3.466 \(\int \frac {\cos (c+d x)}{(a+b \tan ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=114 \[ -\frac {b (4 a-b) \tanh ^{-1}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} d (a-b)^{5/2}}+\frac {b^2 \sin (c+d x)}{2 a d (a-b)^2 \left (a-(a-b) \sin ^2(c+d x)\right )}+\frac {\sin (c+d x)}{d (a-b)^2} \]

[Out]

-1/2*(4*a-b)*b*arctanh(sin(d*x+c)*(a-b)^(1/2)/a^(1/2))/a^(3/2)/(a-b)^(5/2)/d+sin(d*x+c)/(a-b)^2/d+1/2*b^2*sin(
d*x+c)/a/(a-b)^2/d/(a-(a-b)*sin(d*x+c)^2)

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Rubi [A]  time = 0.18, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3676, 390, 385, 208} \[ -\frac {b (4 a-b) \tanh ^{-1}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} d (a-b)^{5/2}}+\frac {b^2 \sin (c+d x)}{2 a d (a-b)^2 \left (a-(a-b) \sin ^2(c+d x)\right )}+\frac {\sin (c+d x)}{d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((4*a - b)*b*ArcTanh[(Sqrt[a - b]*Sin[c + d*x])/Sqrt[a]])/(2*a^(3/2)*(a - b)^(5/2)*d) + Sin[c + d*x]/((a - b)
^2*d) + (b^2*Sin[c + d*x])/(2*a*(a - b)^2*d*(a - (a - b)*Sin[c + d*x]^2))

Rule 208

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

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 3676

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 -
ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n/2] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\cos (c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2}{\left (a-(a-b) x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{(a-b)^2}-\frac {(2 a-b) b-2 (a-b) b x^2}{(a-b)^2 \left (a+(-a+b) x^2\right )^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {\sin (c+d x)}{(a-b)^2 d}-\frac {\operatorname {Subst}\left (\int \frac {(2 a-b) b-2 (a-b) b x^2}{\left (a+(-a+b) x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{(a-b)^2 d}\\ &=\frac {\sin (c+d x)}{(a-b)^2 d}+\frac {b^2 \sin (c+d x)}{2 a (a-b)^2 d \left (a-(a-b) \sin ^2(c+d x)\right )}-\frac {((4 a-b) b) \operatorname {Subst}\left (\int \frac {1}{a+(-a+b) x^2} \, dx,x,\sin (c+d x)\right )}{2 a (a-b)^2 d}\\ &=-\frac {(4 a-b) b \tanh ^{-1}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} (a-b)^{5/2} d}+\frac {\sin (c+d x)}{(a-b)^2 d}+\frac {b^2 \sin (c+d x)}{2 a (a-b)^2 d \left (a-(a-b) \sin ^2(c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.56, size = 119, normalized size = 1.04 \[ \frac {-\frac {\sqrt {a} \sin (c+d x) \left (a^2+a (a-b) \cos (2 (c+d x))+a b+b^2\right )}{(a-b)^2 \left ((a-b) \sin ^2(c+d x)-a\right )}-\frac {b (4 a-b) \tanh ^{-1}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{(a-b)^{5/2}}}{2 a^{3/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(((4*a - b)*b*ArcTanh[(Sqrt[a - b]*Sin[c + d*x])/Sqrt[a]])/(a - b)^(5/2)) - (Sqrt[a]*(a^2 + a*b + b^2 + a*(a
- b)*Cos[2*(c + d*x)])*Sin[c + d*x])/((a - b)^2*(-a + (a - b)*Sin[c + d*x]^2)))/(2*a^(3/2)*d)

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fricas [B]  time = 0.57, size = 451, normalized size = 3.96 \[ \left [-\frac {{\left (4 \, a b^{2} - b^{3} + {\left (4 \, a^{2} b - 5 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a^{2} - a b} \log \left (-\frac {{\left (a - b\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - a b} \sin \left (d x + c\right ) - 2 \, a + b}{{\left (a - b\right )} \cos \left (d x + c\right )^{2} + b}\right ) - 2 \, {\left (2 \, a^{3} b - a^{2} b^{2} - a b^{3} + 2 \, {\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{6} - 4 \, a^{5} b + 6 \, a^{4} b^{2} - 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{5} b - 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} - a^{2} b^{4}\right )} d\right )}}, \frac {{\left (4 \, a b^{2} - b^{3} + {\left (4 \, a^{2} b - 5 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a^{2} + a b} \arctan \left (\frac {\sqrt {-a^{2} + a b} \sin \left (d x + c\right )}{a}\right ) + {\left (2 \, a^{3} b - a^{2} b^{2} - a b^{3} + 2 \, {\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} - 4 \, a^{5} b + 6 \, a^{4} b^{2} - 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{5} b - 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} - a^{2} b^{4}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((4*a*b^2 - b^3 + (4*a^2*b - 5*a*b^2 + b^3)*cos(d*x + c)^2)*sqrt(a^2 - a*b)*log(-((a - b)*cos(d*x + c)^2
- 2*sqrt(a^2 - a*b)*sin(d*x + c) - 2*a + b)/((a - b)*cos(d*x + c)^2 + b)) - 2*(2*a^3*b - a^2*b^2 - a*b^3 + 2*
(a^4 - 2*a^3*b + a^2*b^2)*cos(d*x + c)^2)*sin(d*x + c))/((a^6 - 4*a^5*b + 6*a^4*b^2 - 4*a^3*b^3 + a^2*b^4)*d*c
os(d*x + c)^2 + (a^5*b - 3*a^4*b^2 + 3*a^3*b^3 - a^2*b^4)*d), 1/2*((4*a*b^2 - b^3 + (4*a^2*b - 5*a*b^2 + b^3)*
cos(d*x + c)^2)*sqrt(-a^2 + a*b)*arctan(sqrt(-a^2 + a*b)*sin(d*x + c)/a) + (2*a^3*b - a^2*b^2 - a*b^3 + 2*(a^4
- 2*a^3*b + a^2*b^2)*cos(d*x + c)^2)*sin(d*x + c))/((a^6 - 4*a^5*b + 6*a^4*b^2 - 4*a^3*b^3 + a^2*b^4)*d*cos(d
*x + c)^2 + (a^5*b - 3*a^4*b^2 + 3*a^3*b^3 - a^2*b^4)*d)]

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giac [A]  time = 1.89, size = 152, normalized size = 1.33 \[ -\frac {\frac {b^{2} \sin \left (d x + c\right )}{{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} {\left (a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right )^{2} - a\right )}} + \frac {{\left (4 \, a b - b^{2}\right )} \arctan \left (-\frac {a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )}{\sqrt {-a^{2} + a b}}\right )}{{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \sqrt {-a^{2} + a b}} - \frac {2 \, \sin \left (d x + c\right )}{a^{2} - 2 \, a b + b^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^2*sin(d*x + c)/((a^3 - 2*a^2*b + a*b^2)*(a*sin(d*x + c)^2 - b*sin(d*x + c)^2 - a)) + (4*a*b - b^2)*arc
tan(-(a*sin(d*x + c) - b*sin(d*x + c))/sqrt(-a^2 + a*b))/((a^3 - 2*a^2*b + a*b^2)*sqrt(-a^2 + a*b)) - 2*sin(d*
x + c)/(a^2 - 2*a*b + b^2))/d

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maple [A]  time = 0.62, size = 118, normalized size = 1.04 \[ \frac {\frac {\sin \left (d x +c \right )}{a^{2}-2 a b +b^{2}}+\frac {b \left (-\frac {b \sin \left (d x +c \right )}{2 a \left (a \left (\sin ^{2}\left (d x +c \right )\right )-b \left (\sin ^{2}\left (d x +c \right )\right )-a \right )}-\frac {\left (4 a -b \right ) \arctanh \left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{2 a \sqrt {a \left (a -b \right )}}\right )}{\left (a -b \right )^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is b-a positive or negative?

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mupad [B]  time = 15.55, size = 269, normalized size = 2.36 \[ \frac {\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2+b^2\right )}{a\,{\left (a-b\right )}^2}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,a^2+b^2\right )}{a\,{\left (a-b\right )}^2}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (-2\,a^2+4\,a\,b+b^2\right )}{a\,{\left (a-b\right )}^2}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (4\,b-a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (4\,b-a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}+\frac {b\,\mathrm {atan}\left (\frac {2{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3-6{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b+6{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^2-2{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}{\sqrt {a}\,{\left (a-b\right )}^{5/2}\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}\right )\,\left (4\,a-b\right )\,1{}\mathrm {i}}{2\,a^{3/2}\,d\,{\left (a-b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((tan(c/2 + (d*x)/2)*(2*a^2 + b^2))/(a*(a - b)^2) + (tan(c/2 + (d*x)/2)^5*(2*a^2 + b^2))/(a*(a - b)^2) + (2*ta
n(c/2 + (d*x)/2)^3*(4*a*b - 2*a^2 + b^2))/(a*(a - b)^2))/(d*(a - tan(c/2 + (d*x)/2)^2*(a - 4*b) - tan(c/2 + (d
*x)/2)^4*(a - 4*b) + a*tan(c/2 + (d*x)/2)^6)) + (b*atan((a^3*tan(c/2 + (d*x)/2)*2i - b^3*tan(c/2 + (d*x)/2)*2i
+ a*b^2*tan(c/2 + (d*x)/2)*6i - a^2*b*tan(c/2 + (d*x)/2)*6i)/(a^(1/2)*(a - b)^(5/2)*(tan(c/2 + (d*x)/2)^2 + 1
)))*(4*a - b)*1i)/(2*a^(3/2)*d*(a - b)^(5/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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