\(\int (a+b \sec ^2(e+f x))^{3/2} \sin ^6(e+f x) \, dx\) [86]

Optimal result

Integrand size = 25, antiderivative size = 298 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx=\frac {\left (5 a^3-45 a^2 b+15 a b^2+b^3\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 a^{3/2} f}+\frac {(3 a-5 b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 f}-\frac {\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 a f}+\frac {\left (5 a^2-40 a b+3 b^2\right ) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{48 a f}+\frac {(5 a-3 b) \sin ^4(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 f}-\frac {\cos (e+f x) \sin ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{6 f} \] Output:

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

Mathematica [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx=\int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.04, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 4620, 369, 439, 440, 27, 444, 27, 398, 224, 219, 291, 216}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4620

\(\displaystyle \frac {\int \frac {\tan ^6(e+f x) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}{\left (\tan ^2(e+f x)+1\right )^4}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 369

\(\displaystyle \frac {\frac {1}{6} \int \frac {\tan ^4(e+f x) \sqrt {b \tan ^2(e+f x)+a+b} \left (8 b \tan ^2(e+f x)+5 (a+b)\right )}{\left (\tan ^2(e+f x)+1\right )^3}d\tan (e+f x)-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\)

\(\Big \downarrow \) 439

\(\displaystyle \frac {\frac {1}{6} \left (\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}-\frac {1}{4} \int \frac {\tan ^4(e+f x) \left (2 (5 a-19 b) b \tan ^2(e+f x)+5 (a-7 b) (a+b)\right )}{\left (\tan ^2(e+f x)+1\right )^2 \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\)

\(\Big \downarrow \) 440

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {\int \frac {3 \tan ^2(e+f x) \left (2 b \left (5 a^2-26 b a+b^2\right ) \tan ^2(e+f x)+(a+b) \left (5 a^2-40 b a+3 b^2\right )\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {3 \int \frac {\tan ^2(e+f x) \left (2 b \left (5 a^2-26 b a+b^2\right ) \tan ^2(e+f x)+(a+b) \left (5 a^2-40 b a+3 b^2\right )\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\)

\(\Big \downarrow \) 444

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {3 \left (\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\frac {\int \frac {2 b \left (8 a (3 a-5 b) b \tan ^2(e+f x)+(a+b) \left (5 a^2-26 b a+b^2\right )\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 b}\right )}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {3 \left (\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\int \frac {8 a (3 a-5 b) b \tan ^2(e+f x)+(a+b) \left (5 a^2-26 b a+b^2\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)\right )}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\)

\(\Big \downarrow \) 398

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {3 \left (-\left (5 a^3-45 a^2 b+15 a b^2+b^3\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-8 a b (3 a-5 b) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)+\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {3 \left (-\left (5 a^3-45 a^2 b+15 a b^2+b^3\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-8 a b (3 a-5 b) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}+\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {3 \left (-\left (5 a^3-45 a^2 b+15 a b^2+b^3\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)+\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-8 a \sqrt {b} (3 a-5 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )\right )}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {3 \left (-\left (5 a^3-45 a^2 b+15 a b^2+b^3\right ) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}+\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-8 a \sqrt {b} (3 a-5 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )\right )}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (5 a^2-40 a b+3 b^2\right ) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a \left (\tan ^2(e+f x)+1\right )}-\frac {3 \left (\left (5 a^2-26 a b+b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}-\frac {\left (5 a^3-45 a^2 b+15 a b^2+b^3\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {a}}-8 a \sqrt {b} (3 a-5 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )\right )}{2 a}\right )+\frac {(5 a-3 b) \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {\tan ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\)

Input:

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

Output:

(-1/6*(Tan[e + f*x]^5*(a + b + b*Tan[e + f*x]^2)^(3/2))/(1 + Tan[e + f*x]^ 
2)^3 + (((5*a - 3*b)*Tan[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2])/(4*(1 
+ Tan[e + f*x]^2)^2) + (((5*a^2 - 40*a*b + 3*b^2)*Tan[e + f*x]^3*Sqrt[a + 
b + b*Tan[e + f*x]^2])/(2*a*(1 + Tan[e + f*x]^2)) - (3*(-(((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]])/Sqrt[a]) - 8*a*(3*a - 5*b)*Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x] 
)/Sqrt[a + b + b*Tan[e + f*x]^2]] + (5*a^2 - 26*a*b + b^2)*Tan[e + f*x]*Sq 
rt[a + b + b*Tan[e + f*x]^2]))/(2*a))/4)/6)/f
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

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

rule 219

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

rule 224

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 291

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

rule 369

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

rule 398

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) 
, x_Symbol] :> Simp[f/b   Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ 
b   Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} 
, x]
 

rule 439

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a 
 + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*b*g*(p + 1))), x] + Simp[1/(2*a*b*(p 
+ 1))   Int[(g*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1)*Simp[c*(2*b*e*( 
p + 1) + (b*e - a*f)*(m + 1)) + d*(2*b*e*(p + 1) + (b*e - a*f)*(m + 2*q + 1 
))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && G 
tQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])
 

rule 440

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[g*(b*e - a*f)*(g*x)^(m - 1)*(a + 
 b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] - Simp[ 
g^2/(2*b*(b*c - a*d)*(p + 1))   Int[(g*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + 
d*x^2)^q*Simp[c*(b*e - a*f)*(m - 1) + (d*(b*e - a*f)*(m + 2*q + 1) - b*2*(c 
*f - d*e)*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, q}, x] && 
 LtQ[p, -1] && GtQ[m, 1]
 

rule 444

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q 
_.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[f*g*(g*x)^(m - 1)*(a + b*x^2)^ 
(p + 1)*((c + d*x^2)^(q + 1)/(b*d*(m + 2*(p + q + 1) + 1))), x] - Simp[g^2/ 
(b*d*(m + 2*(p + q + 1) + 1))   Int[(g*x)^(m - 2)*(a + b*x^2)^p*(c + d*x^2) 
^q*Simp[a*f*c*(m - 1) + (a*f*d*(m + 2*q + 1) + b*(f*c*(m + 2*p + 1) - e*d*( 
m + 2*(p + q + 1) + 1)))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, 
q}, x] && GtQ[m, 1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4620

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1171\) vs. \(2(266)=532\).

Time = 37.06 (sec) , antiderivative size = 1172, normalized size of antiderivative = 3.93

method	result	size
default	\(\text {Expression too large to display}\)	\(1172\)

Input:

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

Output:

-1/48/f/(-a)^(1/2)/b/a*(a+b*sec(f*x+e)^2)^(3/2)/((b+a*cos(f*x+e)^2)/(1+cos 
(f*x+e))^2)^(1/2)/(a*cos(f*x+e)^3+a*cos(f*x+e)^2+cos(f*x+e)*b+b)*(60*ln(4* 
(-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+sin(f*x+e 
)*a-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-a-b)/(sin(f*x+e)-1 
))*b^(5/2)*(-a)^(1/2)*a*cos(f*x+e)^3-36*ln(4*(-b^(1/2)*((b+a*cos(f*x+e)^2) 
/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+sin(f*x+e)*a-b^(1/2)*((b+a*cos(f*x+e)^ 
2)/(1+cos(f*x+e))^2)^(1/2)-a-b)/(sin(f*x+e)-1))*b^(3/2)*(-a)^(1/2)*a^2*cos 
(f*x+e)^3+60*ln(-4*(-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*c 
os(f*x+e)+sin(f*x+e)*a-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2) 
+a+b)/(sin(f*x+e)+1))*b^(5/2)*(-a)^(1/2)*a*cos(f*x+e)^3-36*ln(-4*(-b^(1/2) 
*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+sin(f*x+e)*a-b^(1/ 
2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+a+b)/(sin(f*x+e)+1))*b^(3/2 
)*(-a)^(1/2)*a^2*cos(f*x+e)^3-15*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+co 
s(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+ 
e))^2)^(1/2)-4*sin(f*x+e)*a)*a^3*b*cos(f*x+e)^3+135*ln(4*(-a)^(1/2)*((b+a* 
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f* 
x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*a^2*b^2*cos(f*x+e)^3-45*ln 
(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a 
)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*a*b^3* 
cos(f*x+e)^3-3*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1...
 

Fricas [A] (verification not implemented)

Time = 9.54 (sec) , antiderivative size = 1855, normalized size of antiderivative = 6.22 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/384*(3*(5*a^3 - 45*a^2*b + 15*a*b^2 + b^3)*sqrt(-a)*cos(f*x + e)*log(1 
28*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))*sq 
rt(-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*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) + 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)*sqr 
t((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)*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*...
 

Sympy [F(-1)]

Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )^{6} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )^{6} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^6\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^6(e+f x) \, dx=\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{2} \sin \left (f x +e \right )^{6}d x \right ) b +\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sin \left (f x +e \right )^{6}d x \right ) a \] Input:

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

Output:

int(sqrt(sec(e + f*x)**2*b + a)*sec(e + f*x)**2*sin(e + f*x)**6,x)*b + int 
(sqrt(sec(e + f*x)**2*b + a)*sin(e + f*x)**6,x)*a