3.467 \(\int \frac{\cos ^3(c+d x)}{(a+b \tan ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=143 \[ \frac{b^2 (6 a-b) \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} d (a-b)^{7/2}}-\frac{b^3 \sin (c+d x)}{2 a d (a-b)^3 \left (a-(a-b) \sin ^2(c+d x)\right )}-\frac{\sin ^3(c+d x)}{3 d (a-b)^2}+\frac{(a-3 b) \sin (c+d x)}{d (a-b)^3} \]

[Out]

((6*a - b)*b^2*ArcTanh[(Sqrt[a - b]*Sin[c + d*x])/Sqrt[a]])/(2*a^(3/2)*(a - b)^(7/2)*d) + ((a - 3*b)*Sin[c + d
*x])/((a - b)^3*d) - Sin[c + d*x]^3/(3*(a - b)^2*d) - (b^3*Sin[c + d*x])/(2*a*(a - b)^3*d*(a - (a - b)*Sin[c +
d*x]^2))

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Rubi [A]  time = 0.212886, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3676, 390, 385, 208} \[ \frac{b^2 (6 a-b) \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} d (a-b)^{7/2}}-\frac{b^3 \sin (c+d x)}{2 a d (a-b)^3 \left (a-(a-b) \sin ^2(c+d x)\right )}-\frac{\sin ^3(c+d x)}{3 d (a-b)^2}+\frac{(a-3 b) \sin (c+d x)}{d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

((6*a - b)*b^2*ArcTanh[(Sqrt[a - b]*Sin[c + d*x])/Sqrt[a]])/(2*a^(3/2)*(a - b)^(7/2)*d) + ((a - 3*b)*Sin[c + d
*x])/((a - b)^3*d) - Sin[c + d*x]^3/(3*(a - b)^2*d) - (b^3*Sin[c + d*x])/(2*a*(a - b)^3*d*(a - (a - b)*Sin[c +
d*x]^2))

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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{\cos ^3(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 )^3}{\left (a-(a-b) x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a-3 b}{(a-b)^3}-\frac{x^2}{(a-b)^2}+\frac{(3 a-b) b^2-3 (a-b) b^2 x^2}{(a-b)^3 \left (a+(-a+b) x^2\right )^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{(a-3 b) \sin (c+d x)}{(a-b)^3 d}-\frac{\sin ^3(c+d x)}{3 (a-b)^2 d}+\frac{\operatorname{Subst}\left (\int \frac{(3 a-b) b^2-3 (a-b) b^2 x^2}{\left (a+(-a+b) x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{(a-b)^3 d}\\ &=\frac{(a-3 b) \sin (c+d x)}{(a-b)^3 d}-\frac{\sin ^3(c+d x)}{3 (a-b)^2 d}-\frac{b^3 \sin (c+d x)}{2 a (a-b)^3 d \left (a-(a-b) \sin ^2(c+d x)\right )}+\frac{\left ((6 a-b) b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+(-a+b) x^2} \, dx,x,\sin (c+d x)\right )}{2 a (a-b)^3 d}\\ &=\frac{(6 a-b) b^2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} (a-b)^{7/2} d}+\frac{(a-3 b) \sin (c+d x)}{(a-b)^3 d}-\frac{\sin ^3(c+d x)}{3 (a-b)^2 d}-\frac{b^3 \sin (c+d x)}{2 a (a-b)^3 d \left (a-(a-b) \sin ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.5229, size = 147, normalized size = 1.03 \[ \frac{\frac{3 b^2 (b-6 a) \left (\log \left (\sqrt{a}-\sqrt{a-b} \sin (c+d x)\right )-\log \left (\sqrt{a-b} \sin (c+d x)+\sqrt{a}\right )\right )}{a^{3/2} (a-b)^{7/2}}+\frac{3 \sin (c+d x) \left (-\frac{4 b^3}{a ((a-b) \cos (2 (c+d x))+a+b)}+3 a-11 b\right )}{(a-b)^3}+\frac{\sin (3 (c+d x))}{(a-b)^2}}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

((3*b^2*(-6*a + b)*(Log[Sqrt[a] - Sqrt[a - b]*Sin[c + d*x]] - Log[Sqrt[a] + Sqrt[a - b]*Sin[c + d*x]]))/(a^(3/
2)*(a - b)^(7/2)) + (3*(3*a - 11*b - (4*b^3)/(a*(a + b + (a - b)*Cos[2*(c + d*x)])))*Sin[c + d*x])/(a - b)^3 +
Sin[3*(c + d*x)]/(a - b)^2)/(12*d)

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Maple [A]  time = 0.092, size = 164, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ( -{\frac{1}{ \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) } \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}a}{3}}-{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3}}-\sin \left ( dx+c \right ) a+3\,\sin \left ( dx+c \right ) b \right ) }-{\frac{{b}^{2}}{ \left ( a-b \right ) ^{3}} \left ( -{\frac{\sin \left ( dx+c \right ) b}{2\,a \left ( a \left ( \sin \left ( dx+c \right ) \right ) ^{2}-b \left ( \sin \left ( dx+c \right ) \right ) ^{2}-a \right ) }}-{\frac{6\,a-b}{2\,a}{\it Artanh} \left ({ \left ( a-b \right ) \sin \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a+b*tan(d*x+c)^2)^2,x)

[Out]

1/d*(-1/(a^2-2*a*b+b^2)/(a-b)*(1/3*sin(d*x+c)^3*a-1/3*b*sin(d*x+c)^3-sin(d*x+c)*a+3*sin(d*x+c)*b)-b^2/(a-b)^3*
(-1/2*b/a*sin(d*x+c)/(a*sin(d*x+c)^2-b*sin(d*x+c)^2-a)-1/2*(6*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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.08064, size = 1307, normalized size = 9.14 \begin{align*} \left [\frac{3 \,{\left (6 \, a b^{3} - b^{4} +{\left (6 \, a^{2} b^{2} - 7 \, a b^{3} + b^{4}\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 (4 \, a^{4} b - 20 \, a^{3} b^{2} + 13 \, a^{2} b^{3} + 3 \, a b^{4} + 2 \,{\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (2 \, a^{5} - 11 \, a^{4} b + 16 \, a^{3} b^{2} - 7 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \,{\left ({\left (a^{7} - 5 \, a^{6} b + 10 \, a^{5} b^{2} - 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} - a^{2} b^{5}\right )} d \cos \left (d x + c\right )^{2} +{\left (a^{6} b - 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} - 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} d\right )}}, -\frac{3 \,{\left (6 \, a b^{3} - b^{4} +{\left (6 \, a^{2} b^{2} - 7 \, a b^{3} + b^{4}\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 (4 \, a^{4} b - 20 \, a^{3} b^{2} + 13 \, a^{2} b^{3} + 3 \, a b^{4} + 2 \,{\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (2 \, a^{5} - 11 \, a^{4} b + 16 \, a^{3} b^{2} - 7 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \,{\left ({\left (a^{7} - 5 \, a^{6} b + 10 \, a^{5} b^{2} - 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} - a^{2} b^{5}\right )} d \cos \left (d x + c\right )^{2} +{\left (a^{6} b - 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} - 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

[1/12*(3*(6*a*b^3 - b^4 + (6*a^2*b^2 - 7*a*b^3 + b^4)*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*(4*a^4*b - 20*a^3*b^2 + 13*
a^2*b^3 + 3*a*b^4 + 2*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*cos(d*x + c)^4 + 2*(2*a^5 - 11*a^4*b + 16*a^3*b^2
- 7*a^2*b^3)*cos(d*x + c)^2)*sin(d*x + c))/((a^7 - 5*a^6*b + 10*a^5*b^2 - 10*a^4*b^3 + 5*a^3*b^4 - a^2*b^5)*d*
cos(d*x + c)^2 + (a^6*b - 4*a^5*b^2 + 6*a^4*b^3 - 4*a^3*b^4 + a^2*b^5)*d), -1/6*(3*(6*a*b^3 - b^4 + (6*a^2*b^2
- 7*a*b^3 + b^4)*cos(d*x + c)^2)*sqrt(-a^2 + a*b)*arctan(sqrt(-a^2 + a*b)*sin(d*x + c)/a) - (4*a^4*b - 20*a^3
*b^2 + 13*a^2*b^3 + 3*a*b^4 + 2*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*cos(d*x + c)^4 + 2*(2*a^5 - 11*a^4*b + 1
6*a^3*b^2 - 7*a^2*b^3)*cos(d*x + c)^2)*sin(d*x + c))/((a^7 - 5*a^6*b + 10*a^5*b^2 - 10*a^4*b^3 + 5*a^3*b^4 - a
^2*b^5)*d*cos(d*x + c)^2 + (a^6*b - 4*a^5*b^2 + 6*a^4*b^3 - 4*a^3*b^4 + a^2*b^5)*d)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3/(a+b*tan(d*x+c)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.75014, size = 444, normalized size = 3.1 \begin{align*} \frac{\frac{3 \, b^{3} \sin \left (d x + c\right )}{{\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )}{\left (a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right )^{2} - a\right )}} + \frac{3 \,{\left (6 \, a b^{2} - b^{3}\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^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \sqrt{-a^{2} + a b}} - \frac{2 \,{\left (a^{4} \sin \left (d x + c\right )^{3} - 4 \, a^{3} b \sin \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \sin \left (d x + c\right )^{3} - 4 \, a b^{3} \sin \left (d x + c\right )^{3} + b^{4} \sin \left (d x + c\right )^{3} - 3 \, a^{4} \sin \left (d x + c\right ) + 18 \, a^{3} b \sin \left (d x + c\right ) - 36 \, a^{2} b^{2} \sin \left (d x + c\right ) + 30 \, a b^{3} \sin \left (d x + c\right ) - 9 \, b^{4} \sin \left (d x + c\right )\right )}}{a^{6} - 6 \, a^{5} b + 15 \, a^{4} b^{2} - 20 \, a^{3} b^{3} + 15 \, a^{2} b^{4} - 6 \, a b^{5} + b^{6}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/6*(3*b^3*sin(d*x + c)/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*(a*sin(d*x + c)^2 - b*sin(d*x + c)^2 - a)) + 3*(6
*a*b^2 - b^3)*arctan(-(a*sin(d*x + c) - b*sin(d*x + c))/sqrt(-a^2 + a*b))/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)
*sqrt(-a^2 + a*b)) - 2*(a^4*sin(d*x + c)^3 - 4*a^3*b*sin(d*x + c)^3 + 6*a^2*b^2*sin(d*x + c)^3 - 4*a*b^3*sin(d
*x + c)^3 + b^4*sin(d*x + c)^3 - 3*a^4*sin(d*x + c) + 18*a^3*b*sin(d*x + c) - 36*a^2*b^2*sin(d*x + c) + 30*a*b
^3*sin(d*x + c) - 9*b^4*sin(d*x + c))/(a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6))/
d