a+b \tan (c+d x)) \, dx\)

Optimal. Leaf size=24 \[ \frac {a \sin (c+d x)}{d}-\frac {b \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3486, 2637} \[ \frac {a \sin (c+d x)}{d}-\frac {b \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(a + b*Tan[c + d*x]),x]

[Out]

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rubi steps

\begin {align*} \int \cos (c+d x) (a+b \tan (c+d x)) \, dx &=-\frac {b \cos (c+d x)}{d}+a \int \cos (c+d x) \, dx\\ &=-\frac {b \cos (c+d x)}{d}+\frac {a \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 46, normalized size = 1.92 \[ \frac {a \sin (c) \cos (d x)}{d}+\frac {a \cos (c) \sin (d x)}{d}+\frac {b \sin (c) \sin (d x)}{d}-\frac {b \cos (c) \cos (d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(a + b*Tan[c + d*x]),x]

[Out]

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

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fricas [A]  time = 0.68, size = 23, normalized size = 0.96 \[ -\frac {b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*cos(d*x + c) - a*sin(d*x + c))/d

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giac [B]  time = 0.21, size = 129, normalized size = 5.38 \[ -\frac {b \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, a \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) + 2 \, a \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x\right )^{2} - 4 \, b \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a \tan \left (\frac {1}{2} \, d x\right ) - 2 \, a \tan \left (\frac {1}{2} \, c\right ) + b}{d \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + d \tan \left (\frac {1}{2} \, d x\right )^{2} + d \tan \left (\frac {1}{2} \, c\right )^{2} + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.24, size = 23, normalized size = 0.96 \[ \frac {a \sin \left (d x +c \right )-b \cos \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+b*tan(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.33, size = 23, normalized size = 0.96 \[ -\frac {b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*cos(d*x + c) - a*sin(d*x + c))/d

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mupad [B]  time = 3.72, size = 38, normalized size = 1.58 \[ -\frac {2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)*(a + b*tan(c + d*x)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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