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

Optimal. Leaf size=298 \[ -\frac {\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{16 a f}+\frac {\left (5 a^2-40 a b+3 b^2\right ) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{48 a f}+\frac {\left (5 a^3-45 a^2 b+15 a b^2+b^3\right ) \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{16 a^{3/2} f}+\frac {(5 a-3 b) \sin ^4(e+f x) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{24 f}+\frac {\sqrt {b} (3 a-5 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 f}-\frac {\sin ^5(e+f x) \cos (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 f} \]

[Out]

1/16*(5*a^3-45*a^2*b+15*a*b^2+b^3)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(3/2)/f+1/2*(3*a-5*
b)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))*b^(1/2)/f-1/16*(5*a^2-26*a*b+b^2)*(a+b+b*tan(f*x+e)^
2)^(1/2)*tan(f*x+e)/a/f+1/48*(5*a^2-40*a*b+3*b^2)*sin(f*x+e)^2*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/a/f+1/24*
(5*a-3*b)*sin(f*x+e)^4*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/f-1/6*cos(f*x+e)*sin(f*x+e)^5*(a+b+b*tan(f*x+e)^2
)^(3/2)/f

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Rubi [A]  time = 0.47, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4132, 467, 577, 578, 582, 523, 217, 206, 377, 203} \[ \frac {\left (-45 a^2 b+5 a^3+15 a b^2+b^3\right ) \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{16 a^{3/2} f}-\frac {\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{16 a f}+\frac {\left (5 a^2-40 a b+3 b^2\right ) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{48 a f}+\frac {(5 a-3 b) \sin ^4(e+f x) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{24 f}+\frac {\sqrt {b} (3 a-5 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 f}-\frac {\sin ^5(e+f x) \cos (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*a^3 - 45*a^2*b + 15*a*b^2 + b^3)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(16*a^(3/2
)*f) + ((3*a - 5*b)*Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*f) - ((5*a^2 -
26*a*b + b^2)*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(16*a*f) + ((5*a^2 - 40*a*b + 3*b^2)*Sin[e + f*x]^2
*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(48*a*f) + ((5*a - 3*b)*Sin[e + f*x]^4*Tan[e + f*x]*Sqrt[a + b +
b*Tan[e + f*x]^2])/(24*f) - (Cos[e + f*x]*Sin[e + f*x]^5*(a + b + b*Tan[e + f*x]^2)^(3/2))/(6*f)

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 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 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 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 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 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 577

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

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 582

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

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]

Rubi steps

\begin {align*} \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^6 \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 (e+f x) \sin ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{6 f}+\frac {\operatorname {Subst}\left (\int \frac {x^4 \sqrt {a+b+b x^2} \left (5 (a+b)+8 b x^2\right )}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{6 f}\\ &=\frac {(5 a-3 b) \sin ^4(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 f}-\frac {\cos (e+f x) \sin ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{6 f}-\frac {\operatorname {Subst}\left (\int \frac {x^4 \left (5 (a-7 b) (a+b)+2 (5 a-19 b) b x^2\right )}{\left (1+x^2\right )^2 \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{24 f}\\ &=\frac {\left (5 a^2-40 a b+3 b^2\right ) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 a f}+\frac {(5 a-3 b) \sin ^4(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 f}-\frac {\cos (e+f x) \sin ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{6 f}-\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (3 (a+b) \left (5 a^2-40 a b+3 b^2\right )+6 b \left (5 a^2-26 a b+b^2\right ) x^2\right )}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{48 a f}\\ &=-\frac {\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 a f}+\frac {\left (5 a^2-40 a b+3 b^2\right ) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 a f}+\frac {(5 a-3 b) \sin ^4(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 f}-\frac {\cos (e+f x) \sin ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{6 f}+\frac {\operatorname {Subst}\left (\int \frac {6 b (a+b) \left (5 a^2-26 a b+b^2\right )+48 a (3 a-5 b) b^2 x^2}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{96 a b f}\\ &=-\frac {\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 a f}+\frac {\left (5 a^2-40 a b+3 b^2\right ) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 a f}+\frac {(5 a-3 b) \sin ^4(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 f}-\frac {\cos (e+f x) \sin ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{6 f}+\frac {((3 a-5 b) b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac {\left (5 a^3-45 a^2 b+15 a b^2+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 f}\\ &=-\frac {\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 a f}+\frac {\left (5 a^2-40 a b+3 b^2\right ) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 a f}+\frac {(5 a-3 b) \sin ^4(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 f}-\frac {\cos (e+f x) \sin ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{6 f}+\frac {((3 a-5 b) 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 )}{2 f}+\frac {\left (5 a^3-45 a^2 b+15 a b^2+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 f}\\ &=\frac {\left (5 a^3-45 a^2 b+15 a b^2+b^3\right ) \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 {(3 a-5 b) \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 f}-\frac {\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 a f}+\frac {\left (5 a^2-40 a b+3 b^2\right ) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 a f}+\frac {(5 a-3 b) \sin ^4(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 f}-\frac {\cos (e+f x) \sin ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{6 f}\\ \end {align*}

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Mathematica [F]  time = 9.99, size = 0, normalized size = 0.00 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 17.85, size = 1855, normalized size = 6.22 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/384*(3*(5*a^3 - 45*a^2*b + 15*a*b^2 + b^3)*sqrt(-a)*cos(f*x + e)*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)) + 48*(3*a^3 - 5*a^2*b)*sqrt(b)*cos(f*x +
e)*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*co
s(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) + 8*(8*a
^3*cos(f*x + e)^6 - 2*(13*a^3 - 7*a^2*b)*cos(f*x + e)^4 - 24*a^2*b + (33*a^3 - 68*a^2*b + 3*a*b^2)*cos(f*x + e
)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^2*f*cos(f*x + e)), 1/384*(96*(3*a^3 - 5*a^2*
b)*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)))*cos(f*x + e) - 3*(5*a^3 - 45*a^2*b + 15*a*b^2 + b^3)*sqr
t(-a)*cos(f*x + e)*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)^6 - 2*(13*a^3 - 7*a^2*b)*cos(f*x + e)^4 - 24*a^2*b + (33*a^3 - 68*a
^2*b + 3*a*b^2)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^2*f*cos(f*x + e))
, -1/192*(3*(5*a^3 - 45*a^2*b + 15*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)))*cos(f*x + e) + 24*(3*a^3 - 5*a^2*b
)*sqrt(b)*cos(f*x + e)*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) + 4*(8*a^3*cos(f*x + e)^6 - 2*(13*a^3 - 7*a^2*b)*cos(f*x + e)^4 - 24*a^2*b + (33*a^3 - 68*a^2*b +
3*a*b^2)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^2*f*cos(f*x + e)), -1/19
2*(3*(5*a^3 - 45*a^2*b + 15*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)))*cos(f*x + e) - 48*(3*a^3 - 5*a^2*b)*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)))*cos(f*x + e) + 4*(8*a^3*cos(f*x + e)^6 - 2*(13*a^3 - 7*a^2*b)*co
s(f*x + e)^4 - 24*a^2*b + (33*a^3 - 68*a^2*b + 3*a*b^2)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x +
e)^2)*sin(f*x + e))/(a^2*f*cos(f*x + e))]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 2.28, size = 3067, normalized size = 10.29 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48/f*(270*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1
/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(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))*sin(f*x+e)*cos(f*x+e)^2*a^2*b+94*((
2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^4*a^2*b+9*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^
3*a^2*b-68*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^3*a*b^2-9*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2
)*cos(f*x+e)^2*a^2*b+68*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^2*a*b^2-24*((2*I*a^(1/2)*b^(1/2)+a-
b)/(a+b))^(1/2)*cos(f*x+e)*a*b^2-94*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^5*a^2*b-26*((2*I*a^(1/2
)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^7*a^3+26*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^6*a^3+33*((
2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^5*a^3-33*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^4
*a^3+8*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^9*a^3-8*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(
f*x+e)^8*a^3+3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^3*b^3-3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1
/2)*cos(f*x+e)^2*b^3-144*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))*2
^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(
1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*cos(f*x+e)^2*sin(f*x+e)*
a^2*b+240*cos(f*x+e)^2*sin(f*x+e)*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))*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(
-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*a*b^2-63*cos(f*
x+e)^2*sin(f*x+e)*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))*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^
(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*
x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*a^2*b-75*cos(f*x+e)^2*sin(f*x+e)*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))*
2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^
(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*a*b^2-90*2^(1/2)*((I*a^
(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*
cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(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))*sin(f*x+e)*cos(f*x+e)^2*a*b^2+24*((2*I*a^(1/2)*b^(1/2)+a-b)/(a
+b))^(1/2)*a*b^2-17*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^4*a*b^2-22*((2*I*a^(1/2)*b^(1/2)+a-b)/(
a+b))^(1/2)*cos(f*x+e)^6*a^2*b+22*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^7*a^2*b+17*((2*I*a^(1/2)*
b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^5*a*b^2+3*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(
f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+c
os(f*x+e))/(a+b))^(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))*cos(f*x+e)^2*sin(f*x+e)*b^3-30*2^(1/2)*((I*a^
(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*
cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(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))*sin(f*x+e)*cos(f*x+e)^2*a^3-6*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(
f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/
2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(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))*sin(f*x+e)*cos(f*x+e)^2*b^3+15*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*
b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(
f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(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))*sin(f*x+e)*cos(f*x+e)^2*a^3)*co
s(f*x+e)*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(3/2)*sin(f*x+e)/(-1+cos(f*x+e))/(b+a*cos(f*x+e)^2)^2/a/((2*I*a^(1/
2)*b^(1/2)+a-b)/(a+b))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\sin \left (e+f\,x\right )}^6\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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