3.289 \(\int \frac {\sec ^6(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=133 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{b^{5/2} f}-\frac {a (3 a+5 b) \tan (e+f x)}{3 b^2 f (a+b)^2 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {a \tan (e+f x) \sec ^2(e+f x)}{3 b f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]
[Out]
arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/b^(5/2)/f-1/3*a*(3*a+5*b)*tan(f*x+e)/b^2/(a+b)^2/f/(a+b
+b*tan(f*x+e)^2)^(1/2)-1/3*a*sec(f*x+e)^2*tan(f*x+e)/b/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(3/2)
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Rubi [A] time = 0.14, antiderivative size = 133, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used =
{4146, 413, 385, 217, 206} \[ -\frac {a (3 a+5 b) \tan (e+f x)}{3 b^2 f (a+b)^2 \sqrt {a+b \tan ^2(e+f x)+b}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{b^{5/2} f}-\frac {a \tan (e+f x) \sec ^2(e+f x)}{3 b f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(5/2),x]
[Out]
ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(b^(5/2)*f) - (a*Sec[e + f*x]^2*Tan[e + f*x])/(
3*b*(a + b)*f*(a + b + b*Tan[e + f*x]^2)^(3/2)) - (a*(3*a + 5*b)*Tan[e + f*x])/(3*b^2*(a + b)^2*f*Sqrt[a + b +
b*Tan[e + f*x]^2])
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])
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 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 413
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]
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]
Rubi steps
\begin {align*} \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{\left (a+b+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {a \sec ^2(e+f x) \tan (e+f x)}{3 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {a+3 b+3 (a+b) x^2}{\left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 b (a+b) f}\\ &=-\frac {a \sec ^2(e+f x) \tan (e+f x)}{3 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (3 a+5 b) \tan (e+f x)}{3 b^2 (a+b)^2 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^2 f}\\ &=-\frac {a \sec ^2(e+f x) \tan (e+f x)}{3 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (3 a+5 b) \tan (e+f x)}{3 b^2 (a+b)^2 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^2 f}\\ &=\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{b^{5/2} f}-\frac {a \sec ^2(e+f x) \tan (e+f x)}{3 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (3 a+5 b) \tan (e+f x)}{3 b^2 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}}\\ \end {align*}
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Mathematica [C] time = 11.70, size = 357, normalized size = 2.68 \[ \frac {e^{i (e+f x)} \sec ^5(e+f x) \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \left (\frac {i a \sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) \left (3 a^2 \left (1+e^{2 i (e+f x)}\right )^2+a b \left (26 e^{2 i (e+f x)}+5 e^{4 i (e+f x)}+5\right )+24 b^2 e^{2 i (e+f x)}\right )}{(a+b)^2 \left (a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}\right )^2}-\frac {3 \log \left (\frac {4 i f \sqrt {a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}-4 \sqrt {b} f \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}\right )}{\sqrt {a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}}\right ) (a \cos (2 e+2 f x)+a+2 b)^{5/2}}{12 \sqrt {2} b^{5/2} f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(5/2),x]
[Out]
(E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2*I)*(e + f*x))]*(a + 2*b + a*Cos[2*e + 2*f*x]
)^(5/2)*((I*a*Sqrt[b]*(-1 + E^((2*I)*(e + f*x)))*(24*b^2*E^((2*I)*(e + f*x)) + 3*a^2*(1 + E^((2*I)*(e + f*x)))
^2 + a*b*(5 + 26*E^((2*I)*(e + f*x)) + 5*E^((4*I)*(e + f*x)))))/((a + b)^2*(4*b*E^((2*I)*(e + f*x)) + a*(1 + E
^((2*I)*(e + f*x)))^2)^2) - (3*Log[(-4*Sqrt[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])*Sec[e + f*x]^5)/(12*Sqrt[2]*b^(5/2)*f*(a + b*Sec[e + f*x]^2)^(5/2))
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fricas [B] time = 0.93, size = 688, normalized size = 5.17 \[ \left [\frac {3 \, {\left ({\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{2} b^{2} + 2 \, a b^{3} + b^{4} + 2 \, {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )^{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 (3 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (2 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (f x + e\right )\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 )}{12 \, {\left ({\left (a^{4} b^{3} + 2 \, a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{3} b^{4} + 2 \, a^{2} b^{5} + a b^{6}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} b^{5} + 2 \, a b^{6} + b^{7}\right )} f\right )}}, \frac {3 \, {\left ({\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{2} b^{2} + 2 \, a b^{3} + b^{4} + 2 \, {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )^{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 (3 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (2 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (f x + e\right )\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 )}{6 \, {\left ({\left (a^{4} b^{3} + 2 \, a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{3} b^{4} + 2 \, a^{2} b^{5} + a b^{6}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} b^{5} + 2 \, a b^{6} + b^{7}\right )} f\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")
[Out]
[1/12*(3*((a^4 + 2*a^3*b + a^2*b^2)*cos(f*x + e)^4 + a^2*b^2 + 2*a*b^3 + b^4 + 2*(a^3*b + 2*a^2*b^2 + a*b^3)*c
os(f*x + e)^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)*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^3*b + 5*a^2*b^2)*cos(f*x + e)^3 + 2*(2*a^2*b^2 + 3*a*b^3)*cos(f*x + e))*sqrt((a*cos(f*x
+ e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a^4*b^3 + 2*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^4 + 2*(a^3*b^4 + 2*a
^2*b^5 + a*b^6)*f*cos(f*x + e)^2 + (a^2*b^5 + 2*a*b^6 + b^7)*f), 1/6*(3*((a^4 + 2*a^3*b + a^2*b^2)*cos(f*x + e
)^4 + a^2*b^2 + 2*a*b^3 + b^4 + 2*(a^3*b + 2*a^2*b^2 + a*b^3)*cos(f*x + e)^2)*sqrt(-b)*arctan(-1/2*((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)/((a*b*cos(f*x + e)^2 + b
^2)*sin(f*x + e))) - 2*((3*a^3*b + 5*a^2*b^2)*cos(f*x + e)^3 + 2*(2*a^2*b^2 + 3*a*b^3)*cos(f*x + e))*sqrt((a*c
os(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a^4*b^3 + 2*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^4 + 2*(a^3*b^
4 + 2*a^2*b^5 + a*b^6)*f*cos(f*x + e)^2 + (a^2*b^5 + 2*a*b^6 + b^7)*f)]
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")
[Out]
integrate(sec(f*x + e)^6/(b*sec(f*x + e)^2 + a)^(5/2), x)
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maple [C] time = 2.10, size = 3018, normalized size = 22.69 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x)
[Out]
-1/3/f*sin(f*x+e)*(b+a*cos(f*x+e)^2)*(3*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+
b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+
e))/(a+b))^(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))*sin(f*x+e)*cos(f*x+e)^2*a^3+6*cos(f*x+e)^2*sin(f*x+e
)*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))*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+
e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f
*x+e))/(a+b))^(1/2)*a^2*b+3*cos(f*x+e)^2*sin(f*x+e)*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))*2^(1/2)*((I*a^(1/
2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos
(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*a*b^2-6*sin(f*x+e)*cos(f*x+e)^2*2^(1/2)*
((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^
(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(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))*a^3-12*sin(f*x+e)*cos(f*x+e)^2*2^(1/2)*((I*a^(1/2)*b^(1/2
)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I
*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(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))*a^2*b-6*sin(f*x+e)*cos(f*x+e)^2*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^
(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-
a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(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))*a*b^2+3*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+
e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(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))*a^2*b*sin(f*x+e)+6*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2
)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*co
s(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(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))*a*b^2*sin(f*x+e)+3*2^(1/2)*((
I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1
/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(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*sin(f*x+e)-6*sin(f*x+e)*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(
1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/
(a+b))^(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))*a^2*b-12*sin(
f*x+e)*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-
2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticPi((-1+c
os(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))*a*b^2-6*sin(f*x+e)*2^(1/2)*((I*a^(1/2)*b^(
1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e
)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(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+3*cos(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3+5*((2*I*a
^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^3*a^2*b-3*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3
-5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^2*a^2*b+4*cos(f*x+e)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(
1/2)*a^2*b+6*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)*a*b^2-4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2
)*a^2*b-6*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2)/(-1+cos(f*x+e))/cos(f*x+e)^5/((b+a*cos(f*x+e)^2)/cos(
f*x+e)^2)^(5/2)/(a^2+2*a*b+b^2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/b^2
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maxima [B] time = 0.39, size = 275, normalized size = 2.07 \[ -\frac {{\left (\frac {3 \, \tan \left (f x + e\right )^{2}}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b^{2}} + \frac {2}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b}\right )} \tan \left (f x + e\right ) - \frac {3 \, \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {5}{2}}} - \frac {2 \, \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2}} - \frac {\tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}} + \frac {3 \, \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} b^{2}} - \frac {2 \, a \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} b^{2}} + \frac {2 \, \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b} - \frac {4 \, \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} b}}{3 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")
[Out]
-1/3*((3*tan(f*x + e)^2/((b*tan(f*x + e)^2 + a + b)^(3/2)*b) + 2*a/((b*tan(f*x + e)^2 + a + b)^(3/2)*b^2) + 2/
((b*tan(f*x + e)^2 + a + b)^(3/2)*b))*tan(f*x + e) - 3*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/b^(5/2) - 2*tan
(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^2) - tan(f*x + e)/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b))
+ 3*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*b^2) - 2*a*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a +
b)*b^2) + 2*tan(f*x + e)/((b*tan(f*x + e)^2 + a + b)^(3/2)*b) - 4*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b
)*(a + b)*b))/f
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (e+f\,x\right )}^6\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(cos(e + f*x)^6*(a + b/cos(e + f*x)^2)^(5/2)),x)
[Out]
int(1/(cos(e + f*x)^6*(a + b/cos(e + f*x)^2)^(5/2)), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(f*x+e)**6/(a+b*sec(f*x+e)**2)**(5/2),x)
[Out]
Integral(sec(e + f*x)**6/(a + b*sec(e + f*x)**2)**(5/2), x)
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