\(\int (a+b \sec ^2(e+f x))^{3/2} \, dx\) [89]
Optimal result
Integrand size = 16, antiderivative size = 118 \[
\int \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}+\frac {\sqrt {b} (3 a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 f}+\frac {b \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 f}
\]
[Out]
a^(3/2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/f+1/2*(3*a+b)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*
tan(f*x+e)^2)^(1/2))*b^(1/2)/f+1/2*b*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/f
Rubi [A] (verified)
Time = 0.11 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of
steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {4213, 427, 537, 223, 212, 385,
209} \[
\int \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{f}+\frac {\sqrt {b} (3 a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 f}+\frac {b \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 f}
\]
[In]
Int[(a + b*Sec[e + f*x]^2)^(3/2),x]
[Out]
(a^(3/2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/f + (Sqrt[b]*(3*a + b)*ArcTanh[(Sqrt[b
]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*f) + (b*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(2*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 427
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*x*(a + b*x^n)^(p + 1)*((c
+ d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] && !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, 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 4213
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]
Rubi steps \begin{align*}
\text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (a+b+b x^2\right )^{3/2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {b \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 f}+\frac {\text {Subst}\left (\int \frac {(a+b) (2 a+b)+b (3 a+b) x^2}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 f} \\ & = \frac {b \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 f}+\frac {a^2 \text {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 {(b (3 a+b)) \text {Subst}\left (\int \frac {1}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 f} \\ & = \frac {b \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 f}+\frac {a^2 \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 )}{f}+\frac {(b (3 a+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 {a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}+\frac {\sqrt {b} (3 a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 f}+\frac {b \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 f} \\
\end{align*}
Mathematica [C] (warning: unable to verify)
Result contains complex when optimal does not.
Time = 3.46 (sec) , antiderivative size = 527, normalized size of antiderivative = 4.47
\[
\int \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {2} e^{i (e+f x)} \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \cos ^3(e+f x) \left (-\frac {i b \left (-1+e^{2 i (e+f x)}\right )}{\left (1+e^{2 i (e+f x)}\right )^2}+\frac {2 a^{3/2} f x-i a^{3/2} \log \left (a+2 b+a e^{2 i (e+f x)}+\sqrt {a} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )+i a^{3/2} \log \left (a+a e^{2 i (e+f x)}+2 b e^{2 i (e+f x)}+\sqrt {a} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )-3 a \sqrt {b} \log \left (\frac {-2 \sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) f+2 i \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2} f}{b (3 a+b) \left (1+e^{2 i (e+f x)}\right )}\right )-b^{3/2} \log \left (\frac {-2 \sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) f+2 i \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2} f}{b (3 a+b) \left (1+e^{2 i (e+f x)}\right )}\right )}{\sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}}\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{f (a+2 b+a \cos (2 e+2 f x))^{3/2}}
\]
[In]
Integrate[(a + b*Sec[e + f*x]^2)^(3/2),x]
[Out]
(Sqrt[2]*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)
*b*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))^2 + (2*a^(3/2)*f*x - I*a^(3/2)*Log[a + 2*b + a*E^((2*
I)*(e + f*x)) + Sqrt[a]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]] + I*a^(3/2)*Log[a + a*E
^((2*I)*(e + f*x)) + 2*b*E^((2*I)*(e + f*x)) + Sqrt[a]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x
)))^2]] - 3*a*Sqrt[b]*Log[(-2*Sqrt[b]*(-1 + E^((2*I)*(e + f*x)))*f + (2*I)*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1
+ E^((2*I)*(e + f*x)))^2]*f)/(b*(3*a + b)*(1 + E^((2*I)*(e + f*x))))] - b^(3/2)*Log[(-2*Sqrt[b]*(-1 + E^((2*I
)*(e + f*x)))*f + (2*I)*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]*f)/(b*(3*a + b)*(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))/(f*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2))
Maple [B] (warning: unable to verify)
Leaf count of result is larger than twice the leaf count of optimal. \(708\) vs. \(2(100)=200\).
Time = 9.88 (sec) , antiderivative size = 709, normalized size of antiderivative = 6.01
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| method | result | size |
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\(\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \left (\cos \left (f x +e \right )^{3} b^{\frac {5}{2}} \ln \left (\frac {-4 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {b}\, \cos \left (f x +e \right )-4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 \sin \left (f x +e \right ) a -4 a -4 b}{\sin \left (f x +e \right )-1}\right ) \sqrt {-a}+\cos \left (f x +e \right )^{3} b^{\frac {5}{2}} \ln \left (-\frac {4 \left (-\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {b}\, \cos \left (f x +e \right )+\sin \left (f x +e \right ) a -\sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+a +b \right )}{\sin \left (f x +e \right )+1}\right ) \sqrt {-a}+3 \cos \left (f x +e \right )^{3} b^{\frac {3}{2}} \ln \left (\frac {-4 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {b}\, \cos \left (f x +e \right )-4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 \sin \left (f x +e \right ) a -4 a -4 b}{\sin \left (f x +e \right )-1}\right ) \sqrt {-a}\, a +3 \cos \left (f x +e \right )^{3} b^{\frac {3}{2}} \ln \left (-\frac {4 \left (-\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {b}\, \cos \left (f x +e \right )+\sin \left (f x +e \right ) a -\sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+a +b \right )}{\sin \left (f x +e \right )+1}\right ) \sqrt {-a}\, a +2 \sin \left (f x +e \right ) \cos \left (f x +e \right )^{2} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {-a}\, b^{2}+4 \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \cos \left (f x +e \right )^{3} a^{2} b +2 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2} \cos \left (f x +e \right ) \sin \left (f x +e \right )\right )}{4 f \sqrt {-a}\, b \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (b +a \cos \left (f x +e \right )^{2}\right ) \left (1+\cos \left (f x +e \right )\right )}\) |
\(709\) |
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[In]
int((a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
[Out]
1/4/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))*(cos(f*x+e)^3*b^(5/2)*ln(4*(-((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(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)^(1/2)+cos(f*x+e)^3*b^(5/2
)*ln(-4*(-((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(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)^(1/2)+3*cos(f*x+e)^3*b^(3/2)*ln(4*(-((b+a*cos(f*x+e)^2)/
(1+cos(f*x+e))^2)^(1/2)*b^(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)^(1/2)*a+3*cos(f*x+e)^3*b^(3/2)*ln(-4*(-((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(
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)^(1/
2)*a+2*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*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+2*(-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 [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 288 vs. \(2 (100) = 200\).
Time = 0.61 (sec) , antiderivative size = 1457, normalized size of antiderivative = 12.35
\[
\int \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display}
\]
[In]
integrate((a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
[Out]
[1/8*(sqrt(-a)*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)) + (3*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)/co
s(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4) + 4*b*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x
+ e))/(f*cos(f*x + e)), 1/8*(2*(3*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) + sqr
t(-a)*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)) + 4*b*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(f*cos(f*x + e)), -1/8*(2
*a^(3/2)*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))*s
qrt(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)*co
s(f*x + e)^2)*sin(f*x + e)))*cos(f*x + e) - (3*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*b*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)
^2)*sin(f*x + e))/(f*cos(f*x + e)), -1/4*(a^(3/2)*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) - (3*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*c
os(f*x + e)^2 + b^2)*sin(f*x + e)))*cos(f*x + e) - 2*b*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e
))/(f*cos(f*x + e))]
Sympy [F]
\[
\int \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}\, dx
\]
[In]
integrate((a+b*sec(f*x+e)**2)**(3/2),x)
[Out]
Integral((a + b*sec(e + f*x)**2)**(3/2), x)
Maxima [F]
\[
\int \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \,d x }
\]
[In]
integrate((a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
[Out]
integrate((b*sec(f*x + e)^2 + a)^(3/2), x)
Giac [F]
\[
\int \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \,d x }
\]
[In]
integrate((a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
[Out]
integrate((b*sec(f*x + e)^2 + a)^(3/2), x)
Mupad [F(-1)]
Timed out. \[
\int \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int {\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x
\]
[In]
int((a + b/cos(e + f*x)^2)^(3/2),x)
[Out]
int((a + b/cos(e + f*x)^2)^(3/2), x)