3.473 \(\int \frac{\cos ^4(c+d x)}{(a+b \tan ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=212 \[ -\frac{b^{5/2} (7 a-b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} d (a-b)^4}+\frac{x \left (3 a^2-14 a b+35 b^2\right )}{8 (a-b)^4}+\frac{b (a-4 b) (3 a+b) \tan (c+d x)}{8 a d (a-b)^3 \left (a+b \tan ^2(c+d x)\right )}+\frac{\sin (c+d x) \cos ^3(c+d x)}{4 d (a-b) \left (a+b \tan ^2(c+d x)\right )}+\frac{3 (a-3 b) \sin (c+d x) \cos (c+d x)}{8 d (a-b)^2 \left (a+b \tan ^2(c+d x)\right )} \]

[Out]

((3*a^2 - 14*a*b + 35*b^2)*x)/(8*(a - b)^4) - ((7*a - b)*b^(5/2)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(2*a^
(3/2)*(a - b)^4*d) + (3*(a - 3*b)*Cos[c + d*x]*Sin[c + d*x])/(8*(a - b)^2*d*(a + b*Tan[c + d*x]^2)) + (Cos[c +
d*x]^3*Sin[c + d*x])/(4*(a - b)*d*(a + b*Tan[c + d*x]^2)) + ((a - 4*b)*b*(3*a + b)*Tan[c + d*x])/(8*a*(a - b)
^3*d*(a + b*Tan[c + d*x]^2))

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Rubi [A]  time = 0.301012, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3675, 414, 527, 522, 203, 205} \[ -\frac{b^{5/2} (7 a-b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} d (a-b)^4}+\frac{x \left (3 a^2-14 a b+35 b^2\right )}{8 (a-b)^4}+\frac{b (a-4 b) (3 a+b) \tan (c+d x)}{8 a d (a-b)^3 \left (a+b \tan ^2(c+d x)\right )}+\frac{\sin (c+d x) \cos ^3(c+d x)}{4 d (a-b) \left (a+b \tan ^2(c+d x)\right )}+\frac{3 (a-3 b) \sin (c+d x) \cos (c+d x)}{8 d (a-b)^2 \left (a+b \tan ^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

((3*a^2 - 14*a*b + 35*b^2)*x)/(8*(a - b)^4) - ((7*a - b)*b^(5/2)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(2*a^
(3/2)*(a - b)^4*d) + (3*(a - 3*b)*Cos[c + d*x]*Sin[c + d*x])/(8*(a - b)^2*d*(a + b*Tan[c + d*x]^2)) + (Cos[c +
d*x]^3*Sin[c + d*x])/(4*(a - b)*d*(a + b*Tan[c + d*x]^2)) + ((a - 4*b)*b*(3*a + b)*Tan[c + d*x])/(8*a*(a - b)
^3*d*(a + b*Tan[c + d*x]^2))

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\cos ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right )^3 \left (a+b x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\cos ^3(c+d x) \sin (c+d x)}{4 (a-b) d \left (a+b \tan ^2(c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{-3 a+4 b-5 b x^2}{\left (1+x^2\right )^2 \left (a+b x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{4 (a-b) d}\\ &=\frac{3 (a-3 b) \cos (c+d x) \sin (c+d x)}{8 (a-b)^2 d \left (a+b \tan ^2(c+d x)\right )}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 (a-b) d \left (a+b \tan ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{3 a^2-5 a b+8 b^2+9 (a-3 b) b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{8 (a-b)^2 d}\\ &=\frac{3 (a-3 b) \cos (c+d x) \sin (c+d x)}{8 (a-b)^2 d \left (a+b \tan ^2(c+d x)\right )}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 (a-b) d \left (a+b \tan ^2(c+d x)\right )}+\frac{(a-4 b) b (3 a+b) \tan (c+d x)}{8 a (a-b)^3 d \left (a+b \tan ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{2 \left (3 a^3-11 a^2 b+24 a b^2-4 b^3\right )+2 (a-4 b) b (3 a+b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (c+d x)\right )}{16 a (a-b)^3 d}\\ &=\frac{3 (a-3 b) \cos (c+d x) \sin (c+d x)}{8 (a-b)^2 d \left (a+b \tan ^2(c+d x)\right )}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 (a-b) d \left (a+b \tan ^2(c+d x)\right )}+\frac{(a-4 b) b (3 a+b) \tan (c+d x)}{8 a (a-b)^3 d \left (a+b \tan ^2(c+d x)\right )}-\frac{\left ((7 a-b) b^3\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{2 a (a-b)^4 d}+\frac{\left (3 a^2-14 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 (a-b)^4 d}\\ &=\frac{\left (3 a^2-14 a b+35 b^2\right ) x}{8 (a-b)^4}-\frac{(7 a-b) b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} (a-b)^4 d}+\frac{3 (a-3 b) \cos (c+d x) \sin (c+d x)}{8 (a-b)^2 d \left (a+b \tan ^2(c+d x)\right )}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 (a-b) d \left (a+b \tan ^2(c+d x)\right )}+\frac{(a-4 b) b (3 a+b) \tan (c+d x)}{8 a (a-b)^3 d \left (a+b \tan ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 2.0214, size = 148, normalized size = 0.7 \[ \frac{4 \left (3 a^2-14 a b+35 b^2\right ) (c+d x)+\frac{16 b^{5/2} (b-7 a) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{a^{3/2}}-\frac{16 b^3 (a-b) \sin (2 (c+d x))}{a ((a-b) \cos (2 (c+d x))+a+b)}+8 (a-3 b) (a-b) \sin (2 (c+d x))+(a-b)^2 \sin (4 (c+d x))}{32 d (a-b)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

(4*(3*a^2 - 14*a*b + 35*b^2)*(c + d*x) + (16*b^(5/2)*(-7*a + b)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/a^(3/2
) + 8*(a - 3*b)*(a - b)*Sin[2*(c + d*x)] - (16*(a - b)*b^3*Sin[2*(c + d*x)])/(a*(a + b + (a - b)*Cos[2*(c + d*
x)])) + (a - b)^2*Sin[4*(c + d*x)])/(32*(a - b)^4*d)

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Maple [B]  time = 0.088, size = 413, normalized size = 2. \begin{align*} -{\frac{{b}^{3}\tan \left ( dx+c \right ) }{2\,d \left ( a-b \right ) ^{4} \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{{b}^{4}\tan \left ( dx+c \right ) }{2\,d \left ( a-b \right ) ^{4}a \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}-{\frac{7\,{b}^{3}}{2\,d \left ( a-b \right ) ^{4}}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{{b}^{4}}{2\,d \left ( a-b \right ) ^{4}a}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3\, \left ( \tan \left ( dx+c \right ) \right ) ^{3}{a}^{2}}{8\,d \left ( a-b \right ) ^{4} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{7\, \left ( \tan \left ( dx+c \right ) \right ) ^{3}ab}{4\,d \left ( a-b \right ) ^{4} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{11\, \left ( \tan \left ( dx+c \right ) \right ) ^{3}{b}^{2}}{8\,d \left ( a-b \right ) ^{4} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{9\,a\tan \left ( dx+c \right ) b}{4\,d \left ( a-b \right ) ^{4} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{13\,\tan \left ( dx+c \right ){b}^{2}}{8\,d \left ( a-b \right ) ^{4} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{5\,{a}^{2}\tan \left ( dx+c \right ) }{8\,d \left ( a-b \right ) ^{4} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{35\,\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{2}}{8\,d \left ( a-b \right ) ^{4}}}+{\frac{3\,\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}}{8\,d \left ( a-b \right ) ^{4}}}-{\frac{7\,\arctan \left ( \tan \left ( dx+c \right ) \right ) ab}{4\,d \left ( a-b \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a+b*tan(d*x+c)^2)^2,x)

[Out]

-1/2/d*b^3/(a-b)^4*tan(d*x+c)/(a+b*tan(d*x+c)^2)+1/2/d*b^4/(a-b)^4/a*tan(d*x+c)/(a+b*tan(d*x+c)^2)-7/2/d*b^3/(
a-b)^4/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2))+1/2/d*b^4/(a-b)^4/a/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^
(1/2))+3/8/d/(a-b)^4/(tan(d*x+c)^2+1)^2*tan(d*x+c)^3*a^2-7/4/d/(a-b)^4/(tan(d*x+c)^2+1)^2*tan(d*x+c)^3*a*b+11/
8/d/(a-b)^4/(tan(d*x+c)^2+1)^2*tan(d*x+c)^3*b^2-9/4/d/(a-b)^4/(tan(d*x+c)^2+1)^2*tan(d*x+c)*a*b+13/8/d/(a-b)^4
/(tan(d*x+c)^2+1)^2*tan(d*x+c)*b^2+5/8/d/(a-b)^4/(tan(d*x+c)^2+1)^2*tan(d*x+c)*a^2+35/8/d/(a-b)^4*arctan(tan(d
*x+c))*b^2+3/8/d/(a-b)^4*arctan(tan(d*x+c))*a^2-7/4/d/(a-b)^4*arctan(tan(d*x+c))*a*b

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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)^4/(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.14692, size = 1786, normalized size = 8.42 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

[1/8*((3*a^4 - 17*a^3*b + 49*a^2*b^2 - 35*a*b^3)*d*x*cos(d*x + c)^2 + (3*a^3*b - 14*a^2*b^2 + 35*a*b^3)*d*x -
(7*a*b^3 - b^4 + (7*a^2*b^2 - 8*a*b^3 + b^4)*cos(d*x + c)^2)*sqrt(-b/a)*log(((a^2 + 6*a*b + b^2)*cos(d*x + c)^
4 - 2*(3*a*b + b^2)*cos(d*x + c)^2 - 4*((a^2 + a*b)*cos(d*x + c)^3 - a*b*cos(d*x + c))*sqrt(-b/a)*sin(d*x + c)
+ b^2)/((a^2 - 2*a*b + b^2)*cos(d*x + c)^4 + 2*(a*b - b^2)*cos(d*x + c)^2 + b^2)) + (2*(a^4 - 3*a^3*b + 3*a^2
*b^2 - a*b^3)*cos(d*x + c)^5 + 3*(a^4 - 5*a^3*b + 7*a^2*b^2 - 3*a*b^3)*cos(d*x + c)^3 + (3*a^3*b - 14*a^2*b^2
+ 7*a*b^3 + 4*b^4)*cos(d*x + c))*sin(d*x + c))/((a^6 - 5*a^5*b + 10*a^4*b^2 - 10*a^3*b^3 + 5*a^2*b^4 - a*b^5)*
d*cos(d*x + c)^2 + (a^5*b - 4*a^4*b^2 + 6*a^3*b^3 - 4*a^2*b^4 + a*b^5)*d), 1/8*((3*a^4 - 17*a^3*b + 49*a^2*b^2
- 35*a*b^3)*d*x*cos(d*x + c)^2 + (3*a^3*b - 14*a^2*b^2 + 35*a*b^3)*d*x + 2*(7*a*b^3 - b^4 + (7*a^2*b^2 - 8*a*
b^3 + b^4)*cos(d*x + c)^2)*sqrt(b/a)*arctan(1/2*((a + b)*cos(d*x + c)^2 - b)*sqrt(b/a)/(b*cos(d*x + c)*sin(d*x
+ c))) + (2*(a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*cos(d*x + c)^5 + 3*(a^4 - 5*a^3*b + 7*a^2*b^2 - 3*a*b^3)*cos(
d*x + c)^3 + (3*a^3*b - 14*a^2*b^2 + 7*a*b^3 + 4*b^4)*cos(d*x + c))*sin(d*x + c))/((a^6 - 5*a^5*b + 10*a^4*b^2
- 10*a^3*b^3 + 5*a^2*b^4 - a*b^5)*d*cos(d*x + c)^2 + (a^5*b - 4*a^4*b^2 + 6*a^3*b^3 - 4*a^2*b^4 + a*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)**4/(a+b*tan(d*x+c)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.73081, size = 363, normalized size = 1.71 \begin{align*} -\frac{\frac{4 \, b^{3} \tan \left (d x + c\right )}{{\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )}{\left (b \tan \left (d x + c\right )^{2} + a\right )}} - \frac{{\left (3 \, a^{2} - 14 \, a b + 35 \, b^{2}\right )}{\left (d x + c\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} + \frac{4 \,{\left (7 \, a b^{3} - b^{4}\right )}{\left (\pi \left \lfloor \frac{d x + c}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (d x + c\right )}{\sqrt{a b}}\right )\right )}}{{\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} \sqrt{a b}} - \frac{3 \, a \tan \left (d x + c\right )^{3} - 11 \, b \tan \left (d x + c\right )^{3} + 5 \, a \tan \left (d x + c\right ) - 13 \, b \tan \left (d x + c\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")

[Out]

-1/8*(4*b^3*tan(d*x + c)/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*(b*tan(d*x + c)^2 + a)) - (3*a^2 - 14*a*b + 35*b
^2)*(d*x + c)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) + 4*(7*a*b^3 - b^4)*(pi*floor((d*x + c)/pi + 1/2)*sg
n(b) + arctan(b*tan(d*x + c)/sqrt(a*b)))/((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*sqrt(a*b)) - (3*a*ta
n(d*x + c)^3 - 11*b*tan(d*x + c)^3 + 5*a*tan(d*x + c) - 13*b*tan(d*x + c))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(t
an(d*x + c)^2 + 1)^2))/d