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

Optimal. Leaf size=88 \[ \frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a-b)^{5/2}}-\frac{\sin ^3(c+d x)}{3 d (a-b)}+\frac{(a-2 b) \sin (c+d x)}{d (a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.122464, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3676, 390, 208} \[ \frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a-b)^{5/2}}-\frac{\sin ^3(c+d x)}{3 d (a-b)}+\frac{(a-2 b) \sin (c+d x)}{d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.491502, size = 115, normalized size = 1.31 \[ \frac{\frac{6 b^2 \left (\log \left (\sqrt{a-b} \sin (c+d x)+\sqrt{a}\right )-\log \left (\sqrt{a}-\sqrt{a-b} \sin (c+d x)\right )\right )}{\sqrt{a} (a-b)^{5/2}}+\frac{3 (3 a-7 b) \sin (c+d x)}{(a-b)^2}+\frac{\sin (3 (c+d x))}{a-b}}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*b^2*(-Log[Sqrt[a] - Sqrt[a - b]*Sin[c + d*x]] + Log[Sqrt[a] + Sqrt[a - b]*Sin[c + d*x]]))/(Sqrt[a]*(a - b)
^(5/2)) + (3*(3*a - 7*b)*Sin[c + d*x])/(a - b)^2 + Sin[3*(c + d*x)]/(a - b))/(12*d)

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Maple [A]  time = 0.064, size = 98, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( -{\frac{1}{ \left ( a-b \right ) ^{2}} \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+2\,\sin \left ( dx+c \right ) b \right ) }+{\frac{{b}^{2}}{ \left ( a-b \right ) ^{2}}{\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 ) } \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),x)

[Out]

1/d*(-1/(a-b)^2*(1/3*sin(d*x+c)^3*a-1/3*b*sin(d*x+c)^3-sin(d*x+c)*a+2*sin(d*x+c)*b)+b^2/(a-b)^2/(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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.57248, size = 617, normalized size = 7.01 \begin{align*} \left [\frac{3 \, \sqrt{a^{2} - a b} b^{2} \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} - 7 \, a^{2} b + 5 \, a b^{2} +{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} d}, -\frac{3 \, \sqrt{-a^{2} + a b} b^{2} \arctan \left (\frac{\sqrt{-a^{2} + a b} \sin \left (d x + c\right )}{a}\right ) -{\left (2 \, a^{3} - 7 \, a^{2} b + 5 \, a b^{2} +{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \,{\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} d}\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),x, algorithm="fricas")

[Out]

[1/6*(3*sqrt(a^2 - a*b)*b^2*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 - 7*a^2*b + 5*a*b^2 + (a^3 - 2*a^2*b + a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/((
a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d), -1/3*(3*sqrt(-a^2 + a*b)*b^2*arctan(sqrt(-a^2 + a*b)*sin(d*x + c)/a) -
(2*a^3 - 7*a^2*b + 5*a*b^2 + (a^3 - 2*a^2*b + a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/((a^4 - 3*a^3*b + 3*a^2*b^2
- a*b^3)*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),x)

[Out]

Timed out

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Giac [B]  time = 1.65282, size = 217, normalized size = 2.47 \begin{align*} \frac{\frac{3 \, b^{2} \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^{2} - 2 \, a b + b^{2}\right )} \sqrt{-a^{2} + a b}} - \frac{a^{2} \sin \left (d x + c\right )^{3} - 2 \, a b \sin \left (d x + c\right )^{3} + b^{2} \sin \left (d x + c\right )^{3} - 3 \, a^{2} \sin \left (d x + c\right ) + 9 \, a b \sin \left (d x + c\right ) - 6 \, b^{2} \sin \left (d x + c\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}}}{3 \, 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),x, algorithm="giac")

[Out]

1/3*(3*b^2*arctan(-(a*sin(d*x + c) - b*sin(d*x + c))/sqrt(-a^2 + a*b))/((a^2 - 2*a*b + b^2)*sqrt(-a^2 + a*b))
- (a^2*sin(d*x + c)^3 - 2*a*b*sin(d*x + c)^3 + b^2*sin(d*x + c)^3 - 3*a^2*sin(d*x + c) + 9*a*b*sin(d*x + c) -
6*b^2*sin(d*x + c))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3))/d