\(\int \sec ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) [234]

Optimal result

Integrand size = 25, antiderivative size = 186 \[ \int \sec ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {(a+b) \left (a^2-2 a b+5 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 b^{5/2} f}+\frac {\left (a^2-2 a b+5 b^2\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 b^2 f}-\frac {(3 a-5 b) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{24 b^2 f}+\frac {\sec ^2(e+f x) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{6 b f} \]

[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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4231, 427, 396, 201, 223, 212} \[ \int \sec ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {(a+b) \left (a^2-2 a b+5 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{16 b^{5/2} f}+\frac {\left (a^2-2 a b+5 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{16 b^2 f}-\frac {(3 a-5 b) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{24 b^2 f}+\frac {\tan (e+f x) \sec ^2(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 b f} \]

[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 201

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 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 396

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 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 4231

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*} \text {integral}& = \frac {\text {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 {\text {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 ) \text {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 ) \text {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 ) \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 )}{16 b^2 f} \\ & = \frac {(a+b) \left (a^2-2 a b+5 b^2\right ) \text {arctanh}\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*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 15.53 (sec) , antiderivative size = 407, normalized size of antiderivative = 2.19 \[ \int \sec ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {e^{i (e+f x)} \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \cos (e+f x) \left (-\frac {i \sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) \left (-3 a^2 \left (1+e^{2 i (e+f x)}\right )^4+4 a b \left (1+e^{2 i (e+f x)}\right )^2 \left (1+4 e^{2 i (e+f x)}+e^{4 i (e+f x)}\right )+b^2 \left (15+100 e^{2 i (e+f x)}+298 e^{4 i (e+f x)}+100 e^{6 i (e+f x)}+15 e^{8 i (e+f x)}\right )\right )}{\left (1+e^{2 i (e+f x)}\right )^6}-\frac {3 \left (a^3-a^2 b+3 a b^2+5 b^3\right ) \log \left (\frac {-4 \sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) f+4 i \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2} f}{1+e^{2 i (e+f x)}}\right )}{\sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}}\right ) \sqrt {a+b \sec ^2(e+f x)}}{24 \sqrt {2} b^{5/2} f \sqrt {a+2 b+a \cos (2 e+2 f x)}} \]

[In]

Integrate[Sec[e + f*x]^6*Sqrt[a + b*Sec[e + f*x]^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))]*Cos[e + f*x]*(((-I)*Sqrt[b]*(
-1 + E^((2*I)*(e + f*x)))*(-3*a^2*(1 + E^((2*I)*(e + f*x)))^4 + 4*a*b*(1 + E^((2*I)*(e + f*x)))^2*(1 + 4*E^((2
*I)*(e + f*x)) + 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^3 - a^2*b + 3*a*b^2 + 5*b^3)*L
og[(-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])*Sqrt[a + b*S
ec[e + f*x]^2])/(24*Sqrt[2]*b^(5/2)*f*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1374\) vs. \(2(166)=332\).

Time = 17.26 (sec) , antiderivative size = 1375, normalized size of antiderivative = 7.39

method	result	size
default	\(\text {Expression too large to display}\)	\(1375\)

[In]

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

[Out]

1/96/f/b^(11/2)*(a+b*sec(f*x+e)^2)^(1/2)/(1+cos(f*x+e))/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(30*b^(11/
2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*sin(f*x+e)+30*b^(11/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1
/2)*tan(f*x+e)+8*b^(9/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*sin(f*x+e)+20*b^(11/2)*((b+a*cos(f*x+e)
^2)/(1+cos(f*x+e))^2)^(1/2)*tan(f*x+e)*sec(f*x+e)+8*b^(9/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*tan(
f*x+e)-6*sin(f*x+e)*a^2*b^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+20*b^(11/2)*((b+a*cos(f*x+e)^2)/(1
+cos(f*x+e))^2)^(1/2)*tan(f*x+e)*sec(f*x+e)^2+4*b^(9/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*tan(f*x+
e)*sec(f*x+e)-6*b^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*tan(f*x+e)+16*b^(11/2)*((b+a*cos(f*x+e
)^2)/(1+cos(f*x+e))^2)^(1/2)*tan(f*x+e)*sec(f*x+e)^3+4*b^(9/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*t
an(f*x+e)*sec(f*x+e)^2+16*b^(11/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*tan(f*x+e)*sec(f*x+e)^4+3*ln(4*
(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^
(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*cos(f*x+e)*a^3*b^3-3*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)
*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*cos(
f*x+e)*a^2*b^4+9*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)
^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*cos(f*x+e)*a*b^5+15*ln(4*(((b+a*cos(f*x+e)^2)/(1
+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)
/(sin(f*x+e)+1))*cos(f*x+e)*b^6+3*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2
)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*cos(f*x+e)*a^3*b^3-3*ln(-4*(((
b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/
2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*cos(f*x+e)*a^2*b^4+9*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b
^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*cos(f*
x+e)*a*b^5+15*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2
)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*cos(f*x+e)*b^6)

Fricas [A] (verification not implemented)

none

Time = 1.07 (sec) , antiderivative size = 468, normalized size of antiderivative = 2.52 \[ \int \sec ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\left [\frac {3 \, {\left (a^{3} - a^{2} b + 3 \, a b^{2} + 5 \, b^{3}\right )} \sqrt {b} \cos \left (f x + e\right )^{5} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 4 \, {\left ({\left (3 \, a^{2} b - 4 \, a b^{2} - 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 8 \, b^{3} - 2 \, {\left (a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{192 \, b^{3} f \cos \left (f x + e\right )^{5}}, \frac {3 \, {\left (a^{3} - a^{2} b + 3 \, a b^{2} + 5 \, b^{3}\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{5} - 2 \, {\left ({\left (3 \, a^{2} b - 4 \, a b^{2} - 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 8 \, b^{3} - 2 \, {\left (a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{96 \, b^{3} f \cos \left (f x + e\right )^{5}}\right ] \]

[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)]

Sympy [F]

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

[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)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.70 \[ \int \sec ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\frac {8 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{3}}{b} + \frac {3 \, {\left (a + b\right )}^{2} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {5}{2}}} + \frac {3 \, {\left (a + b\right )}^{2} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} - \frac {12 \, {\left (a + b\right )} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} - \frac {12 \, {\left (a + b\right )} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} + \frac {24 \, a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} + 24 \, \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + 24 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} \tan \left (f x + e\right ) - \frac {6 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \tan \left (f x + e\right )}{b^{2}} + \frac {3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2} \tan \left (f x + e\right )}{b^{2}} + \frac {24 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \tan \left (f x + e\right )}{b} - \frac {12 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )}{b}}{48 \, f} \]

[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

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \sec ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \frac {\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{{\cos \left (e+f\,x\right )}^6} \,d x \]

[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)