3.455 \(\int \frac{\cos ^5(c+d x)}{a+b \tan ^2(c+d x)} \, dx\)

Optimal. Leaf size=126 \[ \frac{\left (a^2-3 a b+3 b^2\right ) \sin (c+d x)}{d (a-b)^3}-\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a-b)^{7/2}}+\frac{\sin ^5(c+d x)}{5 d (a-b)}-\frac{(2 a-3 b) \sin ^3(c+d x)}{3 d (a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.147802, antiderivative size = 126, 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} \[ \frac{\left (a^2-3 a b+3 b^2\right ) \sin (c+d x)}{d (a-b)^3}-\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a-b)^{7/2}}+\frac{\sin ^5(c+d x)}{5 d (a-b)}-\frac{(2 a-3 b) \sin ^3(c+d x)}{3 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5/(a + b*Tan[c + d*x]^2),x]

[Out]

-((b^3*ArcTanh[(Sqrt[a - b]*Sin[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)^(7/2)*d)) + ((a^2 - 3*a*b + 3*b^2)*Sin[c
+ d*x])/((a - b)^3*d) - ((2*a - 3*b)*Sin[c + d*x]^3)/(3*(a - b)^2*d) + Sin[c + d*x]^5/(5*(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 ^5(c+d x)}{a+b \tan ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^3}{a-(a-b) x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a^2-3 a b+3 b^2}{(a-b)^3}-\frac{(2 a-3 b) x^2}{(a-b)^2}+\frac{x^4}{a-b}-\frac{b^3}{(a-b)^3 \left (a-(a-b) x^2\right )}\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{\left (a^2-3 a b+3 b^2\right ) \sin (c+d x)}{(a-b)^3 d}-\frac{(2 a-3 b) \sin ^3(c+d x)}{3 (a-b)^2 d}+\frac{\sin ^5(c+d x)}{5 (a-b) d}-\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{a-(a-b) x^2} \, dx,x,\sin (c+d x)\right )}{(a-b)^3 d}\\ &=-\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} (a-b)^{7/2} d}+\frac{\left (a^2-3 a b+3 b^2\right ) \sin (c+d x)}{(a-b)^3 d}-\frac{(2 a-3 b) \sin ^3(c+d x)}{3 (a-b)^2 d}+\frac{\sin ^5(c+d x)}{5 (a-b) d}\\ \end{align*}

Mathematica [A]  time = 1.71354, size = 148, normalized size = 1.17 \[ \frac{\frac{30 \left (5 a^2-16 a b+19 b^2\right ) \sin (c+d x)}{(a-b)^3}+\frac{120 b^3 \left (\log \left (\sqrt{a}-\sqrt{a-b} \sin (c+d x)\right )-\log \left (\sqrt{a-b} \sin (c+d x)+\sqrt{a}\right )\right )}{\sqrt{a} (a-b)^{7/2}}+\frac{5 (5 a-9 b) \sin (3 (c+d x))}{(a-b)^2}+\frac{3 \sin (5 (c+d x))}{a-b}}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5/(a + b*Tan[c + d*x]^2),x]

[Out]

((120*b^3*(Log[Sqrt[a] - Sqrt[a - b]*Sin[c + d*x]] - Log[Sqrt[a] + Sqrt[a - b]*Sin[c + d*x]]))/(Sqrt[a]*(a - b
)^(7/2)) + (30*(5*a^2 - 16*a*b + 19*b^2)*Sin[c + d*x])/(a - b)^3 + (5*(5*a - 9*b)*Sin[3*(c + d*x)])/(a - b)^2
+ (3*Sin[5*(c + d*x)])/(a - b))/(240*d)

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Maple [A]  time = 0.071, size = 165, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({\frac{1}{ \left ( a-b \right ) ^{3}} \left ({\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5}}-{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}ab}{5}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5}}-{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}{a}^{2}}{3}}+{\frac{5\,ab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3}}- \left ( \sin \left ( dx+c \right ) \right ) ^{3}{b}^{2}+{a}^{2}\sin \left ( dx+c \right ) -3\,\sin \left ( dx+c \right ) ab+3\,{b}^{2}\sin \left ( dx+c \right ) \right ) }-{\frac{{b}^{3}}{ \left ( a-b \right ) ^{3}}{\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)^5/(a+b*tan(d*x+c)^2),x)

[Out]

1/d*(1/(a-b)^3*(1/5*a^2*sin(d*x+c)^5-2/5*sin(d*x+c)^5*a*b+1/5*b^2*sin(d*x+c)^5-2/3*sin(d*x+c)^3*a^2+5/3*a*b*si
n(d*x+c)^3-sin(d*x+c)^3*b^2+a^2*sin(d*x+c)-3*sin(d*x+c)*a*b+3*b^2*sin(d*x+c))-b^3/(a-b)^3/(a*(a-b))^(1/2)*arct
anh((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)^5/(a+b*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.76536, size = 892, normalized size = 7.08 \begin{align*} \left [-\frac{15 \, \sqrt{a^{2} - a b} b^{3} \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 (3 \,{\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{4} - 34 \, a^{3} b + 59 \, a^{2} b^{2} - 33 \, a b^{3} +{\left (4 \, a^{4} - 17 \, a^{3} b + 22 \, a^{2} b^{2} - 9 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{30 \,{\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} d}, \frac{15 \, \sqrt{-a^{2} + a b} b^{3} \arctan \left (\frac{\sqrt{-a^{2} + a b} \sin \left (d x + c\right )}{a}\right ) +{\left (3 \,{\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{4} - 34 \, a^{3} b + 59 \, a^{2} b^{2} - 33 \, a b^{3} +{\left (4 \, a^{4} - 17 \, a^{3} b + 22 \, a^{2} b^{2} - 9 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \,{\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+b*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

[-1/30*(15*sqrt(a^2 - a*b)*b^3*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*(3*(a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*cos(d*x + c)^4 + 8*a^4 - 34*a^3*b + 59*a^2*
b^2 - 33*a*b^3 + (4*a^4 - 17*a^3*b + 22*a^2*b^2 - 9*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))/((a^5 - 4*a^4*b + 6*a
^3*b^2 - 4*a^2*b^3 + a*b^4)*d), 1/15*(15*sqrt(-a^2 + a*b)*b^3*arctan(sqrt(-a^2 + a*b)*sin(d*x + c)/a) + (3*(a^
4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*cos(d*x + c)^4 + 8*a^4 - 34*a^3*b + 59*a^2*b^2 - 33*a*b^3 + (4*a^4 - 17*a^3*b
+ 22*a^2*b^2 - 9*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))/((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*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)**5/(a+b*tan(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.76499, size = 431, normalized size = 3.42 \begin{align*} -\frac{\frac{15 \, b^{3} \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} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt{-a^{2} + a b}} - \frac{3 \, a^{4} \sin \left (d x + c\right )^{5} - 12 \, a^{3} b \sin \left (d x + c\right )^{5} + 18 \, a^{2} b^{2} \sin \left (d x + c\right )^{5} - 12 \, a b^{3} \sin \left (d x + c\right )^{5} + 3 \, b^{4} \sin \left (d x + c\right )^{5} - 10 \, a^{4} \sin \left (d x + c\right )^{3} + 45 \, a^{3} b \sin \left (d x + c\right )^{3} - 75 \, a^{2} b^{2} \sin \left (d x + c\right )^{3} + 55 \, a b^{3} \sin \left (d x + c\right )^{3} - 15 \, b^{4} \sin \left (d x + c\right )^{3} + 15 \, a^{4} \sin \left (d x + c\right ) - 75 \, a^{3} b \sin \left (d x + c\right ) + 150 \, a^{2} b^{2} \sin \left (d x + c\right ) - 135 \, a b^{3} \sin \left (d x + c\right ) + 45 \, b^{4} \sin \left (d x + c\right )}{a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+b*tan(d*x+c)^2),x, algorithm="giac")

[Out]

-1/15*(15*b^3*arctan(-(a*sin(d*x + c) - b*sin(d*x + c))/sqrt(-a^2 + a*b))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqr
t(-a^2 + a*b)) - (3*a^4*sin(d*x + c)^5 - 12*a^3*b*sin(d*x + c)^5 + 18*a^2*b^2*sin(d*x + c)^5 - 12*a*b^3*sin(d*
x + c)^5 + 3*b^4*sin(d*x + c)^5 - 10*a^4*sin(d*x + c)^3 + 45*a^3*b*sin(d*x + c)^3 - 75*a^2*b^2*sin(d*x + c)^3
+ 55*a*b^3*sin(d*x + c)^3 - 15*b^4*sin(d*x + c)^3 + 15*a^4*sin(d*x + c) - 75*a^3*b*sin(d*x + c) + 150*a^2*b^2*
sin(d*x + c) - 135*a*b^3*sin(d*x + c) + 45*b^4*sin(d*x + c))/(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^
4 - b^5))/d