3.3.31 \(\int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\) [231]

Optimal. Leaf size=232 \[ -\frac {3 (5 A+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a d}-\frac {(3 A+5 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a d}+\frac {3 (5 A+7 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 a d}-\frac {(3 A+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}+\frac {(5 A+7 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {(A+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \]

[Out]

-1/3*(3*A+5*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d+1/5*(5*A+7*C)*sec(d*x+c)^(5/2)*sin(d*x+c)/a/d-(A+C)*sec(d*x+c)^
(7/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))+3/5*(5*A+7*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d-3/5*(5*A+7*C)*(cos(1/2*d*x+1
/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d-
1/3*(3*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)
^(1/2)*sec(d*x+c)^(1/2)/a/d

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Rubi [A]
time = 0.17, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4170, 3872, 3853, 3856, 2720, 2719} \begin {gather*} -\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}+\frac {(5 A+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 a d}-\frac {(3 A+5 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a d}+\frac {3 (5 A+7 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 a d}-\frac {(3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}-\frac {3 (5 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(5*A + 7*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a*d) - ((3*A + 5*C)*Sqrt[C
os[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a*d) + (3*(5*A + 7*C)*Sqrt[Sec[c + d*x]]*Sin[c +
d*x])/(5*a*d) - ((3*A + 5*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*a*d) + ((5*A + 7*C)*Sec[c + d*x]^(5/2)*Sin[c
+ d*x])/(5*a*d) - ((A + C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4170

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(-a)*(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*
(2*m + 1))), x] + Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*C*n + A*b
*(2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x
] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx &=-\frac {(A+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {\int \sec ^{\frac {5}{2}}(c+d x) \left (\frac {1}{2} a (3 A+5 C)-\frac {1}{2} a (5 A+7 C) \sec (c+d x)\right ) \, dx}{a^2}\\ &=-\frac {(A+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {(3 A+5 C) \int \sec ^{\frac {5}{2}}(c+d x) \, dx}{2 a}+\frac {(5 A+7 C) \int \sec ^{\frac {7}{2}}(c+d x) \, dx}{2 a}\\ &=-\frac {(3 A+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}+\frac {(5 A+7 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {(A+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {(3 A+5 C) \int \sqrt {\sec (c+d x)} \, dx}{6 a}+\frac {(3 (5 A+7 C)) \int \sec ^{\frac {3}{2}}(c+d x) \, dx}{10 a}\\ &=\frac {3 (5 A+7 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 a d}-\frac {(3 A+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}+\frac {(5 A+7 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {(A+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {(3 (5 A+7 C)) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{10 a}-\frac {\left ((3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a}\\ &=-\frac {(3 A+5 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a d}+\frac {3 (5 A+7 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 a d}-\frac {(3 A+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}+\frac {(5 A+7 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {(A+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {\left (3 (5 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{10 a}\\ &=-\frac {3 (5 A+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a d}-\frac {(3 A+5 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a d}+\frac {3 (5 A+7 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 a d}-\frac {(3 A+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}+\frac {(5 A+7 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {(A+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.61, size = 342, normalized size = 1.47 \begin {gather*} -\frac {e^{-i d x} \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \left (-3 i (5 A+7 C) e^{-2 i (c+d x)} \left (1+e^{i (c+d x)}\right ) \left (1+e^{2 i (c+d x)}\right )^{5/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+40 (3 A+5 C) \cos \left (\frac {1}{2} (c+d x)\right ) \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-i \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 i (30 A+54 C+2 (45 A+56 C) \cos (c+d x)+6 (5 A+7 C) \cos (2 (c+d x))+30 A \cos (3 (c+d x))+44 C \cos (3 (c+d x))+15 i A \sin (c+d x)+31 i C \sin (c+d x)-4 i C \sin (2 (c+d x))+15 i A \sin (3 (c+d x))+19 i C \sin (3 (c+d x)))\right ) \left (\cos \left (\frac {1}{2} (c+3 d x)\right )+i \sin \left (\frac {1}{2} (c+3 d x)\right )\right )}{60 a d (1+\sec (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/60*(Cos[(c + d*x)/2]*Sec[c + d*x]^(7/2)*(((-3*I)*(5*A + 7*C)*(1 + E^(I*(c + d*x)))*(1 + E^((2*I)*(c + d*x))
)^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^((2*I)*(c + d*x)) + 40*(3*A + 5*C)*Cos[(c +
d*x)/2]*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2]*(Cos[(c + d*x)/2] - I*Sin[(c + d*x)/2]) + (2*I)*(30*A + 5
4*C + 2*(45*A + 56*C)*Cos[c + d*x] + 6*(5*A + 7*C)*Cos[2*(c + d*x)] + 30*A*Cos[3*(c + d*x)] + 44*C*Cos[3*(c +
d*x)] + (15*I)*A*Sin[c + d*x] + (31*I)*C*Sin[c + d*x] - (4*I)*C*Sin[2*(c + d*x)] + (15*I)*A*Sin[3*(c + d*x)] +
(19*I)*C*Sin[3*(c + d*x)]))*(Cos[(c + 3*d*x)/2] + I*Sin[(c + 3*d*x)/2]))/(a*d*E^(I*d*x)*(1 + Sec[c + d*x]))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(775\) vs. \(2(258)=516\).
time = 10.21, size = 776, normalized size = 3.34

method	result	size
default	\(\text {Expression too large to display}\)	\(776\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/5*C/sin(1/2*d*x+1/2*c)^2/(8*sin(1/2*d*x+1/2*
c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(2*sin(1
/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)
^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c)
,2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(2*sin
(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*(-2*sin(1/2*d*x
+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(-A-C)*(cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1
/2*c)^2-1)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*sin(1/2*d*x+1
/2*c)^4+sin(1/2*d*x+1/2*c)^2)/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-2*C*(-1/
6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(si
n(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/
2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+(2*A+2*C)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/
2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(2*sin(1/2*d*x+1/2*c)^2-
1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*
x+1/2*c)^2-1)^(1/2)/d

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*sec(d*x + c)^2 + A)*sec(d*x + c)^(5/2)/(a*sec(d*x + c) + a), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.72, size = 333, normalized size = 1.44 \begin {gather*} -\frac {5 \, {\left (\sqrt {2} {\left (-3 i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-3 i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (3 i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (3 i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 \, {\left (\sqrt {2} {\left (5 i \, A + 7 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (5 i \, A + 7 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 \, {\left (\sqrt {2} {\left (-5 i \, A - 7 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-5 i \, A - 7 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (9 \, {\left (5 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (15 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{2} - 4 \, C \cos \left (d x + c\right ) + 6 \, C\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(5*(sqrt(2)*(-3*I*A - 5*I*C)*cos(d*x + c)^3 + sqrt(2)*(-3*I*A - 5*I*C)*cos(d*x + c)^2)*weierstrassPInver
se(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(3*I*A + 5*I*C)*cos(d*x + c)^3 + sqrt(2)*(3*I*A + 5*I*C)
*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 9*(sqrt(2)*(5*I*A + 7*I*C)*cos(d*
x + c)^3 + sqrt(2)*(5*I*A + 7*I*C)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x +
c) + I*sin(d*x + c))) + 9*(sqrt(2)*(-5*I*A - 7*I*C)*cos(d*x + c)^3 + sqrt(2)*(-5*I*A - 7*I*C)*cos(d*x + c)^2)
*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(9*(5*A + 7*C)*cos(d*x
+ c)^3 + 2*(15*A + 19*C)*cos(d*x + c)^2 - 4*C*cos(d*x + c) + 6*C)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*
x + c)^3 + a*d*cos(d*x + c)^2)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(5/2)*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c)),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3063 deep

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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)^(5/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*sec(d*x + c)^(5/2)/(a*sec(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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