3.247 \(\int \sec ^6(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=243 \[ \frac {\left (3 a^2-10 a b+35 b^2\right ) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{192 b^2 f}+\frac {(a+b) \left (3 a^2-10 a b+35 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{128 b^2 f}+\frac {(a+b)^2 \left (3 a^2-10 a b+35 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{128 b^{5/2} f}-\frac {(3 a-7 b) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{48 b^2 f}+\frac {\tan (e+f x) \sec ^2(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{8 b f} \]

[Out]

1/128*(a+b)^2*(3*a^2-10*a*b+35*b^2)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/b^(5/2)/f+1/128*(a+
b)*(3*a^2-10*a*b+35*b^2)*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/b^2/f+1/192*(3*a^2-10*a*b+35*b^2)*tan(f*x+e)*(a
+b+b*tan(f*x+e)^2)^(3/2)/b^2/f-1/48*(3*a-7*b)*tan(f*x+e)*(a+b+b*tan(f*x+e)^2)^(5/2)/b^2/f+1/8*sec(f*x+e)^2*tan
(f*x+e)*(a+b+b*tan(f*x+e)^2)^(5/2)/b/f

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Rubi [A]  time = 0.22, antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {\left (3 a^2-10 a b+35 b^2\right ) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{192 b^2 f}+\frac {(a+b) \left (3 a^2-10 a b+35 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{128 b^2 f}+\frac {(a+b)^2 \left (3 a^2-10 a b+35 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{128 b^{5/2} f}-\frac {(3 a-7 b) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{48 b^2 f}+\frac {\tan (e+f x) \sec ^2(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{8 b f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^6*(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

((a + b)^2*(3*a^2 - 10*a*b + 35*b^2)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(128*b^(5
/2)*f) + ((a + b)*(3*a^2 - 10*a*b + 35*b^2)*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(128*b^2*f) + ((3*a^2
- 10*a*b + 35*b^2)*Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2))/(192*b^2*f) - ((3*a - 7*b)*Tan[e + f*x]*(a
+ b + b*Tan[e + f*x]^2)^(5/2))/(48*b^2*f) + (Sec[e + f*x]^2*Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(5/2))/(8*
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) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx &=\frac {\operatorname {Subst}\left (\int \left (1+x^2\right )^2 \left (a+b+b x^2\right )^{3/2} \, 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 )^{5/2}}{8 b f}+\frac {\operatorname {Subst}\left (\int \left (-a+7 b-(3 a-7 b) x^2\right ) \left (a+b+b x^2\right )^{3/2} \, dx,x,\tan (e+f x)\right )}{8 b f}\\ &=-\frac {(3 a-7 b) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{48 b^2 f}+\frac {\sec ^2(e+f x) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{8 b f}+\frac {\left (3 a^2-10 a b+35 b^2\right ) \operatorname {Subst}\left (\int \left (a+b+b x^2\right )^{3/2} \, dx,x,\tan (e+f x)\right )}{48 b^2 f}\\ &=\frac {\left (3 a^2-10 a b+35 b^2\right ) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{192 b^2 f}-\frac {(3 a-7 b) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{48 b^2 f}+\frac {\sec ^2(e+f x) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{8 b f}+\frac {\left ((a+b) \left (3 a^2-10 a b+35 b^2\right )\right ) \operatorname {Subst}\left (\int \sqrt {a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{64 b^2 f}\\ &=\frac {(a+b) \left (3 a^2-10 a b+35 b^2\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{128 b^2 f}+\frac {\left (3 a^2-10 a b+35 b^2\right ) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{192 b^2 f}-\frac {(3 a-7 b) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{48 b^2 f}+\frac {\sec ^2(e+f x) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{8 b f}+\frac {\left ((a+b)^2 \left (3 a^2-10 a b+35 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{128 b^2 f}\\ &=\frac {(a+b) \left (3 a^2-10 a b+35 b^2\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{128 b^2 f}+\frac {\left (3 a^2-10 a b+35 b^2\right ) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{192 b^2 f}-\frac {(3 a-7 b) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{48 b^2 f}+\frac {\sec ^2(e+f x) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{8 b f}+\frac {\left ((a+b)^2 \left (3 a^2-10 a b+35 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 )}{128 b^2 f}\\ &=\frac {(a+b)^2 \left (3 a^2-10 a b+35 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{128 b^{5/2} f}+\frac {(a+b) \left (3 a^2-10 a b+35 b^2\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{128 b^2 f}+\frac {\left (3 a^2-10 a b+35 b^2\right ) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{192 b^2 f}-\frac {(3 a-7 b) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{48 b^2 f}+\frac {\sec ^2(e+f x) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{8 b f}\\ \end {align*}

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Mathematica [C]  time = 10.85, size = 512, normalized size = 2.11 \[ \frac {e^{i (e+f x)} \cos ^3(e+f x) \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \left (-\frac {3 \left (3 a^2-10 a b+35 b^2\right ) (a+b)^2 \log \left (\frac {4 i f \sqrt {a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}-4 \sqrt {b} f \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}\right )}{\sqrt {a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}}-\frac {i \sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) \left (-9 a^3 \left (1+e^{2 i (e+f x)}\right )^6+3 a^2 b \left (18 e^{2 i (e+f x)}+5 e^{4 i (e+f x)}+5\right ) \left (1+e^{2 i (e+f x)}\right )^4+a b^2 \left (948 e^{2 i (e+f x)}+2758 e^{4 i (e+f x)}+948 e^{6 i (e+f x)}+145 e^{8 i (e+f x)}+145\right ) \left (1+e^{2 i (e+f x)}\right )^2+b^3 \left (910 e^{2 i (e+f x)}+3591 e^{4 i (e+f x)}+8644 e^{6 i (e+f x)}+3591 e^{8 i (e+f x)}+910 e^{10 i (e+f x)}+105 e^{12 i (e+f x)}+105\right )\right )}{\left (1+e^{2 i (e+f x)}\right )^8}\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{96 \sqrt {2} b^{5/2} f (a \cos (2 e+2 f x)+a+2 b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^6*(a + b*Sec[e + f*x]^2)^(3/2),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]^3*(((-I)*Sqrt[b]
*(-1 + E^((2*I)*(e + f*x)))*(-9*a^3*(1 + E^((2*I)*(e + f*x)))^6 + 3*a^2*b*(1 + E^((2*I)*(e + f*x)))^4*(5 + 18*
E^((2*I)*(e + f*x)) + 5*E^((4*I)*(e + f*x))) + a*b^2*(1 + E^((2*I)*(e + f*x)))^2*(145 + 948*E^((2*I)*(e + f*x)
) + 2758*E^((4*I)*(e + f*x)) + 948*E^((6*I)*(e + f*x)) + 145*E^((8*I)*(e + f*x))) + b^3*(105 + 910*E^((2*I)*(e
+ f*x)) + 3591*E^((4*I)*(e + f*x)) + 8644*E^((6*I)*(e + f*x)) + 3591*E^((8*I)*(e + f*x)) + 910*E^((10*I)*(e +
f*x)) + 105*E^((12*I)*(e + f*x)))))/(1 + E^((2*I)*(e + f*x)))^8 - (3*(a + b)^2*(3*a^2 - 10*a*b + 35*b^2)*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])*(a + b*Sec[e +
f*x]^2)^(3/2))/(96*Sqrt[2]*b^(5/2)*f*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2))

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fricas [A]  time = 8.55, size = 566, normalized size = 2.33 \[ \left [\frac {3 \, {\left (3 \, a^{4} - 4 \, a^{3} b + 18 \, a^{2} b^{2} + 60 \, a b^{3} + 35 \, b^{4}\right )} \sqrt {b} \cos \left (f x + e\right )^{7} \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 (9 \, a^{3} b - 15 \, a^{2} b^{2} - 145 \, a b^{3} - 105 \, b^{4}\right )} \cos \left (f x + e\right )^{6} - 2 \, {\left (3 \, a^{2} b^{2} + 46 \, a b^{3} + 35 \, b^{4}\right )} \cos \left (f x + e\right )^{4} - 48 \, b^{4} - 8 \, {\left (9 \, a b^{3} + 7 \, b^{4}\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 )}{1536 \, b^{3} f \cos \left (f x + e\right )^{7}}, \frac {3 \, {\left (3 \, a^{4} - 4 \, a^{3} b + 18 \, a^{2} b^{2} + 60 \, a b^{3} + 35 \, b^{4}\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 )^{7} - 2 \, {\left ({\left (9 \, a^{3} b - 15 \, a^{2} b^{2} - 145 \, a b^{3} - 105 \, b^{4}\right )} \cos \left (f x + e\right )^{6} - 2 \, {\left (3 \, a^{2} b^{2} + 46 \, a b^{3} + 35 \, b^{4}\right )} \cos \left (f x + e\right )^{4} - 48 \, b^{4} - 8 \, {\left (9 \, a b^{3} + 7 \, b^{4}\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 )}{768 \, b^{3} f \cos \left (f x + e\right )^{7}}\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)^(3/2),x, algorithm="fricas")

[Out]

[1/1536*(3*(3*a^4 - 4*a^3*b + 18*a^2*b^2 + 60*a*b^3 + 35*b^4)*sqrt(b)*cos(f*x + e)^7*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*((9*a^3*b - 15*a^2*b^2 - 145*a*b
^3 - 105*b^4)*cos(f*x + e)^6 - 2*(3*a^2*b^2 + 46*a*b^3 + 35*b^4)*cos(f*x + e)^4 - 48*b^4 - 8*(9*a*b^3 + 7*b^4)
*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)^7), 1/768*(3*(3
*a^4 - 4*a^3*b + 18*a^2*b^2 + 60*a*b^3 + 35*b^4)*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)^7 - 2*((9*a^3*b - 15*a^2*b^2 - 145*a*b^3 - 105*b^4)*cos(f*x + e)^6 - 2*(3*a^2*b^2 + 46*a*b^3 + 35*b^4)*cos(
f*x + e)^4 - 48*b^4 - 8*(9*a*b^3 + 7*b^4)*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)^7)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sec \left (f x + e\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e)^2 + a)^(3/2)*sec(f*x + e)^6, x)

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maple [C]  time = 2.80, size = 3343, normalized size = 13.76 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x)

[Out]

1/384/f*sin(f*x+e)*(-24*sin(f*x+e)*cos(f*x+e)^8*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*b+70*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^5*b^4-70*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*co
s(f*x+e)^4*b^4-56*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^2*b^4+18*sin(f*x+e)*cos(f*x+e)^8*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^4+210*sin(f*x+e)*cos(f*x+e)^8*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^4-9*sin(f*x+e)*cos(f*x+e)^8*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)/s
in(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^4-105*sin(f*x+e)*cos(f*x
+e)^8*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^4+48*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)*b^4-148*((2*I*a^(1/2)*b^(1/2
)+a-b)/(a+b))^(1/2)*cos(f*x+e)^4*a*b^3-120*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^2*a*b^3+148*((2*
I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^5*a*b^3+215*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^
7*a*b^3+3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^6*a^3*b-107*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/
2)*cos(f*x+e)^6*a^2*b^2-215*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^6*a*b^3-3*((2*I*a^(1/2)*b^(1/2)
+a-b)/(a+b))^(1/2)*cos(f*x+e)^7*a^3*b+107*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^7*a^2*b^2+78*((2*
I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^5*a^2*b^2-78*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)
^4*a^2*b^2+120*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^3*a*b^3-9*cos(f*x+e)^9*((2*I*a^(1/2)*b^(1/2)
+a-b)/(a+b))^(1/2)*a^4+9*cos(f*x+e)^8*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^4+105*cos(f*x+e)^7*((2*I*a^(1/
2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^4-105*cos(f*x+e)^6*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^4+56*cos(f*x+e)^3*
((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^4+108*sin(f*x+e)*cos(f*x+e)^8*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^2+360*sin(f*x+e)*cos(f*x+e)^8*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^3+12*sin(f*x+e)*cos(f*x+e)^8*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*b-54*sin(f*x+e)*cos(f*x+e)^8*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)*co
s(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^2-180*sin(f*x+e)*cos(f*x+e)^8*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*b^3-48*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^4+1
5*cos(f*x+e)^9*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3*b+145*cos(f*x+e)^9*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b)
)^(1/2)*a^2*b^2+105*cos(f*x+e)^9*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^3-15*cos(f*x+e)^8*((2*I*a^(1/2)*b
^(1/2)+a-b)/(a+b))^(1/2)*a^3*b-145*cos(f*x+e)^8*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b^2-105*cos(f*x+e)
^8*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^3)*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(3/2)/(-1+cos(f*x+e))/(b+a
*cos(f*x+e)^2)^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 = 415, normalized size = 1.71 \[ \frac {\frac {48 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {5}{2}} \tan \left (f x + e\right )^{3}}{b} + \frac {9 \, {\left (a + b\right )}^{3} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {5}{2}}} + \frac {9 \, {\left (a + b\right )}^{3} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} - \frac {48 \, {\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 {3}{2}}} - \frac {48 \, {\left (a + b\right )}^{2} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} + \frac {144 \, {\left (a + b\right )} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} + 144 \, {\left (a + b\right )} \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + 96 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \tan \left (f x + e\right ) + 144 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right ) - \frac {24 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {5}{2}} {\left (a + b\right )} \tan \left (f x + e\right )}{b^{2}} + \frac {6 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{2} \tan \left (f x + e\right )}{b^{2}} + \frac {9 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{3} \tan \left (f x + e\right )}{b^{2}} + \frac {128 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {5}{2}} \tan \left (f x + e\right )}{b} - \frac {32 \, {\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} - \frac {48 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2} \tan \left (f x + e\right )}{b}}{384 \, 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)^(3/2),x, algorithm="maxima")

[Out]

1/384*(48*(b*tan(f*x + e)^2 + a + b)^(5/2)*tan(f*x + e)^3/b + 9*(a + b)^3*a*arcsinh(b*tan(f*x + e)/sqrt((a + b
)*b))/b^(5/2) + 9*(a + b)^3*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/b^(3/2) - 48*(a + b)^2*a*arcsinh(b*tan(f*x
+ e)/sqrt((a + b)*b))/b^(3/2) - 48*(a + b)^2*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/sqrt(b) + 144*(a + b)*a*
arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/sqrt(b) + 144*(a + b)*sqrt(b)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))
+ 96*(b*tan(f*x + e)^2 + a + b)^(3/2)*tan(f*x + e) + 144*sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)*tan(f*x + e) -
24*(b*tan(f*x + e)^2 + a + b)^(5/2)*(a + b)*tan(f*x + e)/b^2 + 6*(b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)^2*t
an(f*x + e)/b^2 + 9*sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^3*tan(f*x + e)/b^2 + 128*(b*tan(f*x + e)^2 + a + b)
^(5/2)*tan(f*x + e)/b - 32*(b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)*tan(f*x + e)/b - 48*sqrt(b*tan(f*x + e)^2
+ a + b)*(a + b)^2*tan(f*x + e)/b)/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}}{{\cos \left (e+f\,x\right )}^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(e + f*x)^2)^(3/2)/cos(e + f*x)^6,x)

[Out]

int((a + b/cos(e + f*x)^2)^(3/2)/cos(e + f*x)^6, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**6*(a+b*sec(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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