3.234 \(\int \sec ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\)

Optimal. Leaf size=186 \[ \frac {\left (a^2-2 a b+5 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{16 b^2 f}+\frac {(a+b) \left (a^2-2 a b+5 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{16 b^{5/2} f}-\frac {(3 a-5 b) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{24 b^2 f}+\frac {\tan (e+f x) \sec ^2(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 b f} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4146, 416, 388, 195, 217, 206} \[ \frac {\left (a^2-2 a b+5 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{16 b^2 f}+\frac {(a+b) \left (a^2-2 a b+5 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{16 b^{5/2} f}-\frac {(3 a-5 b) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{24 b^2 f}+\frac {\tan (e+f x) \sec ^2(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 b f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^6*Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

((a + b)*(a^2 - 2*a*b + 5*b^2)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(16*b^(5/2)*f)
+ ((a^2 - 2*a*b + 5*b^2)*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(16*b^2*f) - ((3*a - 5*b)*Tan[e + f*x]*(
a + b + b*Tan[e + f*x]^2)^(3/2))/(24*b^2*f) + (Sec[e + f*x]^2*Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2))/(
6*b*f)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 388

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

Rule 416

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

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Mathematica [C]  time = 11.40, size = 968, normalized size = 5.20 \[ -\frac {i e^{i (e+f x)} \sqrt {a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2+4 b} \left (\frac {15 (a+b) \tan ^{-1}\left (\frac {\sqrt {b} \left (-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^2}{8 b^{3/2}}+\frac {15 \left (-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)}} a^2}{8 b \left (1+e^{2 i (e+f x)}\right )^2}-\frac {15 (a+b) \tan ^{-1}\left (\frac {\sqrt {b} \left (-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}{4 \sqrt {b}}-\frac {15 \left (-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)}} a}{4 \left (1+e^{2 i (e+f x)}\right )^2}+\frac {7 (7 a+15 b) \left (a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}\right )^{3/2}}{4 b \left (1+e^{2 i (e+f x)}\right )^3}-\frac {2 (8 a+35 b) \left (a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}\right )^{3/2}}{b \left (1+e^{2 i (e+f x)}\right )^3}+\frac {50 \left (a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}\right )^{3/2}}{\left (1+e^{2 i (e+f x)}\right )^3}+\frac {3 (5 a+21 b) \left (a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}\right )^{3/2}}{2 b \left (1+e^{2 i (e+f x)}\right )^4}-\frac {24 \left (a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}\right )^{3/2}}{\left (1+e^{2 i (e+f x)}\right )^4}-\frac {60 \left (a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}\right )^{3/2}}{\left (1+e^{2 i (e+f x)}\right )^5}+\frac {40 \left (a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}\right )^{3/2}}{\left (1+e^{2 i (e+f x)}\right )^6}+\frac {75}{8} \sqrt {b} (a+b) \tan ^{-1}\left (\frac {\sqrt {b} \left (-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 )+\frac {75 b \left (-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)}}}{8 \left (1+e^{2 i (e+f x)}\right )^2}\right ) \cos (e+f x) \sqrt {b \sec ^2(e+f x)+a}}{15 \sqrt {2} b \sqrt {a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}} f \sqrt {\cos (2 e+2 f x) a+a+2 b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^6*Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

((-1/15*I)*E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2*I)*(e + f*x))]*((-15*a*(-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])/(4*(1 + E^((2*I)*(e + f*x)))^2)
+ (15*a^2*(-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])/(8*b*(1 +
E^((2*I)*(e + f*x)))^2) + (75*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])/(8*(1 + E^((2*I)*(e + f*x)))^2) + (40*(4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2)^(
3/2))/(1 + E^((2*I)*(e + f*x)))^6 - (60*(4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2)^(3/2))/(1 +
E^((2*I)*(e + f*x)))^5 - (24*(4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2)^(3/2))/(1 + E^((2*I)*(e
+ f*x)))^4 + (3*(5*a + 21*b)*(4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2)^(3/2))/(2*b*(1 + E^((2
*I)*(e + f*x)))^4) + (50*(4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2)^(3/2))/(1 + E^((2*I)*(e + f
*x)))^3 + (7*(7*a + 15*b)*(4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2)^(3/2))/(4*b*(1 + E^((2*I)*
(e + f*x)))^3) - (2*(8*a + 35*b)*(4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2)^(3/2))/(b*(1 + E^((
2*I)*(e + f*x)))^3) + (15*a^2*(a + b)*ArcTan[(Sqrt[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]])/(8*b^(3/2)) - (15*a*(a + b)*ArcTan[(Sqrt[b]*(-1 + E^((2*I)*(e + f*x))))/Sq
rt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]])/(4*Sqrt[b]) + (75*Sqrt[b]*(a + b)*ArcTan[(Sqrt[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]])/8)*Cos[e + f*x]*
Sqrt[a + b*Sec[e + f*x]^2])/(Sqrt[2]*b*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]*f*Sqrt[a
+ 2*b + a*Cos[2*e + 2*f*x]])

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fricas [A]  time = 1.98, size = 468, normalized size = 2.52 \[ \left [\frac {3 \, {\left (a^{3} - a^{2} b + 3 \, a b^{2} + 5 \, b^{3}\right )} \sqrt {b} \cos \left (f x + e\right )^{5} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 4 \, {\left ({\left (3 \, a^{2} b - 4 \, a b^{2} - 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 8 \, b^{3} - 2 \, {\left (a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{192 \, b^{3} f \cos \left (f x + e\right )^{5}}, \frac {3 \, {\left (a^{3} - a^{2} b + 3 \, a b^{2} + 5 \, b^{3}\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{5} - 2 \, {\left ({\left (3 \, a^{2} b - 4 \, a b^{2} - 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 8 \, b^{3} - 2 \, {\left (a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{96 \, b^{3} f \cos \left (f x + e\right )^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sec \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sec(f*x + e)^2 + a)*sec(f*x + e)^6, x)

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maple [C]  time = 2.12, size = 2518, normalized size = 13.54 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x)

[Out]

-1/48/f*sin(f*x+e)*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(1/2)*(-18*sin(f*x+e)*cos(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*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^2-10*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^3*a*b^2+((2
*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^5*a^2*b-((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^4*a
^2*b+10*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^2*a*b^2+14*cos(f*x+e)^4*((2*I*a^(1/2)*b^(1/2)+a-b)/
(a+b))^(1/2)*a*b^2-14*cos(f*x+e)^5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2-3*sin(f*x+e)*cos(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*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+9*sin(f*x+e)*cos(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*cos(f*x+e)-b)/(1+c
os(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+6*sin(f*x+e)*cos(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*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-3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^6*a
^3+3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^7*a^3-8*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*
x+e)*b^3-10*cos(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^3+10*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b
)/(a+b))^(1/2)*b^3-4*cos(f*x+e)^7*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b-15*cos(f*x+e)^7*((2*I*a^(1/2)*
b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2+4*cos(f*x+e)^6*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b+15*cos(f*x+e)^6*(
(2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2-15*cos(f*x+e)^5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^3+15*co
s(f*x+e)^4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^3+8*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^3-6*sin(f*x
+e)*cos(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*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*Ellipt
icPi((-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^3-30*sin(f*x+e)*cos(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*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*sin(f*x+e)*cos(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*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^
3+15*sin(f*x+e)*cos(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*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)/(-1+cos(f*x+e))/(b+a*cos(f*x+e)^2)/cos(f*x+e)^5/((2*I*a^(1
/2)*b^(1/2)+a-b)/(a+b))^(1/2)/b^2

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maxima [A]  time = 0.36, size = 317, normalized size = 1.70 \[ \frac {\frac {8 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{3}}{b} + \frac {3 \, {\left (a + b\right )}^{2} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {5}{2}}} + \frac {3 \, {\left (a + b\right )}^{2} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} - \frac {12 \, {\left (a + b\right )} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} - \frac {12 \, {\left (a + b\right )} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} + \frac {24 \, a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} + 24 \, \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + 24 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} \tan \left (f x + e\right ) - \frac {6 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \tan \left (f x + e\right )}{b^{2}} + \frac {3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2} \tan \left (f x + e\right )}{b^{2}} + \frac {24 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \tan \left (f x + e\right )}{b} - \frac {12 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )}{b}}{48 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

1/48*(8*(b*tan(f*x + e)^2 + a + b)^(3/2)*tan(f*x + e)^3/b + 3*(a + b)^2*a*arcsinh(b*tan(f*x + e)/sqrt((a + b)*
b))/b^(5/2) + 3*(a + b)^2*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/b^(3/2) - 12*(a + b)*a*arcsinh(b*tan(f*x + e
)/sqrt((a + b)*b))/b^(3/2) - 12*(a + b)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/sqrt(b) + 24*a*arcsinh(b*tan(f
*x + e)/sqrt((a + b)*b))/sqrt(b) + 24*sqrt(b)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b)) + 24*sqrt(b*tan(f*x + e)
^2 + a + b)*tan(f*x + e) - 6*(b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)*tan(f*x + e)/b^2 + 3*sqrt(b*tan(f*x + e)
^2 + a + b)*(a + b)^2*tan(f*x + e)/b^2 + 24*(b*tan(f*x + e)^2 + a + b)^(3/2)*tan(f*x + e)/b - 12*sqrt(b*tan(f*
x + e)^2 + a + b)*(a + b)*tan(f*x + e)/b)/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{{\cos \left (e+f\,x\right )}^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(e + f*x)^2)^(1/2)/cos(e + f*x)^6,x)

[Out]

int((a + b/cos(e + f*x)^2)^(1/2)/cos(e + f*x)^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \sec ^{2}{\left (e + f x \right )}} \sec ^{6}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**6*(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

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

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