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

Optimal. Leaf size=129 \[ \frac{x \left (3 a^2-10 a b+15 b^2\right )}{8 (a-b)^3}-\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a-b)^3}+\frac{\sin (c+d x) \cos ^3(c+d x)}{4 d (a-b)}+\frac{(3 a-7 b) \sin (c+d x) \cos (c+d x)}{8 d (a-b)^2} \]

[Out]

((3*a^2 - 10*a*b + 15*b^2)*x)/(8*(a - b)^3) - (b^(5/2)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b
)^3*d) + ((3*a - 7*b)*Cos[c + d*x]*Sin[c + d*x])/(8*(a - b)^2*d) + (Cos[c + d*x]^3*Sin[c + d*x])/(4*(a - b)*d)

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Rubi [A]  time = 0.166607, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{x \left (3 a^2-10 a b+15 b^2\right )}{8 (a-b)^3}-\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a-b)^3}+\frac{\sin (c+d x) \cos ^3(c+d x)}{4 d (a-b)}+\frac{(3 a-7 b) \sin (c+d x) \cos (c+d x)}{8 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^2 - 10*a*b + 15*b^2)*x)/(8*(a - b)^3) - (b^(5/2)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b
)^3*d) + ((3*a - 7*b)*Cos[c + d*x]*Sin[c + d*x])/(8*(a - b)^2*d) + (Cos[c + d*x]^3*Sin[c + d*x])/(4*(a - b)*d)

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)}{a+b \tan ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right )^3 \left (a+b x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\cos ^3(c+d x) \sin (c+d x)}{4 (a-b) d}-\frac{\operatorname{Subst}\left (\int \frac{-3 a+4 b-3 b x^2}{\left (1+x^2\right )^2 \left (a+b x^2\right )} \, dx,x,\tan (c+d x)\right )}{4 (a-b) d}\\ &=\frac{(3 a-7 b) \cos (c+d x) \sin (c+d x)}{8 (a-b)^2 d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 (a-b) d}+\frac{\operatorname{Subst}\left (\int \frac{3 a^2-7 a b+8 b^2+(3 a-7 b) b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (c+d x)\right )}{8 (a-b)^2 d}\\ &=\frac{(3 a-7 b) \cos (c+d x) \sin (c+d x)}{8 (a-b)^2 d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 (a-b) d}-\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{(a-b)^3 d}+\frac{\left (3 a^2-10 a b+15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 (a-b)^3 d}\\ &=\frac{\left (3 a^2-10 a b+15 b^2\right ) x}{8 (a-b)^3}-\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} (a-b)^3 d}+\frac{(3 a-7 b) \cos (c+d x) \sin (c+d x)}{8 (a-b)^2 d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 (a-b) d}\\ \end{align*}

Mathematica [A]  time = 0.427649, size = 113, normalized size = 0.88 \[ \frac{\sqrt{a} \left (4 \left (3 a^2-10 a b+15 b^2\right ) (c+d x)+8 \left (a^2-3 a b+2 b^2\right ) \sin (2 (c+d x))+(a-b)^2 \sin (4 (c+d x))\right )-32 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{32 \sqrt{a} d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*b^(5/2)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]] + Sqrt[a]*(4*(3*a^2 - 10*a*b + 15*b^2)*(c + d*x) + 8*(a^2
- 3*a*b + 2*b^2)*Sin[2*(c + d*x)] + (a - b)^2*Sin[4*(c + d*x)]))/(32*Sqrt[a]*(a - b)^3*d)

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Maple [B]  time = 0.077, size = 303, normalized size = 2.4 \begin{align*} -{\frac{{b}^{3}}{d \left ( a-b \right ) ^{3}}\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 ) ^{3} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{5\, \left ( \tan \left ( dx+c \right ) \right ) ^{3}ab}{4\,d \left ( a-b \right ) ^{3} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{7\, \left ( \tan \left ( dx+c \right ) \right ) ^{3}{b}^{2}}{8\,d \left ( a-b \right ) ^{3} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{7\,a\tan \left ( dx+c \right ) b}{4\,d \left ( a-b \right ) ^{3} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{9\,\tan \left ( dx+c \right ){b}^{2}}{8\,d \left ( a-b \right ) ^{3} \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 ) ^{3} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{15\,\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{2}}{8\,d \left ( a-b \right ) ^{3}}}+{\frac{3\,\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}}{8\,d \left ( a-b \right ) ^{3}}}-{\frac{5\,\arctan \left ( \tan \left ( dx+c \right ) \right ) ab}{4\,d \left ( a-b \right ) ^{3}}} \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),x)

[Out]

-1/d/(a-b)^3*b^3/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2))+3/8/d/(a-b)^3/(tan(d*x+c)^2+1)^2*tan(d*x+c)^3*a^
2-5/4/d/(a-b)^3/(tan(d*x+c)^2+1)^2*tan(d*x+c)^3*a*b+7/8/d/(a-b)^3/(tan(d*x+c)^2+1)^2*tan(d*x+c)^3*b^2-7/4/d/(a
-b)^3/(tan(d*x+c)^2+1)^2*tan(d*x+c)*a*b+9/8/d/(a-b)^3/(tan(d*x+c)^2+1)^2*tan(d*x+c)*b^2+5/8/d/(a-b)^3/(tan(d*x
+c)^2+1)^2*tan(d*x+c)*a^2+15/8/d/(a-b)^3*arctan(tan(d*x+c))*b^2+3/8/d/(a-b)^3*arctan(tan(d*x+c))*a^2-5/4/d/(a-
b)^3*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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.74992, size = 929, normalized size = 7.2 \begin{align*} \left [-\frac{2 \, b^{2} \sqrt{-\frac{b}{a}} \log \left (\frac{{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \,{\left ({\left (a^{2} + a b\right )} \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \sqrt{-\frac{b}{a}} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) -{\left (3 \, a^{2} - 10 \, a b + 15 \, b^{2}\right )} d x -{\left (2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} +{\left (3 \, a^{2} - 10 \, a b + 7 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} d}, \frac{4 \, b^{2} \sqrt{\frac{b}{a}} \arctan \left (\frac{{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt{\frac{b}{a}}}{2 \, b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) +{\left (3 \, a^{2} - 10 \, a b + 15 \, b^{2}\right )} d x +{\left (2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} +{\left (3 \, a^{2} - 10 \, a b + 7 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} d}\right ] \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),x, algorithm="fricas")

[Out]

[-1/8*(2*b^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)) - (3*a^2 - 10*a*b + 15*b^2)*d*x - (2*(a^2 - 2*a*b + b^2)*cos(d*x + c)^3 + (
3*a^2 - 10*a*b + 7*b^2)*cos(d*x + c))*sin(d*x + c))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d), 1/8*(4*b^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))) + (3*a^2 - 10*a*b + 15*b^2)*d
*x + (2*(a^2 - 2*a*b + b^2)*cos(d*x + c)^3 + (3*a^2 - 10*a*b + 7*b^2)*cos(d*x + c))*sin(d*x + c))/((a^3 - 3*a^
2*b + 3*a*b^2 - 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)**4/(a+b*tan(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.65097, size = 247, normalized size = 1.91 \begin{align*} -\frac{\frac{8 \,{\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 )} b^{3}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt{a b}} - \frac{{\left (3 \, a^{2} - 10 \, a b + 15 \, b^{2}\right )}{\left (d x + c\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{3 \, a \tan \left (d x + c\right )^{3} - 7 \, b \tan \left (d x + c\right )^{3} + 5 \, a \tan \left (d x + c\right ) - 9 \, b \tan \left (d x + c\right )}{{\left (a^{2} - 2 \, a b + b^{2}\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),x, algorithm="giac")

[Out]

-1/8*(8*(pi*floor((d*x + c)/pi + 1/2)*sgn(b) + arctan(b*tan(d*x + c)/sqrt(a*b)))*b^3/((a^3 - 3*a^2*b + 3*a*b^2
- b^3)*sqrt(a*b)) - (3*a^2 - 10*a*b + 15*b^2)*(d*x + c)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (3*a*tan(d*x + c)^3
- 7*b*tan(d*x + c)^3 + 5*a*tan(d*x + c) - 9*b*tan(d*x + c))/((a^2 - 2*a*b + b^2)*(tan(d*x + c)^2 + 1)^2))/d