\(\int \sec ^4(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\) [248]

Optimal result

Integrand size = 25, antiderivative size = 165 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {(a-5 b) (a+b)^2 \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 b^{3/2} f}-\frac {(a-5 b) (a+b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 b f}-\frac {(a-5 b) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{24 b f}+\frac {\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{6 b f} \] Output:

-1/16*(a-5*b)*(a+b)^2*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2 
))/b^(3/2)/f-1/16*(a-5*b)*(a+b)*tan(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/b/f- 
1/24*(a-5*b)*tan(f*x+e)*(a+b+b*tan(f*x+e)^2)^(3/2)/b/f+1/6*tan(f*x+e)*(a+b 
+b*tan(f*x+e)^2)^(5/2)/b/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 7.84 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.42 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {e^{i (e+f x)} \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \cos ^3(e+f x) \left (-\frac {i \sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) \left (3 a^2 \left (1+e^{2 i (e+f x)}\right )^4+2 a b \left (1+e^{2 i (e+f x)}\right )^2 \left (11+50 e^{2 i (e+f x)}+11 e^{4 i (e+f x)}\right )+b^2 \left (15+100 e^{2 i (e+f x)}+298 e^{4 i (e+f x)}+100 e^{6 i (e+f x)}+15 e^{8 i (e+f x)}\right )\right )}{\left (1+e^{2 i (e+f x)}\right )^6}+\frac {3 (a-5 b) (a+b)^2 \log \left (\frac {-4 \sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) f+4 i \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2} f}{1+e^{2 i (e+f x)}}\right )}{\sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}}\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{12 \sqrt {2} b^{3/2} f (a+2 b+a \cos (2 e+2 f x))^{3/2}} \] Input:

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

Output:

(E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2*I)*(e + 
f*x))]*Cos[e + f*x]^3*(((-I)*Sqrt[b]*(-1 + E^((2*I)*(e + f*x)))*(3*a^2*(1 
+ E^((2*I)*(e + f*x)))^4 + 2*a*b*(1 + E^((2*I)*(e + f*x)))^2*(11 + 50*E^(( 
2*I)*(e + f*x)) + 11*E^((4*I)*(e + f*x))) + b^2*(15 + 100*E^((2*I)*(e + f* 
x)) + 298*E^((4*I)*(e + f*x)) + 100*E^((6*I)*(e + f*x)) + 15*E^((8*I)*(e + 
 f*x)))))/(1 + E^((2*I)*(e + f*x)))^6 + (3*(a - 5*b)*(a + b)^2*Log[(-4*Sqr 
t[b]*(-1 + E^((2*I)*(e + f*x)))*f + (4*I)*Sqrt[4*b*E^((2*I)*(e + f*x)) + a 
*(1 + E^((2*I)*(e + f*x)))^2]*f)/(1 + E^((2*I)*(e + f*x)))])/Sqrt[4*b*E^(( 
2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2])*(a + b*Sec[e + f*x]^2)^( 
3/2))/(12*Sqrt[2]*b^(3/2)*f*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4634, 299, 211, 211, 224, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 299

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

((Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(5/2))/(6*b) - ((a - 5*b)*((Tan[ 
e + f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2))/4 + (3*(a + b)*(((a + b)*ArcTan 
h[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*Sqrt[b]) + (T 
an[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/2))/4))/(6*b))/f
 

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 3042

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

rule 4634

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1150\) vs. \(2(145)=290\).

Time = 48.30 (sec) , antiderivative size = 1151, normalized size of antiderivative = 6.98

method	result	size
default	\(\text {Expression too large to display}\)	\(1151\)

Input:

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

Output:

-1/96/f/b^(9/2)*(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)*(3*cos(f*x+e)^3* 
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))*a^3*b^3-9*cos(f*x+e)^3*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))*a^2*b^4-27*cos(f*x+e)^3*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))*a*b 
^5-15*cos(f*x+e)^3*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^6+3*cos(f*x+e)^3*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))*a^3*b^3-9*cos(f*x+e 
)^3*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)/(si 
n(f*x+e)+1))*a^2*b^4-27*cos(f*x+e)^3*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))*a*b^5-15*cos(f*x+e)^3*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*...
 

Fricas [A] (verification not implemented)

Time = 0.83 (sec) , antiderivative size = 470, normalized size of antiderivative = 2.85 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\left [-\frac {3 \, {\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} \sqrt {b} \cos \left (f x + e\right )^{5} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 4 \, {\left ({\left (3 \, a^{2} b + 22 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + 8 \, b^{3} + 2 \, {\left (7 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{192 \, b^{2} f \cos \left (f x + e\right )^{5}}, -\frac {3 \, {\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{5} - 2 \, {\left ({\left (3 \, a^{2} b + 22 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + 8 \, b^{3} + 2 \, {\left (7 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{96 \, b^{2} f \cos \left (f x + e\right )^{5}}\right ] \] Input:

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

Output:

[-1/192*(3*(a^3 - 3*a^2*b - 9*a*b^2 - 5*b^3)*sqrt(b)*cos(f*x + e)^5*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*((3*a^2*b + 22* 
a*b^2 + 15*b^3)*cos(f*x + e)^4 + 8*b^3 + 2*(7*a*b^2 + 5*b^3)*cos(f*x + e)^ 
2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(b^2*f*cos(f* 
x + e)^5), -1/96*(3*(a^3 - 3*a^2*b - 9*a*b^2 - 5*b^3)*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)^5 - 2*((3*a^2*b + 22*a*b^2 + 15*b^3)*cos(f*x + e)^4 + 8*b^3 + 2*(7*a 
*b^2 + 5*b^3)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)* 
sin(f*x + e))/(b^2*f*cos(f*x + e)^5)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.47 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {\frac {3 \, {\left (a + b\right )}^{2} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} + \frac {3 \, {\left (a + b\right )}^{2} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} - \frac {18 \, {\left (a + b\right )} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} - 18 \, {\left (a + b\right )} \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) - 12 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \tan \left (f x + e\right ) - 18 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right ) - \frac {8 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {5}{2}} \tan \left (f x + e\right )}{b} + \frac {2 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \tan \left (f x + e\right )}{b} + \frac {3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2} \tan \left (f x + e\right )}{b}}{48 \, f} \] Input:

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

Output:

-1/48*(3*(a + b)^2*a*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/b^(3/2) + 3*( 
a + b)^2*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/sqrt(b) - 18*(a + b)*a*ar 
csinh(b*tan(f*x + e)/sqrt((a + b)*b))/sqrt(b) - 18*(a + b)*sqrt(b)*arcsinh 
(b*tan(f*x + e)/sqrt((a + b)*b)) - 12*(b*tan(f*x + e)^2 + a + b)^(3/2)*tan 
(f*x + e) - 18*sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)*tan(f*x + e) - 8*(b* 
tan(f*x + e)^2 + a + b)^(5/2)*tan(f*x + e)/b + 2*(b*tan(f*x + e)^2 + a + b 
)^(3/2)*(a + b)*tan(f*x + e)/b + 3*sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^ 
2*tan(f*x + e)/b)/f
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{6}d x \right ) b +\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{4}d x \right ) a \] Input:

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

Output:

int(sqrt(sec(e + f*x)**2*b + a)*sec(e + f*x)**6,x)*b + int(sqrt(sec(e + f* 
x)**2*b + a)*sec(e + f*x)**4,x)*a