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

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

[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 = 0.24, 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 = {4085, 3787, 3768, 3771, 2641, 2639} \[ -\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}+\frac {(5 A+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 a d}-\frac {(3 A+5 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a d}+\frac {3 (5 A+7 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 a d}-\frac {(3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}-\frac {3 (5 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d} \]

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 2639

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

Rule 2641

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

Rule 3768

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 3771

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 3787

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 4085

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]  time = 6.43, size = 342, normalized size = 1.47 \[ -\frac {e^{-i d x} \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \left (\cos \left (\frac {1}{2} (c+3 d x)\right )+i \sin \left (\frac {1}{2} (c+3 d x)\right )\right ) \left (-3 i (5 A+7 C) e^{-2 i (c+d x)} \left (1+e^{i (c+d x)}\right ) \left (1+e^{2 i (c+d x)}\right )^{5/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+2 i (2 (45 A+56 C) \cos (c+d x)+6 (5 A+7 C) \cos (2 (c+d x))+15 i A \sin (c+d x)+15 i A \sin (3 (c+d x))+30 A \cos (3 (c+d x))+30 A+31 i C \sin (c+d x)-4 i C \sin (2 (c+d x))+19 i C \sin (3 (c+d x))+44 C \cos (3 (c+d x))+54 C)+40 (3 A+5 C) \cos \left (\frac {1}{2} (c+d x)\right ) \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-i \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{60 a d (\sec (c+d x)+1)} \]

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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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{4} + A \sec \left (d x + c\right )^{2}\right )} \sqrt {\sec \left (d x + c\right )}}{a \sec \left (d x + c\right ) + a}, x\right ) \]

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]

integral((C*sec(d*x + c)^4 + A*sec(d*x + c)^2)*sqrt(sec(d*x + c))/(a*sec(d*x + c) + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{a \sec \left (d x + c\right ) + a}\,{d x} \]

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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maple [B]  time = 16.02, size = 803, normalized size = 3.46 \[ \text {result too large to display} \]

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)

[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/(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)/sin(1/2*d*x+1/2*c)^2*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d
*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^6-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin
(1/2*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(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))-8*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*A+2*C)*(-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(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))+2*(-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)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/
2*d*x+1/2*c)^2-1)+(-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)*(El
lipticF(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)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2+1/3*(sin(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))))/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(-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(d*x+c)^(5/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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

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

[Out]

Timed out

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