3.253 \(\int \cos ^6(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=193 \[ \frac{(5 a-b) (a+b)^2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{16 a^{3/2} f}+\frac{\sin (e+f x) \cos ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 a f}+\frac{(5 a-b) \sin (e+f x) \cos ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{24 a f}+\frac{(5 a-b) (a+b) \sin (e+f x) \cos (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{16 a f} \]

[Out]

((5*a - b)*(a + b)^2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(16*a^(3/2)*f) + ((5*a - b
)*(a + b)*Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(16*a*f) + ((5*a - b)*Cos[e + f*x]^3*Sin[e
+ f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2))/(24*a*f) + (Cos[e + f*x]^5*Sin[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(
5/2))/(6*a*f)

________________________________________________________________________________________

Rubi [A]  time = 0.162324, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4146, 382, 378, 377, 203} \[ \frac{(5 a-b) (a+b)^2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{16 a^{3/2} f}+\frac{\sin (e+f x) \cos ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 a f}+\frac{(5 a-b) \sin (e+f x) \cos ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{24 a f}+\frac{(5 a-b) (a+b) \sin (e+f x) \cos (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{16 a f} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^6*(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

((5*a - b)*(a + b)^2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(16*a^(3/2)*f) + ((5*a - b
)*(a + b)*Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(16*a*f) + ((5*a - b)*Cos[e + f*x]^3*Sin[e
+ f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2))/(24*a*f) + (Cos[e + f*x]^5*Sin[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(
5/2))/(6*a*f)

Rule 4146

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/
2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]

Rule 382

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[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 378

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

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

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])

Rubi steps

\begin{align*} \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b+b x^2\right )^{3/2}}{\left (1+x^2\right )^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cos ^5(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{6 a f}+\frac{(5 a-b) \operatorname{Subst}\left (\int \frac{\left (a+b+b x^2\right )^{3/2}}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{6 a f}\\ &=\frac{(5 a-b) \cos ^3(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{24 a f}+\frac{\cos ^5(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{6 a f}+\frac{((5 a-b) (a+b)) \operatorname{Subst}\left (\int \frac{\sqrt{a+b+b x^2}}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{8 a f}\\ &=\frac{(5 a-b) (a+b) \cos (e+f x) \sin (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{16 a f}+\frac{(5 a-b) \cos ^3(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{24 a f}+\frac{\cos ^5(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{6 a f}+\frac{\left ((5 a-b) (a+b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{16 a f}\\ &=\frac{(5 a-b) (a+b) \cos (e+f x) \sin (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{16 a f}+\frac{(5 a-b) \cos ^3(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{24 a f}+\frac{\cos ^5(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{6 a f}+\frac{\left ((5 a-b) (a+b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{16 a f}\\ &=\frac{(5 a-b) (a+b)^2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{16 a^{3/2} f}+\frac{(5 a-b) (a+b) \cos (e+f x) \sin (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{16 a f}+\frac{(5 a-b) \cos ^3(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{24 a f}+\frac{\cos ^5(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{6 a f}\\ \end{align*}

Mathematica [A]  time = 1.84457, size = 165, normalized size = 0.85 \[ \frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)} \left (\frac{3 \sqrt{2} \sqrt{a+b} \left (5 a^2+4 a b-b^2\right ) \sin ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{\sqrt{\frac{a \cos (2 (e+f x))+a+2 b}{a+b}}}+\sqrt{a} \sin (e+f x) \left (a^2 \cos (4 (e+f x))+23 a^2+a (9 a+7 b) \cos (2 (e+f x))+29 a b+3 b^2\right )\right )}{48 a^{3/2} f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^6*(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

(Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2]*((3*Sqrt[2]*Sqrt[a + b]*(5*a^2 + 4*a*b - b^2)*ArcSin[(Sqrt[a]*Sin[e +
f*x])/Sqrt[a + b]])/Sqrt[(a + 2*b + a*Cos[2*(e + f*x)])/(a + b)] + Sqrt[a]*(23*a^2 + 29*a*b + 3*b^2 + a*(9*a
+ 7*b)*Cos[2*(e + f*x)] + a^2*Cos[4*(e + f*x)])*Sin[e + f*x]))/(48*a^(3/2)*f)

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Maple [C]  time = 0.49, size = 2439, normalized size = 12.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x)

[Out]

-1/48/f/a/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*sin(f*x+e)*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(3/2)*cos(f*x+e
)^3*(-3*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)
*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1
+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2
+6*a*b-b^2)/(a+b)^2)^(1/2))*b^3*sin(f*x+e)+15*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)
+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-
b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a
^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a^3*sin(f*x+e)-15*cos(f*x+e)*((2*I*a^(1/2)*b
^(1/2)+a-b)/(a+b))^(1/2)*a^2*b+15*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b+27*2^(1/2)*(1/(a+b)*(I*cos(f*x
+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^
(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+
a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a^2*b*s
in(f*x+e)+9*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(
1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF
((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)
-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*b^2*sin(f*x+e)-54*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b
^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f
*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-
1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(
1/2))*a^2*b*sin(f*x+e)-18*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+
cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(
1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b
)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a*b^2*sin(f*x+e)+3*(
(2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^3+6*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+
a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b
)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*
a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*b
^3*sin(f*x+e)-8*cos(f*x+e)^7*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3+8*cos(f*x+e)^6*((2*I*a^(1/2)*b^(1/2)+
a-b)/(a+b))^(1/2)*a^3-10*cos(f*x+e)^5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3+10*cos(f*x+e)^4*((2*I*a^(1/2
)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3-3*cos(f*x+e)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^3+22*((2*I*a^(1/2)*b^(1
/2)+a-b)/(a+b))^(1/2)*a*b^2-22*cos(f*x+e)^5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b+22*cos(f*x+e)^4*((2*
I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b-32*cos(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b-17*cos
(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2+32*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2
)*a^2*b+17*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2-22*cos(f*x+e)*((2*I*a^(1/2)*b^(1/2)+a-b)
/(a+b))^(1/2)*a*b^2-15*cos(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3+15*cos(f*x+e)^2*((2*I*a^(1/2)*
b^(1/2)+a-b)/(a+b))^(1/2)*a^3-30*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)
+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x
+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1
/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a^3*sin(f*x+e
))/(-1+cos(f*x+e))/(b+a*cos(f*x+e)^2)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \cos \left (f x + e\right )^{6}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e)^2 + a)^(3/2)*cos(f*x + e)^6, x)

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Fricas [A]  time = 4.04061, size = 1539, normalized size = 7.97 \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)^6*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/384*(3*(5*a^3 + 9*a^2*b + 3*a*b^2 - b^3)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x +
e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a
^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5
+ 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt(
(a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 8*(8*a^3*cos(f*x + e)^5 + 2*(5*a^3 + 7*a^2*b)*cos(f*x +
e)^3 + (15*a^3 + 22*a^2*b + 3*a*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/
(a^2*f), -1/192*(3*(5*a^3 + 9*a^2*b + 3*a*b^2 - b^3)*sqrt(a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*
cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3
*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e))) - 4*(8*a^3*cos(f*x + e)^5 + 2
*(5*a^3 + 7*a^2*b)*cos(f*x + e)^3 + (15*a^3 + 22*a^2*b + 3*a*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/co
s(f*x + e)^2)*sin(f*x + e))/(a^2*f)]

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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(f*x+e)**6*(a+b*sec(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \cos \left (f x + e\right )^{6}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e)^2 + a)^(3/2)*cos(f*x + e)^6, x)