\(\int (a+b \sec ^2(e+f x))^{3/2} \sin ^4(e+f x) \, dx\) [87]

Optimal result

Integrand size = 25, antiderivative size = 217 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^4(e+f x) \, dx=\frac {3 \left (a^2-6 a b+b^2\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 \sqrt {a} f}+\frac {3 (a-b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 f}-\frac {3 (a-3 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 f}+\frac {3 (a-b) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 f}-\frac {\cos (e+f x) \sin ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{4 f} \] Output:

3/8*(a^2-6*a*b+b^2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/ 
a^(1/2)/f+3/2*(a-b)*b^(1/2)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2 
)^(1/2))/f-3/8*(a-3*b)*tan(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/f+3/8*(a-b)*s 
in(f*x+e)^2*tan(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/f-1/4*cos(f*x+e)*sin(f*x 
+e)^3*(a+b+b*tan(f*x+e)^2)^(3/2)/f
 

Mathematica [A] (verified)

Time = 3.13 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.04 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^4(e+f x) \, dx=\frac {3 \left (\left (a^2-6 a b+b^2\right ) \arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )+4 \sqrt {a} (a-b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )\right ) \cos (e+f x) \left (b+a \cos ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)}}{2 \sqrt {2} \sqrt {a} f (a+2 b+a \cos (2 (e+f x)))^{3/2}}+\frac {(-7 a+26 b+(-6 a+10 b) \cos (2 (e+f x))+a \cos (4 (e+f x))) \sqrt {a+b \sec ^2(e+f x)} \tan (e+f x)}{32 f} \] Input:

Integrate[(a + b*Sec[e + f*x]^2)^(3/2)*Sin[e + f*x]^4,x]
 

Output:

(3*((a^2 - 6*a*b + b^2)*ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e 
 + f*x]^2]] + 4*Sqrt[a]*(a - b)*Sqrt[b]*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqr 
t[a + b - a*Sin[e + f*x]^2]])*Cos[e + f*x]*(b + a*Cos[e + f*x]^2)*Sqrt[a + 
 b*Sec[e + f*x]^2])/(2*Sqrt[2]*Sqrt[a]*f*(a + 2*b + a*Cos[2*(e + f*x)])^(3 
/2)) + ((-7*a + 26*b + (-6*a + 10*b)*Cos[2*(e + f*x)] + a*Cos[4*(e + f*x)] 
)*Sqrt[a + b*Sec[e + f*x]^2]*Tan[e + f*x])/(32*f)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 4620, 369, 27, 439, 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.

\(\displaystyle \int \sin ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (e+f x)^4 \left (a+b \sec (e+f x)^2\right )^{3/2}dx\)

\(\Big \downarrow \) 4620

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

\(\Big \downarrow \) 369

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 439

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

\(\Big \downarrow \) 444

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

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

Input:

Int[(a + b*Sec[e + f*x]^2)^(3/2)*Sin[e + f*x]^4,x]
 

Output:

(-1/4*(Tan[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^(3/2))/(1 + Tan[e + f*x]^ 
2)^2 + (3*(((a - b)*Tan[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/(2*(1 + 
 Tan[e + f*x]^2)) + (((a^2 - 6*a*b + b^2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sq 
rt[a + b + b*Tan[e + f*x]^2]])/Sqrt[a] + 4*(a - b)*Sqrt[b]*ArcTanh[(Sqrt[b 
]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]] - (a - 3*b)*Tan[e + f*x]*S 
qrt[a + b + b*Tan[e + f*x]^2])/2))/4)/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 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. \(960\) vs. \(2(189)=378\).

Time = 36.35 (sec) , antiderivative size = 961, normalized size of antiderivative = 4.43

method	result	size
default	\(\text {Expression too large to display}\)	\(961\)

Input:

int((a+b*sec(f*x+e)^2)^(3/2)*sin(f*x+e)^4,x,method=_RETURNVERBOSE)
 

Output:

1/8/f/b/(-a)^(1/2)*(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)*(-6*ln(4*(b^( 
1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+b^(1/2)*((b+a* 
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*b^ 
(5/2)*(-a)^(1/2)*cos(f*x+e)^3+6*ln(4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f 
*x+e))^2)^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^( 
1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*b^(3/2)*(-a)^(1/2)*a*cos(f*x+e)^3-6 
*ln(-4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+b^( 
1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f* 
x+e)-1))*b^(5/2)*(-a)^(1/2)*cos(f*x+e)^3+6*ln(-4*(b^(1/2)*((b+a*cos(f*x+e) 
^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos( 
f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*b^(3/2)*(-a)^(1/2)*a*co 
s(f*x+e)^3+3*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*c 
os(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*cos(f*x+e)^3-18*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^2*cos(f*x+e)^3+3*ln(4*(-a)^(1/2)*((b+a*cos( 
f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*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)*b^3*cos(f*x+e)^3+sin(f*x+e)*co 
s(f*x+e)^3*(2*cos(f*x+e)^3+2*cos(f*x+e)^2-5*cos(f*x+e)-5)*((b+a*cos(f*x...
 

Fricas [A] (verification not implemented)

Time = 2.92 (sec) , antiderivative size = 1667, normalized size of antiderivative = 7.68 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^4(e+f x) \, dx=\text {Too large to display} \] Input:

integrate((a+b*sec(f*x+e)^2)^(3/2)*sin(f*x+e)^4,x, algorithm="fricas")
 

Output:

[-1/64*(3*(a^2 - 6*a*b + b^2)*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*co 
s(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 24*(a^2 - a*b)*sqrt(b)*c 
os(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*c 
os(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4) - 
 8*(2*a^2*cos(f*x + e)^4 - 5*(a^2 - a*b)*cos(f*x + e)^2 + 4*a*b)*sqrt((a*c 
os(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a*f*cos(f*x + e)), 1/64* 
(48*(a^2 - a*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*(a^2 - 6*a*b + b^2)*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 -...
 

Sympy [F(-1)]

Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^4(e+f x) \, dx=\text {Timed out} \] Input:

integrate((a+b*sec(f*x+e)**2)**(3/2)*sin(f*x+e)**4,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^4(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )^{4} \,d x } \] Input:

integrate((a+b*sec(f*x+e)^2)^(3/2)*sin(f*x+e)^4,x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^(3/2)*sin(f*x + e)^4, x)
 

Giac [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^4(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )^{4} \,d x } \] Input:

integrate((a+b*sec(f*x+e)^2)^(3/2)*sin(f*x+e)^4,x, algorithm="giac")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^(3/2)*sin(f*x + e)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^4(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^4\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \] Input:

int(sin(e + f*x)^4*(a + b/cos(e + f*x)^2)^(3/2),x)
 

Output:

int(sin(e + f*x)^4*(a + b/cos(e + f*x)^2)^(3/2), x)
 

Reduce [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^4(e+f x) \, dx=\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{2} \sin \left (f x +e \right )^{4}d x \right ) b +\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sin \left (f x +e \right )^{4}d x \right ) a \] Input:

int((a+b*sec(f*x+e)^2)^(3/2)*sin(f*x+e)^4,x)
 

Output:

int(sqrt(sec(e + f*x)**2*b + a)*sec(e + f*x)**2*sin(e + f*x)**4,x)*b + int 
(sqrt(sec(e + f*x)**2*b + a)*sin(e + f*x)**4,x)*a