3.210 \(\int \frac{\cos ^3(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=181 \[ \frac{b^4 \sin (e+f x)}{4 a^4 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2}-\frac{b^3 (16 a+13 b) \sin (e+f x)}{8 a^4 f (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}+\frac{b^2 \left (48 a^2+80 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{8 a^{9/2} f (a+b)^{5/2}}+\frac{(a-3 b) \sin (e+f x)}{a^4 f}-\frac{\sin ^3(e+f x)}{3 a^3 f} \]

[Out]

(b^2*(48*a^2 + 80*a*b + 35*b^2)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(8*a^(9/2)*(a + b)^(5/2)*f) + ((a
- 3*b)*Sin[e + f*x])/(a^4*f) - Sin[e + f*x]^3/(3*a^3*f) + (b^4*Sin[e + f*x])/(4*a^4*(a + b)*f*(a + b - a*Sin[
e + f*x]^2)^2) - (b^3*(16*a + 13*b)*Sin[e + f*x])/(8*a^4*(a + b)^2*f*(a + b - a*Sin[e + f*x]^2))

________________________________________________________________________________________

Rubi [A]  time = 0.242773, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4147, 390, 1157, 385, 208} \[ \frac{b^4 \sin (e+f x)}{4 a^4 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2}-\frac{b^3 (16 a+13 b) \sin (e+f x)}{8 a^4 f (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}+\frac{b^2 \left (48 a^2+80 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{8 a^{9/2} f (a+b)^{5/2}}+\frac{(a-3 b) \sin (e+f x)}{a^4 f}-\frac{\sin ^3(e+f x)}{3 a^3 f} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^3/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

(b^2*(48*a^2 + 80*a*b + 35*b^2)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(8*a^(9/2)*(a + b)^(5/2)*f) + ((a
- 3*b)*Sin[e + f*x])/(a^4*f) - Sin[e + f*x]^3/(3*a^3*f) + (b^4*Sin[e + f*x])/(4*a^4*(a + b)*f*(a + b - a*Sin[
e + f*x]^2)^2) - (b^3*(16*a + 13*b)*Sin[e + f*x])/(8*a^4*(a + b)^2*f*(a + b - a*Sin[e + f*x]^2))

Rule 4147

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fr
eeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b + 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 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

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(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^4}{\left (a+b-a x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a-3 b}{a^4}-\frac{x^2}{a^3}+\frac{b^2 \left (6 a^2+8 a b+3 b^2\right )-4 a b^2 (3 a+2 b) x^2+6 a^2 b^2 x^4}{a^4 \left (a+b-a x^2\right )^3}\right ) \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{(a-3 b) \sin (e+f x)}{a^4 f}-\frac{\sin ^3(e+f x)}{3 a^3 f}+\frac{\operatorname{Subst}\left (\int \frac{b^2 \left (6 a^2+8 a b+3 b^2\right )-4 a b^2 (3 a+2 b) x^2+6 a^2 b^2 x^4}{\left (a+b-a x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{a^4 f}\\ &=\frac{(a-3 b) \sin (e+f x)}{a^4 f}-\frac{\sin ^3(e+f x)}{3 a^3 f}+\frac{b^4 \sin (e+f x)}{4 a^4 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}-\frac{\operatorname{Subst}\left (\int \frac{-b^2 \left (24 a^2+32 a b+11 b^2\right )+24 a b^2 (a+b) x^2}{\left (a+b-a x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{4 a^4 (a+b) f}\\ &=\frac{(a-3 b) \sin (e+f x)}{a^4 f}-\frac{\sin ^3(e+f x)}{3 a^3 f}+\frac{b^4 \sin (e+f x)}{4 a^4 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}-\frac{b^3 (16 a+13 b) \sin (e+f x)}{8 a^4 (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )}+\frac{\left (b^2 \left (48 a^2+80 a b+35 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b-a x^2} \, dx,x,\sin (e+f x)\right )}{8 a^4 (a+b)^2 f}\\ &=\frac{b^2 \left (48 a^2+80 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{8 a^{9/2} (a+b)^{5/2} f}+\frac{(a-3 b) \sin (e+f x)}{a^4 f}-\frac{\sin ^3(e+f x)}{3 a^3 f}+\frac{b^4 \sin (e+f x)}{4 a^4 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}-\frac{b^3 (16 a+13 b) \sin (e+f x)}{8 a^4 (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 4.45507, size = 194, normalized size = 1.07 \[ \frac{-\frac{3 b^2 \left (48 a^2+80 a b+35 b^2\right ) \left (\log \left (\sqrt{a+b}-\sqrt{a} \sin (e+f x)\right )-\log \left (\sqrt{a+b}+\sqrt{a} \sin (e+f x)\right )\right )}{(a+b)^{5/2}}+4 a^{3/2} \sin (3 (e+f x))+12 \sqrt{a} \sin (e+f x) \left (-\frac{b^4 (13 a \cos (2 (e+f x))+9 a+22 b)}{(a+b)^2 (a \cos (2 (e+f x))+a+2 b)^2}+a \left (3-\frac{16 b^3}{(a+b)^2 (a \cos (2 (e+f x))+a+2 b)}\right )-12 b\right )}{48 a^{9/2} f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^3/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

((-3*b^2*(48*a^2 + 80*a*b + 35*b^2)*(Log[Sqrt[a + b] - Sqrt[a]*Sin[e + f*x]] - Log[Sqrt[a + b] + Sqrt[a]*Sin[e
+ f*x]]))/(a + b)^(5/2) + 12*Sqrt[a]*(-12*b - (b^4*(9*a + 22*b + 13*a*Cos[2*(e + f*x)]))/((a + b)^2*(a + 2*b
+ a*Cos[2*(e + f*x)])^2) + a*(3 - (16*b^3)/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)]))))*Sin[e + f*x] + 4*a^(3/
2)*Sin[3*(e + f*x)])/(48*a^(9/2)*f)

________________________________________________________________________________________

Maple [A]  time = 0.102, size = 177, normalized size = 1. \begin{align*}{\frac{1}{f} \left ( -{\frac{1}{{a}^{4}} \left ({\frac{a \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{3}}-\sin \left ( fx+e \right ) a+3\,\sin \left ( fx+e \right ) b \right ) }-{\frac{{b}^{2}}{{a}^{4}} \left ({\frac{1}{ \left ( -a-b+a \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}} \left ( -{\frac{ab \left ( 16\,a+13\,b \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{8\,{a}^{2}+16\,ab+8\,{b}^{2}}}+{\frac{ \left ( 16\,a+11\,b \right ) b\sin \left ( fx+e \right ) }{8\,a+8\,b}} \right ) }-{\frac{48\,{a}^{2}+80\,ab+35\,{b}^{2}}{8\,{a}^{2}+16\,ab+8\,{b}^{2}}{\it Artanh} \left ({\sin \left ( fx+e \right ) a{\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x)

[Out]

1/f*(-1/a^4*(1/3*a*sin(f*x+e)^3-sin(f*x+e)*a+3*sin(f*x+e)*b)-1/a^4*b^2*((-1/8*a*b*(16*a+13*b)/(a^2+2*a*b+b^2)*
sin(f*x+e)^3+1/8*(16*a+11*b)*b/(a+b)*sin(f*x+e))/(-a-b+a*sin(f*x+e)^2)^2-1/8*(48*a^2+80*a*b+35*b^2)/(a^2+2*a*b
+b^2)/((a+b)*a)^(1/2)*arctanh(a*sin(f*x+e)/((a+b)*a)^(1/2))))

________________________________________________________________________________________

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(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 0.778927, size = 1908, normalized size = 10.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[1/48*(3*(48*a^2*b^4 + 80*a*b^5 + 35*b^6 + (48*a^4*b^2 + 80*a^3*b^3 + 35*a^2*b^4)*cos(f*x + e)^4 + 2*(48*a^3*b
^3 + 80*a^2*b^4 + 35*a*b^5)*cos(f*x + e)^2)*sqrt(a^2 + a*b)*log(-(a*cos(f*x + e)^2 - 2*sqrt(a^2 + a*b)*sin(f*x
+ e) - 2*a - b)/(a*cos(f*x + e)^2 + b)) + 2*(16*a^5*b^2 - 24*a^4*b^3 - 210*a^3*b^4 - 275*a^2*b^5 - 105*a*b^6
+ 8*(a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3)*cos(f*x + e)^6 + 8*(2*a^7 - a^6*b - 15*a^5*b^2 - 19*a^4*b^3 - 7*a^3*
b^4)*cos(f*x + e)^4 + (32*a^6*b - 40*a^5*b^2 - 360*a^4*b^3 - 463*a^3*b^4 - 175*a^2*b^5)*cos(f*x + e)^2)*sin(f*
x + e))/((a^10 + 3*a^9*b + 3*a^8*b^2 + a^7*b^3)*f*cos(f*x + e)^4 + 2*(a^9*b + 3*a^8*b^2 + 3*a^7*b^3 + a^6*b^4)
*f*cos(f*x + e)^2 + (a^8*b^2 + 3*a^7*b^3 + 3*a^6*b^4 + a^5*b^5)*f), -1/24*(3*(48*a^2*b^4 + 80*a*b^5 + 35*b^6 +
(48*a^4*b^2 + 80*a^3*b^3 + 35*a^2*b^4)*cos(f*x + e)^4 + 2*(48*a^3*b^3 + 80*a^2*b^4 + 35*a*b^5)*cos(f*x + e)^2
)*sqrt(-a^2 - a*b)*arctan(sqrt(-a^2 - a*b)*sin(f*x + e)/(a + b)) - (16*a^5*b^2 - 24*a^4*b^3 - 210*a^3*b^4 - 27
5*a^2*b^5 - 105*a*b^6 + 8*(a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3)*cos(f*x + e)^6 + 8*(2*a^7 - a^6*b - 15*a^5*b^2
- 19*a^4*b^3 - 7*a^3*b^4)*cos(f*x + e)^4 + (32*a^6*b - 40*a^5*b^2 - 360*a^4*b^3 - 463*a^3*b^4 - 175*a^2*b^5)*
cos(f*x + e)^2)*sin(f*x + e))/((a^10 + 3*a^9*b + 3*a^8*b^2 + a^7*b^3)*f*cos(f*x + e)^4 + 2*(a^9*b + 3*a^8*b^2
+ 3*a^7*b^3 + a^6*b^4)*f*cos(f*x + e)^2 + (a^8*b^2 + 3*a^7*b^3 + 3*a^6*b^4 + a^5*b^5)*f)]

________________________________________________________________________________________

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(f*x+e)**3/(a+b*sec(f*x+e)**2)**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.29686, size = 323, normalized size = 1.78 \begin{align*} -\frac{\frac{3 \,{\left (48 \, a^{2} b^{2} + 80 \, a b^{3} + 35 \, b^{4}\right )} \arctan \left (\frac{a \sin \left (f x + e\right )}{\sqrt{-a^{2} - a b}}\right )}{{\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} \sqrt{-a^{2} - a b}} - \frac{3 \,{\left (16 \, a^{2} b^{3} \sin \left (f x + e\right )^{3} + 13 \, a b^{4} \sin \left (f x + e\right )^{3} - 16 \, a^{2} b^{3} \sin \left (f x + e\right ) - 27 \, a b^{4} \sin \left (f x + e\right ) - 11 \, b^{5} \sin \left (f x + e\right )\right )}}{{\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )}{\left (a \sin \left (f x + e\right )^{2} - a - b\right )}^{2}} + \frac{8 \,{\left (a^{6} \sin \left (f x + e\right )^{3} - 3 \, a^{6} \sin \left (f x + e\right ) + 9 \, a^{5} b \sin \left (f x + e\right )\right )}}{a^{9}}}{24 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")

[Out]

-1/24*(3*(48*a^2*b^2 + 80*a*b^3 + 35*b^4)*arctan(a*sin(f*x + e)/sqrt(-a^2 - a*b))/((a^6 + 2*a^5*b + a^4*b^2)*s
qrt(-a^2 - a*b)) - 3*(16*a^2*b^3*sin(f*x + e)^3 + 13*a*b^4*sin(f*x + e)^3 - 16*a^2*b^3*sin(f*x + e) - 27*a*b^4
*sin(f*x + e) - 11*b^5*sin(f*x + e))/((a^6 + 2*a^5*b + a^4*b^2)*(a*sin(f*x + e)^2 - a - b)^2) + 8*(a^6*sin(f*x
+ e)^3 - 3*a^6*sin(f*x + e) + 9*a^5*b*sin(f*x + e))/a^9)/f