3.99 \(\int \frac{\sin ^6(e+f x)}{\sqrt{a+b \sec ^2(e+f x)}} \, dx\)

Optimal. Leaf size=193 \[ -\frac{\left (33 a^2+40 a b+15 b^2\right ) \sin (e+f x) \cos (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{48 a^3 f}+\frac{5 (a+b)^3 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{16 a^{7/2} f}+\frac{(9 a+5 b) \sin (e+f x) \cos ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{24 a^2 f}+\frac{\sin ^3(e+f x) \cos ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{6 a f} \]

[Out]

(5*(a + b)^3*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(16*a^(7/2)*f) - ((33*a^2 + 40*a*b
+ 15*b^2)*Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(48*a^3*f) + ((9*a + 5*b)*Cos[e + f*x]^3*
Sin[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(24*a^2*f) + (Cos[e + f*x]^3*Sin[e + f*x]^3*Sqrt[a + b + b*Tan[e
+ f*x]^2])/(6*a*f)

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Rubi [A]  time = 0.278349, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {4132, 470, 578, 527, 12, 377, 203} \[ -\frac{\left (33 a^2+40 a b+15 b^2\right ) \sin (e+f x) \cos (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{48 a^3 f}+\frac{5 (a+b)^3 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{16 a^{7/2} f}+\frac{(9 a+5 b) \sin (e+f x) \cos ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{24 a^2 f}+\frac{\sin ^3(e+f x) \cos ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{6 a f} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^6/Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

(5*(a + b)^3*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(16*a^(7/2)*f) - ((33*a^2 + 40*a*b
+ 15*b^2)*Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(48*a^3*f) + ((9*a + 5*b)*Cos[e + f*x]^3*
Sin[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(24*a^2*f) + (Cos[e + f*x]^3*Sin[e + f*x]^3*Sqrt[a + b + b*Tan[e
+ f*x]^2])/(6*a*f)

Rule 4132

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

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 578

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c -
a*d)*(p + 1)), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 \frac{\sin ^6(e+f x)}{\sqrt{a+b \sec ^2(e+f x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^4 \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x) \sin ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{6 a f}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 (a+b)-2 (3 a+b) x^2\right )}{\left (1+x^2\right )^3 \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{6 a f}\\ &=\frac{(9 a+5 b) \cos ^3(e+f x) \sin (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{24 a^2 f}+\frac{\cos ^3(e+f x) \sin ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{6 a f}-\frac{\operatorname{Subst}\left (\int \frac{(a+b) (9 a+5 b)-2 \left (12 a^2+13 a b+5 b^2\right ) x^2}{\left (1+x^2\right )^2 \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{24 a^2 f}\\ &=-\frac{\left (33 a^2+40 a b+15 b^2\right ) \cos (e+f x) \sin (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{48 a^3 f}+\frac{(9 a+5 b) \cos ^3(e+f x) \sin (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{24 a^2 f}+\frac{\cos ^3(e+f x) \sin ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{6 a f}+\frac{\operatorname{Subst}\left (\int \frac{15 (a+b)^3}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{48 a^3 f}\\ &=-\frac{\left (33 a^2+40 a b+15 b^2\right ) \cos (e+f x) \sin (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{48 a^3 f}+\frac{(9 a+5 b) \cos ^3(e+f x) \sin (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{24 a^2 f}+\frac{\cos ^3(e+f x) \sin ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{6 a f}+\frac{\left (5 (a+b)^3\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^3 f}\\ &=-\frac{\left (33 a^2+40 a b+15 b^2\right ) \cos (e+f x) \sin (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{48 a^3 f}+\frac{(9 a+5 b) \cos ^3(e+f x) \sin (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{24 a^2 f}+\frac{\cos ^3(e+f x) \sin ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{6 a f}+\frac{\left (5 (a+b)^3\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^3 f}\\ &=\frac{5 (a+b)^3 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{16 a^{7/2} f}-\frac{\left (33 a^2+40 a b+15 b^2\right ) \cos (e+f x) \sin (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{48 a^3 f}+\frac{(9 a+5 b) \cos ^3(e+f x) \sin (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{24 a^2 f}+\frac{\cos ^3(e+f x) \sin ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{6 a f}\\ \end{align*}

Mathematica [A]  time = 1.6466, size = 163, normalized size = 0.84 \[ \frac{\sec (e+f x) \sqrt{a \cos (2 (e+f x))+a+2 b} \left (15 (a+b)^3 \tan ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{-a \sin ^2(e+f x)+a+b}}\right )-\sqrt{a} \sin (e+f x) \sqrt{-a \sin ^2(e+f x)+a+b} \left (8 a^2 \sin ^4(e+f x)+10 a (a+b) \sin ^2(e+f x)+15 (a+b)^2\right )\right )}{48 \sqrt{2} a^{7/2} f \sqrt{a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^6/Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

(Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*Sec[e + f*x]*(15*(a + b)^3*ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Si
n[e + f*x]^2]] - Sqrt[a]*Sin[e + f*x]*Sqrt[a + b - a*Sin[e + f*x]^2]*(15*(a + b)^2 + 10*a*(a + b)*Sin[e + f*x]
^2 + 8*a^2*Sin[e + f*x]^4)))/(48*Sqrt[2]*a^(7/2)*f*Sqrt[a + b*Sec[e + f*x]^2])

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Maple [C]  time = 0.62, size = 2425, 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(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x)

[Out]

-1/48/f/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/a^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))*b^3*sin(f*x+e)-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)+45*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*sin(f*x
+e)+45*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)-90*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)-90*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)-15*((2*I
*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^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*c
os(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)+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)*Ellipti
cF((-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)+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^3-26*cos(f*x+e)^5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/
2)*a^3+26*cos(f*x+e)^4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3+33*cos(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/
(a+b))^(1/2)*a^3-33*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3+15*cos(f*x+e)*((2*I*a^(1/2)*b^(1/
2)+a-b)/(a+b))^(1/2)*b^3-33*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b-40*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))
^(1/2)*a*b^2-2*cos(f*x+e)^5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b+2*cos(f*x+e)^4*((2*I*a^(1/2)*b^(1/2)
+a-b)/(a+b))^(1/2)*a^2*b+14*cos(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b+5*cos(f*x+e)^3*((2*I*a^
(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2-14*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b-5*cos(f*x+
e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2+33*cos(f*x+e)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b
+40*cos(f*x+e)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2)/(-1+cos(f*x+e))/((b+a*cos(f*x+e)^2)/cos(f*x+e)^2
)^(1/2)/cos(f*x+e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 4.38913, size = 1543, normalized size = 7.99 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/384*(15*(a^3 + 3*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*(13*a^3 + 5*a^2*b)*cos(f*x
+ e)^3 + (33*a^3 + 40*a^2*b + 15*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^4*f), -1/192*(15*(a^3 + 3*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*(13*a^3 + 5*a^2*b)*cos(f*x + e)^3 + (33*a^3 + 40*a^2*b + 15*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^4*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(sin(f*x+e)**6/(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

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