3.277 \(\int \frac{\sec ^4(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=77 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{b^{3/2} f}-\frac{a \tan (e+f x)}{b f (a+b) \sqrt{a+b \tan ^2(e+f x)+b}} \]
[Out]
ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(b^(3/2)*f) - (a*Tan[e + f*x])/(b*(a + b)*f*Sqr
t[a + b + b*Tan[e + f*x]^2])
________________________________________________________________________________________
Rubi [A] time = 0.0969246, antiderivative size = 77, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used =
{4146, 385, 217, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{b^{3/2} f}-\frac{a \tan (e+f x)}{b f (a+b) \sqrt{a+b \tan ^2(e+f x)+b}} \]
Antiderivative was successfully verified.
[In]
Int[Sec[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(3/2),x]
[Out]
ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(b^(3/2)*f) - (a*Tan[e + f*x])/(b*(a + b)*f*Sqr
t[a + b + b*Tan[e + f*x]^2])
Rule 4146
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[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]
Rule 385
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])
Rule 217
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 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{\left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{a \tan (e+f x)}{b (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{b f}\\ &=-\frac{a \tan (e+f x)}{b (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}+\frac{\operatorname{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 )}{b f}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{b^{3/2} f}-\frac{a \tan (e+f x)}{b (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}\\ \end{align*}
Mathematica [C] time = 7.10777, size = 405, normalized size = 5.26 \[ -\frac{\tan (e+f x) \sec ^4(e+f x) \sqrt{1-\frac{2 a \sin ^2(e+f x)}{2 a+2 b}} (a \cos (2 e+2 f x)+a+2 b)^{3/2} \left (4 b (a+b) \sin ^2(e+f x) \text{Hypergeometric2F1}\left (2,2,\frac{7}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \left (-\frac{b \tan ^2(e+f x) \sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{(a+b)^2}\right )^{3/2}+15 \sec ^2(e+f x) \left (a^2 \left (2 \sin ^4(e+f x)-5 \sin ^2(e+f x)+3\right )+a b \left (6-5 \sin ^2(e+f x)\right )+3 b^2\right ) \sin ^{-1}\left (\sqrt{-\frac{b \tan ^2(e+f x)}{a+b}}\right )+15 (a+b) \left (a \left (2 \sin ^2(e+f x)-3\right )-3 b\right ) \sqrt{-\frac{b \tan ^2(e+f x) \sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{(a+b)^2}}\right )}{15 f (a+b)^2 (2 a+2 b) \sqrt{-2 a \sin ^2(e+f x)+2 a+2 b} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \left (-\frac{b \tan ^2(e+f x)}{a+b}\right )^{3/2} \left (a+b \sec ^2(e+f x)\right )^{3/2} \sqrt{\frac{a+b \sec ^2(e+f x)}{a+b}}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sec[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(3/2),x]
[Out]
-((a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)*Sec[e + f*x]^4*Sqrt[1 - (2*a*Sin[e + f*x]^2)/(2*a + 2*b)]*Tan[e + f*x]*
(15*ArcSin[Sqrt[-((b*Tan[e + f*x]^2)/(a + b))]]*Sec[e + f*x]^2*(3*b^2 + a*b*(6 - 5*Sin[e + f*x]^2) + a^2*(3 -
5*Sin[e + f*x]^2 + 2*Sin[e + f*x]^4)) + 15*(a + b)*(-3*b + a*(-3 + 2*Sin[e + f*x]^2))*Sqrt[-((b*Sec[e + f*x]^2
*(a + b - a*Sin[e + f*x]^2)*Tan[e + f*x]^2)/(a + b)^2)] + 4*b*(a + b)*Hypergeometric2F1[2, 2, 7/2, -((b*Tan[e
+ f*x]^2)/(a + b))]*Sin[e + f*x]^2*(-((b*Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)*Tan[e + f*x]^2)/(a + b)^2))
^(3/2)))/(15*(a + b)^2*(2*a + 2*b)*f*(a + b*Sec[e + f*x]^2)^(3/2)*Sqrt[(a + b*Sec[e + f*x]^2)/(a + b)]*Sqrt[2*
a + 2*b - 2*a*Sin[e + f*x]^2]*Sqrt[1 - (a*Sin[e + f*x]^2)/(a + b)]*(-((b*Tan[e + f*x]^2)/(a + b)))^(3/2))
________________________________________________________________________________________
Maple [C] time = 0.472, size = 3068, normalized size = 39.8 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x)
[Out]
-1/2/f/b^(3/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/(a+b)*(2*2^(1/2)*sin(f*x+e)*cos(f*x+e)^2*(1/(a+b)*(I*co
s(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/
2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(
1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*b^
(3/2)*a-4*2^(1/2)*sin(f*x+e)*cos(f*x+e)^2*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e
)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*
x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1
/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*b^(3/2)*a+2*2
^(1/2)*sin(f*x+e)*cos(f*x+e)^2*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos
(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2
)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1
/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*b^(1/2)*a^2-4*2^(1/2)*sin(f*x+e)*cos(f*x+e)^2*(1/(a+b)*(I*cos(f*x+e
)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1
/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a
-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a
^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*b^(1/2)*a^2+2*2^(1/2)*sin(f*x+e)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^
(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-
a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*
x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*b^(5/2)-4*2^(1/2)*sin(f*x+e)*(1
/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(
f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*
I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(
a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*b^(5/2)+2*2^(1/2)*sin(f*x+e)*(1/(a+b)*(I*cos(f*x+e)*a^(1/
2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a
^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b
))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*b^(3/2)*a-4*2^(1
/2)*sin(f*x+e)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*
(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1
+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/
2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*b^(3/2)*a-sin(f*x+e)*cos(f*x+e)^2*((b+a*
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*arctanh(1/8*b^(1/2)*4^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*4^(1/2)-2*cos(f*
x+e)-4^(1/2)-2)/sin(f*x+e)^2/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(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)*arctanh(1/8*b^(1/2)*4^(1/2)*(-1+co
s(f*x+e))*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)
)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b+sin(f*x+e)*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1
/2)*arctanh(1/4*b^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/((b+a*cos(f*x
+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(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)*arctanh(1/4*b^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)
-2)/sin(f*x+e)^2/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b+2*co
s(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^(1/2)*a^2-sin(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2
)^(1/2)*arctanh(1/8*b^(1/2)*4^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/(
(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b-sin(f*x+e)*((b+a*cos(f
*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*arctanh(1/8*b^(1/2)*4^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-
4^(1/2)-2)/sin(f*x+e)^2/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b
^2+sin(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*arctanh(1/4*b^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*4^(1
/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((2*I*a^(1/2)*b^(1/2)+a-
b)/(a+b))^(1/2)*a*b+sin(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*arctanh(1/4*b^(1/2)*(-1+cos(f*x+e))
*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((2*I*a
^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^2-2*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^(1/2)*a^2+2*cos(
f*x+e)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^(3/2)*a-2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^(3/2)*a)*
sin(f*x+e)/(-1+cos(f*x+e))/cos(f*x+e)^3/((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(3/2)
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 0.805869, size = 1007, normalized size = 13.08 \begin{align*} \left [-\frac{4 \, a b \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) -{\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + a b + b^{2}\right )} \sqrt{b} \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 (a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} +{\left (a b^{3} + b^{4}\right )} f\right )}}, -\frac{2 \, a b \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) -{\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + a b + b^{2}\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 )}{2 \,{\left ({\left (a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} +{\left (a b^{3} + b^{4}\right )} f\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
[Out]
[-1/4*(4*a*b*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)*sin(f*x + e) - ((a^2 + a*b)*cos(f*x + e)
^2 + a*b + b^2)*sqrt(b)*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)*co
s(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))/((a^2*b^2 + a*b^3)*f*cos(f*x + e)^2 + (a*b^3 + b^4)*f), -1/2*(2*a*b*sqrt((a*cos(f*x + e)^2 + b)/
cos(f*x + e)^2)*cos(f*x + e)*sin(f*x + e) - ((a^2 + a*b)*cos(f*x + e)^2 + a*b + b^2)*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))))/((a^2*b^2 + a*b^3)*f*cos(f*x + e)^2 + (a*b^3 + b^4)*f)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{4}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)**4/(a+b*sec(f*x+e)**2)**(3/2),x)
[Out]
Integral(sec(e + f*x)**4/(a + b*sec(e + f*x)**2)**(3/2), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (f x + e\right )^{4}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
[Out]
integrate(sec(f*x + e)^4/(b*sec(f*x + e)^2 + a)^(3/2), x)