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

Optimal. Leaf size=290 \[ -\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{f}+\frac {\left (3 a^2-50 a b-5 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{192 b f}-\frac {\left (3 a^3+17 a^2 b-55 a b^2-5 b^3\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{128 b^2 f}+\frac {\left (3 a^4+20 a^3 b+90 a^2 b^2-60 a b^3-5 b^4\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{128 b^{5/2} f}+\frac {b \tan ^7(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{8 f}+\frac {(9 a+b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{48 f} \]

[Out]

-a^(3/2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/f+1/128*(3*a^4+20*a^3*b+90*a^2*b^2-60*a*b^3-5*b
^4)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/b^(5/2)/f-1/128*(3*a^3+17*a^2*b-55*a*b^2-5*b^3)*(a+
b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/b^2/f+1/192*(3*a^2-50*a*b-5*b^2)*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^3/b/
f+1/48*(9*a+b)*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^5/f+1/8*b*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^7/f

________________________________________________________________________________________

Rubi [A]  time = 0.57, antiderivative size = 290, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {4141, 1975, 477, 582, 523, 217, 206, 377, 203} \[ \frac {\left (3 a^2-50 a b-5 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{192 b f}-\frac {\left (17 a^2 b+3 a^3-55 a b^2-5 b^3\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{128 b^2 f}+\frac {\left (90 a^2 b^2+20 a^3 b+3 a^4-60 a b^3-5 b^4\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{128 b^{5/2} f}-\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{f}+\frac {b \tan ^7(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{8 f}+\frac {(9 a+b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{48 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^(3/2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/f) + ((3*a^4 + 20*a^3*b + 90*a^2*b^2
- 60*a*b^3 - 5*b^4)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(128*b^(5/2)*f) - ((3*a^3
+ 17*a^2*b - 55*a*b^2 - 5*b^3)*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(128*b^2*f) + ((3*a^2 - 50*a*b -
5*b^2)*Tan[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/(192*b*f) + ((9*a + b)*Tan[e + f*x]^5*Sqrt[a + b + b*Tan
[e + f*x]^2])/(48*f) + (b*Tan[e + f*x]^7*Sqrt[a + b + b*Tan[e + f*x]^2])/(8*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 477

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*(e*x)
^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(b*e*(m + n*(p + q) + 1)), x] + Dist[1/(b*(m + n*(p + q) + 1
)), Int[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Simp[c*((c*b - a*d)*(m + 1) + c*b*n*(p + q)) + (d*(c*b - a*d
)*(m + 1) + d*n*(q - 1)*(b*c - a*d) + c*b*d*n*(p + q))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && N
eQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 1] && 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 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 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 4141

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

Rubi steps

\begin {align*} \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \tan ^6(e+f x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^6 \left (a+b \left (1+x^2\right )\right )^{3/2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^6 \left (a+b+b x^2\right )^{3/2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {b \tan ^7(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 f}+\frac {\operatorname {Subst}\left (\int \frac {x^6 \left ((a+b) (8 a+b)+b (9 a+b) x^2\right )}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac {(9 a+b) \tan ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 f}+\frac {b \tan ^7(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 f}-\frac {\operatorname {Subst}\left (\int \frac {x^4 \left (5 b (a+b) (9 a+b)-b \left (3 a^2-50 a b-5 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 b f}\\ &=\frac {\left (3 a^2-50 a b-5 b^2\right ) \tan ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{192 b f}+\frac {(9 a+b) \tan ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 f}+\frac {b \tan ^7(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 f}+\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (-3 b (a+b) \left (3 a^2-50 a b-5 b^2\right )-3 b \left (3 a^3+17 a^2 b-55 a b^2-5 b^3\right ) x^2\right )}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{192 b^2 f}\\ &=-\frac {\left (3 a^3+17 a^2 b-55 a b^2-5 b^3\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{128 b^2 f}+\frac {\left (3 a^2-50 a b-5 b^2\right ) \tan ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{192 b f}+\frac {(9 a+b) \tan ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 f}+\frac {b \tan ^7(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 f}-\frac {\operatorname {Subst}\left (\int \frac {-3 b (a+b) \left (3 a^3+17 a^2 b-55 a b^2-5 b^3\right )-3 b \left (3 a^4+20 a^3 b+90 a^2 b^2-60 a b^3-5 b^4\right ) x^2}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{384 b^3 f}\\ &=-\frac {\left (3 a^3+17 a^2 b-55 a b^2-5 b^3\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{128 b^2 f}+\frac {\left (3 a^2-50 a b-5 b^2\right ) \tan ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{192 b f}+\frac {(9 a+b) \tan ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 f}+\frac {b \tan ^7(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 f}-\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}+\frac {\left (3 a^4+20 a^3 b+90 a^2 b^2-60 a b^3-5 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{128 b^2 f}\\ &=-\frac {\left (3 a^3+17 a^2 b-55 a b^2-5 b^3\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{128 b^2 f}+\frac {\left (3 a^2-50 a b-5 b^2\right ) \tan ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{192 b f}+\frac {(9 a+b) \tan ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 f}+\frac {b \tan ^7(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 f}-\frac {a^2 \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 )}{f}+\frac {\left (3 a^4+20 a^3 b+90 a^2 b^2-60 a b^3-5 b^4\right ) \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 )}{128 b^2 f}\\ &=-\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}+\frac {\left (3 a^4+20 a^3 b+90 a^2 b^2-60 a b^3-5 b^4\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{128 b^{5/2} f}-\frac {\left (3 a^3+17 a^2 b-55 a b^2-5 b^3\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{128 b^2 f}+\frac {\left (3 a^2-50 a b-5 b^2\right ) \tan ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{192 b f}+\frac {(9 a+b) \tan ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 f}+\frac {b \tan ^7(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 6.19, size = 353, normalized size = 1.22 \[ -\frac {\tan (e+f x) \sec ^6(e+f x) \left (9 a^3 \cos (6 (e+f x))+90 a^3+57 a^2 b \cos (6 (e+f x))+498 a^2 b+\left (135 a^3+759 a^2 b-2303 a b^2+513 b^3\right ) \cos (2 (e+f x))+2 \left (27 a^3+159 a^2 b-523 a b^2-191 b^3\right ) \cos (4 (e+f x))-337 a b^2 \cos (6 (e+f x))-1594 a b^2+15 b^3 \cos (6 (e+f x))-626 b^3\right ) \sqrt {a+b \sec ^2(e+f x)}}{12288 b^2 f}-\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \left (128 a^{3/2} b^2 \tan ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {-a \sin ^2(e+f x)+a+b}}\right )-\frac {\left (3 a^4+20 a^3 b+90 a^2 b^2-60 a b^3-5 b^4\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {-a \sin ^2(e+f x)+a+b}}\right )}{\sqrt {b}}\right )}{32 \sqrt {2} b^2 f (a \cos (2 e+2 f x)+a+2 b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*((128*a^(3/2)*b^2*ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]] - ((3*a^4 + 20*a^3*b + 9
0*a^2*b^2 - 60*a*b^3 - 5*b^4)*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]])/Sqrt[b])*Cos[e +
f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2))/(Sqrt[2]*b^2*f*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)) - ((90*a^3 + 498*a^
2*b - 1594*a*b^2 - 626*b^3 + (135*a^3 + 759*a^2*b - 2303*a*b^2 + 513*b^3)*Cos[2*(e + f*x)] + 2*(27*a^3 + 159*a
^2*b - 523*a*b^2 - 191*b^3)*Cos[4*(e + f*x)] + 9*a^3*Cos[6*(e + f*x)] + 57*a^2*b*Cos[6*(e + f*x)] - 337*a*b^2*
Cos[6*(e + f*x)] + 15*b^3*Cos[6*(e + f*x)])*Sec[e + f*x]^6*Sqrt[a + b*Sec[e + f*x]^2]*Tan[e + f*x])/(12288*b^2
*f)

________________________________________________________________________________________

fricas [A]  time = 25.23, size = 1973, normalized size = 6.80 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/1536*(192*sqrt(-a)*a*b^3*cos(f*x + e)^7*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)) - 3*(3*a^4 + 20*a^3*b + 90*a^2*b^2 - 60*a*b^3 - 5*b^4)*sqrt(b)*cos(
f*x + e)^7*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*((9*a^3*b + 57*a^2*b^2 - 337*a*b^3 + 15*b^4)*cos(f*x + e)^6 - 2*(3*a^2*b^2 - 122*a*b^3 + 59*b^4)*cos(f*x +
e)^4 - 48*b^4 - 8*(9*a*b^3 - 17*b^4)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)
)/(b^3*f*cos(f*x + e)^7), 1/768*(96*sqrt(-a)*a*b^3*cos(f*x + e)^7*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)*co
s(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))*s
qrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 3*(3*a^4 + 20*a^3*b + 90*a^2*b^2 - 60*a*b^
3 - 5*b^4)*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)^7 - 2*((9*a^3*b + 57*a^2*b^2 - 337*
a*b^3 + 15*b^4)*cos(f*x + e)^6 - 2*(3*a^2*b^2 - 122*a*b^3 + 59*b^4)*cos(f*x + e)^4 - 48*b^4 - 8*(9*a*b^3 - 17*
b^4)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(b^3*f*cos(f*x + e)^7), 1/1536*
(384*a^(3/2)*b^3*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)^7 - 3*(3*a^4 + 20*a^3*b + 90*a^2*b^2 - 60*a*b^3 - 5*b^4)*sqr
t(b)*cos(f*x + e)^7*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*((9*a^3*b + 57*a^2*b^2 - 337*a*b^3 + 15*b^4)*cos(f*x + e)^6 - 2*(3*a^2*b^2 - 122*a*b^3 + 59*b^4)*
cos(f*x + e)^4 - 48*b^4 - 8*(9*a*b^3 - 17*b^4)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin
(f*x + e))/(b^3*f*cos(f*x + e)^7), 1/768*(192*a^(3/2)*b^3*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*co
s(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)^7 + 3*(3*a^4 + 20*a
^3*b + 90*a^2*b^2 - 60*a*b^3 - 5*b^4)*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)^7 - 2*((
9*a^3*b + 57*a^2*b^2 - 337*a*b^3 + 15*b^4)*cos(f*x + e)^6 - 2*(3*a^2*b^2 - 122*a*b^3 + 59*b^4)*cos(f*x + e)^4
- 48*b^4 - 8*(9*a*b^3 - 17*b^4)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(b^3
*f*cos(f*x + e)^7)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 2.44, size = 3583, normalized size = 12.36 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384/f*sin(f*x+e)*(60*sin(f*x+e)*cos(f*x+e)^8*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))*a^3*b-48*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(
1/2)*cos(f*x+e)*b^4+380*cos(f*x+e)^5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^3-380*cos(f*x+e)^4*((2*I*a^(1
/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^3-57*cos(f*x+e)^8*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3*b+337*cos(f*x+
e)^8*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b^2-15*cos(f*x+e)^8*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a
*b^3+301*cos(f*x+e)^7*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b^2-455*cos(f*x+e)^7*((2*I*a^(1/2)*b^(1/2)+a
-b)/(a+b))^(1/2)*a*b^3+57*cos(f*x+e)^9*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3*b-337*cos(f*x+e)^9*((2*I*a^
(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b^2+15*cos(f*x+e)^9*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^3-3*cos(f*
x+e)^6*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3*b-301*cos(f*x+e)^6*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*
a^2*b^2+455*cos(f*x+e)^6*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^3+120*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(
1/2)*cos(f*x+e)^2*a*b^3+3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^7*a^3*b-78*((2*I*a^(1/2)*b^(1/2)+
a-b)/(a+b))^(1/2)*cos(f*x+e)^5*a^2*b^2+78*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^4*a^2*b^2-120*((2
*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^3*a*b^3-15*sin(f*x+e)*cos(f*x+e)^8*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))*b^4-18*si
n(f*x+e)*cos(f*x+e)^8*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)*E
llipticPi((-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^4+30*sin(f*x+e)*cos(f*x+e
)^8*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))*b^4+9*sin(f*x+e)*cos(f*x+e)^8*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)*c
os(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
))*a^4+9*cos(f*x+e)^9*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^4-9*cos(f*x+e)^8*((2*I*a^(1/2)*b^(1/2)+a-b)/(a
+b))^(1/2)*a^4+15*cos(f*x+e)^7*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^4-15*cos(f*x+e)^6*((2*I*a^(1/2)*b^(1/
2)+a-b)/(a+b))^(1/2)*b^4-118*cos(f*x+e)^5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^4+118*cos(f*x+e)^4*((2*I*a
^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^4+136*cos(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^4-136*cos(f*x+
e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^4+48*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^4-114*sin(f*x+e)
*cos(f*x+e)^8*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))*a^2*b^2-180*sin(f*x+e)*cos(f*x+e)^8*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))*a*b^3-120*sin(f*x+e)*cos
(f*x+e)^8*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))*a^3*b-540*sin(f*x+e)*cos(f*x+e)^8*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))*a^2*b^2+360*sin(f*x+e)*cos(f*x+e)^8*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))*a*b^3+768*sin(f*x+e)*cos(f*x+e)^8*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))*a^2*b^2)*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(3/2)/(-1+cos(f*x+e))/(b+a*cos(f*x+e)^
2)^2/cos(f*x+e)^5/b^2/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)*tan(f*x+e)**6,x)

[Out]

Timed out

________________________________________________________________________________________