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

Optimal. Leaf size=240 \[ \frac{\left (-15 a^2 b+5 a^3-5 a b^2-b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{16 a^{5/2} f}-\frac{(a-b) (5 a+b) \sin (e+f x) \cos (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{16 a^2 f}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{f}-\frac{\sin ^5(e+f x) \cos (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{6 f}-\frac{(5 a-b) \sin ^3(e+f x) \cos (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{24 a f} \]

[Out]

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

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Rubi [A]  time = 0.384422, antiderivative size = 240, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {4132, 467, 578, 523, 217, 206, 377, 203} \[ \frac{\left (-15 a^2 b+5 a^3-5 a b^2-b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{16 a^{5/2} f}-\frac{(a-b) (5 a+b) \sin (e+f x) \cos (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{16 a^2 f}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{f}-\frac{\sin ^5(e+f x) \cos (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{6 f}-\frac{(5 a-b) \sin ^3(e+f x) \cos (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{24 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*a^3 - 15*a^2*b - 5*a*b^2 - b^3)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(16*a^(5/2)
*f) + (Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/f - ((a - b)*(5*a + b)*Cos[e +
f*x]*Sin[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(16*a^2*f) - ((5*a - b)*Cos[e + f*x]*Sin[e + f*x]^3*Sqrt[a +
b + b*Tan[e + f*x]^2])/(24*a*f) - (Cos[e + f*x]*Sin[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2])/(6*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 467

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*n*(p + 1)), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& 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 523

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

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

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

Mathematica [F]  time = 9.03791, size = 0, normalized size = 0. \[ \int \sqrt{a+b \sec ^2(e+f x)} \sin ^6(e+f x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.752, size = 2665, normalized size = 11.1 \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/a^2/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*(-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*c
os(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)+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))*a^2*b*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*b^2*sin(f*x+e)-96*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^2*b*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*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)+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)+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^3-3*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+14*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2+10*cos(f*x+e)^5*((2*I*a^(1/2)
*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b-10*cos(f*x+e)^4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b-40*cos(f*x+e)^3
*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b-cos(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2+40*c
os(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b+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-14*cos(f*x+e)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a
+b))^(1/2)*a*b^2)*cos(f*x+e)*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(1/2)*sin(f*x+e)/(-1+cos(f*x+e))/(b+a*cos(f*x+e
)^2)

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

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Fricas [A]  time = 9.96628, size = 4146, normalized size = 17.27 \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*(96*a^3*sqrt(b)*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 + 4*((a - b)*cos
(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(
f*x + e)^4) + 3*(5*a^3 - 15*a^2*b - 5*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 - a^2*b)*cos
(f*x + e)^3 + (33*a^3 - 14*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^3*f), 1/384*(192*a^3*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt((
a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e))) + 3*(5*a^3 - 15*a^2*b - 5*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)*c
os(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 - a^2*b)*cos(f*x + e)^3 + (33*a^3 - 14*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^3*f), 1/192*(48*a^3*sqrt(b)*log(
((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 + 4*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x +
e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4) - 3*(5*a^3 - 15
*a^2*b - 5*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 - a^2*b)*cos(f*x + e
)^3 + (33*a^3 - 14*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
^3*f), 1/192*(96*a^3*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt((a*cos(f*x
+ e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e))) - 3*(5*a^3 - 15*a^2*b - 5*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))*s
qrt(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)*co
s(f*x + e)^2)*sin(f*x + e))) - 4*(8*a^3*cos(f*x + e)^5 - 2*(13*a^3 - a^2*b)*cos(f*x + e)^3 + (33*a^3 - 14*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^3*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 \sqrt{b \sec \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )^{6}\,{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(sqrt(b*sec(f*x + e)^2 + a)*sin(f*x + e)^6, x)