3.6.62 \(\int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [562]

Optimal. Leaf size=399 \[ -\frac {2 \left (16 a^4-8 a^2 b^2-3 b^4\right ) \cot (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 b^5 \sqrt {a+b} d}-\frac {2 (4 a+3 b) \left (4 a^2+b^2\right ) \cot (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 b^4 \sqrt {a+b} d}-\frac {2 a^2 \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 a \left (8 a^2-3 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (6 a^2-b^2\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d} \]

[Out]

-2/5*(16*a^4-8*a^2*b^2-3*b^4)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*
(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^5/d/(a+b)^(1/2)-2/5*(4*a+3*b)*(4*a^2+b^2)*cot(d*
x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+s
ec(d*x+c))/(a-b))^(1/2)/b^4/d/(a+b)^(1/2)-2*a^2*sec(d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)-2
/5*a*(8*a^2-3*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^3/(a^2-b^2)/d+2/5*(6*a^2-b^2)*sec(d*x+c)*(a+b*sec(d*x+c
))^(1/2)*tan(d*x+c)/b^2/(a^2-b^2)/d

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Rubi [A]
time = 0.53, antiderivative size = 399, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3930, 4177, 4167, 4090, 3917, 4089} \begin {gather*} -\frac {2 (4 a+3 b) \left (4 a^2+b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{5 b^4 d \sqrt {a+b}}-\frac {2 a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (6 a^2-b^2\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b^2 d \left (a^2-b^2\right )}-\frac {2 a \left (8 a^2-3 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b^3 d \left (a^2-b^2\right )}-\frac {2 \left (16 a^4-8 a^2 b^2-3 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{5 b^5 d \sqrt {a+b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(-2*(16*a^4 - 8*a^2*b^2 - 3*b^4)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/
(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(5*b^5*Sqrt[a + b]*d) -
(2*(4*a + 3*b)*(4*a^2 + b^2)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a
- b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(5*b^4*Sqrt[a + b]*d) - (2
*a^2*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) - (2*a*(8*a^2 - 3*b^2)*Sqrt[a + b
*Sec[c + d*x]]*Tan[c + d*x])/(5*b^3*(a^2 - b^2)*d) + (2*(6*a^2 - b^2)*Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Ta
n[c + d*x])/(5*b^2*(a^2 - b^2)*d)

Rule 3917

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(Rt[a + b, 2]/(b*
f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin
[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3930

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a^2)
*d^3*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 3)/(b*f*(m + 1)*(a^2 - b^2))), x] + Dist
[d^3/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3)*Simp[a^2*(n - 3) + a*b
*(m + 1)*Csc[e + f*x] - (a^2*(n - 2) + b^2*(m + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] &&
NeQ[a^2 - b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n, 2]))

Rule 4089

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4090

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, Int[Csc[e + f*x]*((1 +
Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && NeQ[A
^2 - B^2, 0]

Rule 4167

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m
+ 2))), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*
B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 4177

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(
e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^
(m + 1)/(b*f*(m + 3))), x] + Dist[1/(b*(m + 3)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m +
2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B*(m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C,
m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx &=-\frac {2 a^2 \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 \int \frac {\sec ^2(c+d x) \left (2 a^2-\frac {1}{2} a b \sec (c+d x)-\frac {1}{2} \left (6 a^2-b^2\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac {2 a^2 \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (6 a^2-b^2\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {4 \int \frac {\sec (c+d x) \left (-\frac {1}{2} a \left (6 a^2-b^2\right )+\frac {1}{4} b \left (2 a^2+3 b^2\right ) \sec (c+d x)+\frac {3}{4} a \left (8 a^2-3 b^2\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{5 b^2 \left (a^2-b^2\right )}\\ &=-\frac {2 a^2 \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 a \left (8 a^2-3 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (6 a^2-b^2\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {8 \int \frac {\sec (c+d x) \left (-\frac {3}{8} a b \left (4 a^2+b^2\right )-\frac {3}{8} \left (16 a^4-8 a^2 b^2-3 b^4\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 b^3 \left (a^2-b^2\right )}\\ &=-\frac {2 a^2 \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 a \left (8 a^2-3 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (6 a^2-b^2\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {\left ((4 a+3 b) \left (4 a^2+b^2\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{5 b^3 (a+b)}+\frac {\left (16 a^4-8 a^2 b^2-3 b^4\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{5 b^3 \left (a^2-b^2\right )}\\ &=-\frac {2 \left (16 a^4-8 a^2 b^2-3 b^4\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 b^5 \sqrt {a+b} d}-\frac {2 (4 a+3 b) \left (4 a^2+b^2\right ) \cot (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 b^4 \sqrt {a+b} d}-\frac {2 a^2 \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 a \left (8 a^2-3 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (6 a^2-b^2\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}\\ \end {align*}

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Mathematica [A]
time = 10.20, size = 455, normalized size = 1.14 \begin {gather*} \frac {(b+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \left (\frac {2 \left (4 a^2+b^2\right ) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 \left (4 a^3+4 a^2 b-3 a b^2-3 b^3\right ) E\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {a+b \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}+2 b \left (-4 a^2-a b+3 b^2\right ) F\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {a+b \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}+\left (4 a^2-3 b^2\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^4 \left (-a^2+b^2\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}}+\frac {2 \sqrt {\sec (c+d x)} \left (\left (-8 a^4 b+5 a^2 b^3+3 b^5\right ) \sin (c+d x)+\left (-8 a^5+4 a^3 b^2+\frac {3 a b^4}{2}\right ) \sin (2 (c+d x))+b^2 \left (-a^2+b^2\right ) (b-2 a \cos (c+d x)) \sec (c+d x) \tan (c+d x)\right )}{-a^2 b^4+b^6}\right )}{5 d (a+b \sec (c+d x))^{3/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^5/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]^(3/2)*((2*(4*a^2 + b^2)*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(4*a^3 + 4
*a^2*b - 3*a*b^2 - 3*b^3)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)]*S
qrt[(a + b*Sec[c + d*x])/((a + b)*(1 + Sec[c + d*x]))] + 2*b*(-4*a^2 - a*b + 3*b^2)*EllipticF[ArcSin[Tan[(c +
d*x)/2]], (a - b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqrt[(a + b*Sec[c + d*x])/((a + b)*(1 + Sec[c + d*x])
)] + (4*a^2 - 3*b^2)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(b^4*(-a^2 + b^2)
*Sqrt[Sec[(c + d*x)/2]^2]) + (2*Sqrt[Sec[c + d*x]]*((-8*a^4*b + 5*a^2*b^3 + 3*b^5)*Sin[c + d*x] + (-8*a^5 + 4*
a^3*b^2 + (3*a*b^4)/2)*Sin[2*(c + d*x)] + b^2*(-a^2 + b^2)*(b - 2*a*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x]))/
(-(a^2*b^4) + b^6)))/(5*d*(a + b*Sec[c + d*x])^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2476\) vs. \(2(367)=734\).
time = 0.54, size = 2477, normalized size = 6.21

method	result	size
default	\(\text {Expression too large to display}\)	\(2477\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/5/d*4^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-3*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2
))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*b^5
-8*cos(d*x+c)^4*a^4*b+3*cos(d*x+c)^4*a^2*b^3+6*cos(d*x+c)^3*a^3*b^2+5*cos(d*x+c)^3*a*b^4+6*cos(d*x+c)^2*a^2*b^
3-2*cos(d*x+c)*a*b^4+b^5-16*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*a^5+3*EllipticE((-1+cos(d*x+c)
)/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1
/2)*cos(d*x+c)^3*sin(d*x+c)*b^5-3*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*b^5-16*EllipticE((-1+cos
(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a
+b))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*a^5+3*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)
/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*b^5-8*cos(d*x+c)^
4*a^3*b^2-3*cos(d*x+c)^4*a*b^4+16*cos(d*x+c)^3*a^4*b-8*cos(d*x+c)^3*a^2*b^3-8*cos(d*x+c)^2*a^4*b+2*cos(d*x+c)*
a^3*b^2+2*cos(d*x+c)^2*b^5+16*cos(d*x+c)^4*a^5-16*cos(d*x+c)^3*a^5-3*cos(d*x+c)^3*b^5+EllipticF((-1+cos(d*x+c)
)/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1
/2)*cos(d*x+c)^3*sin(d*x+c)*a*b^4-16*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+
cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*a^4*b+8*EllipticE((-1
+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c)
)/(a+b))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*a^3*b^2+8*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*a^2*b^3+3
*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))
/(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*a*b^4+16*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b
))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+c)^2*sin(d*x
+c)*a^4*b+4*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*
cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*a^3*b^2-8*EllipticF((-1+cos(d*x+c))/sin(d*x+c)
,((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+
c)^2*sin(d*x+c)*a^2*b^3+EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*a*b^4-16*EllipticE((-1+cos(d*x+c))
/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/
2)*cos(d*x+c)^2*sin(d*x+c)*a^4*b+8*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*a^3*b^2+8*EllipticE((-1
+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c)
)/(a+b))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*a^2*b^3+3*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*a*b^4+16*
EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/
(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*a^4*b+4*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))
^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+c)^3*sin(d*x+c
)*a^3*b^2-8*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*
cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*a^2*b^3-a^2*b^3)/(b+a*cos(d*x+c))/cos(d*x+c)^2
/sin(d*x+c)/(a-b)/(a+b)/b^4

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(d*x + c) + a)*sec(d*x + c)^5/(b^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(a+b*sec(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)**5/(a + b*sec(c + d*x))**(3/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^5/(b*sec(d*x + c) + a)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\cos \left (c+d\,x\right )}^5\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*(a + b/cos(c + d*x))^(3/2)),x)

[Out]

int(1/(cos(c + d*x)^5*(a + b/cos(c + d*x))^(3/2)), x)

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