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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {4217, 478, 591, 596, 537, 223, 212, 385, 209} \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^4(e+f x) \, dx=\frac {3 \left (a^2-6 a b+b^2\right ) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{8 \sqrt {a} f}+\frac {3 \sqrt {b} (a-b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 f}-\frac {3 (a-3 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{8 f}+\frac {3 (a-b) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{8 f}-\frac {\sin ^3(e+f x) \cos (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 f} \]

[In]

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

[Out]

(3*(a^2 - 6*a*b + b^2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(8*Sqrt[a]*f) + (3*(a -
b)*Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*f) - (3*(a - 3*b)*Tan[e + f*x]*S
qrt[a + b + b*Tan[e + f*x]^2])/(8*f) + (3*(a - b)*Sin[e + f*x]^2*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/
(8*f) - (Cos[e + f*x]*Sin[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^(3/2))/(4*f)

Rule 209

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 385

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 478

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 537

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 591

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] + Dis
t[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 596

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 4217

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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^4 \left (a+b+b x^2\right )^{3/2}}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{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}+\frac {\text {Subst}\left (\int \frac {x^2 \sqrt {a+b+b x^2} \left (3 (a+b)+6 b x^2\right )}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 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}-\frac {\text {Subst}\left (\int \frac {x^2 \left (3 (a-5 b) (a+b)+6 (a-3 b) 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 {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}+\frac {\text {Subst}\left (\int \frac {6 (a-3 b) b (a+b)+24 (a-b) b^2 x^2}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{16 b 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}+\frac {(3 (a-b) b) \text {Subst}\left (\int \frac {1}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac {\left (3 \left (a^2-6 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 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}+\frac {(3 (a-b) b) \text {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 (3 \left (a^2-6 a b+b^2\right )\right ) \text {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 )}{8 f} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.57 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.04 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^4(e+f x) \, dx=\frac {3 \left (\left (a^2-6 a b+b^2\right ) \arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )+4 \sqrt {a} (a-b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )\right ) \cos (e+f x) \left (b+a \cos ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)}}{2 \sqrt {2} \sqrt {a} f (a+2 b+a \cos (2 (e+f x)))^{3/2}}+\frac {(-7 a+26 b+(-6 a+10 b) \cos (2 (e+f x))+a \cos (4 (e+f x))) \sqrt {a+b \sec ^2(e+f x)} \tan (e+f x)}{32 f} \]

[In]

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

[Out]

(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])/Sqrt[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)

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1185\) vs. \(2(189)=378\).

Time = 16.91 (sec) , antiderivative size = 1186, normalized size of antiderivative = 5.47

method	result	size
default	\(\text {Expression too large to display}\)	\(1186\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 3.35 (sec) , antiderivative size = 1667, normalized size of antiderivative = 7.68 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^4(e+f x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

[-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*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 24*(a^2 - a*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)*sqr
t((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*(2*a^2*cos(f*x + e)^4 - 5*(
a^2 - a*b)*cos(f*x + e)^2 + 4*a*b)*sqrt((a*cos(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 - 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*(2*a^2*cos(f*x + e)^4 - 5*(a^2 - a*b)*cos(f*x + e)^2 + 4*a*b)*sqrt((a*cos(f*x + e)
^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a*f*cos(f*x + e)), -1/32*(3*(a^2 - 6*a*b + b^2)*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) + 12*(a^2 - a*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*(2*a^2*cos(f*x + e)^4 - 5*(a^2 - a*b)*cos(f*x + e)^2 + 4*a*b)
*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a*f*cos(f*x + e)), -1/32*(3*(a^2 - 6*a*b + b^2)*sq
rt(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*(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) - 4*(2*a^2*cos(f*x + e)^4 - 5*(a^2 - a*b)*cos(f*x + e)^2 + 4*a*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*
x + e)^2)*sin(f*x + e))/(a*f*cos(f*x + e))]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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