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 \[ -\frac{(3 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a d}-\frac{(A+C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}+\frac{(5 A+7 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 a d}-\frac{(3 A+5 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 a d}+\frac{3 (5 A+7 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 a d}-\frac{3 (5 A+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a d} \]

[Out]

(-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]))

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Rubi [A]  time = 0.236459, antiderivative size = 232, normalized size of antiderivative = 1., 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 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)]

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 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 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 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]

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*}

Mathematica [C]  time = 6.24486, size = 342, normalized size = 1.47 \[ -\frac{e^{-i d x} \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{7}{2}}(c+d x) \left (\cos \left (\frac{1}{2} (c+3 d x)\right )+i \sin \left (\frac{1}{2} (c+3 d x)\right )\right ) \left (-3 i (5 A+7 C) e^{-2 i (c+d x)} \left (1+e^{i (c+d x)}\right ) \left (1+e^{2 i (c+d x)}\right )^{5/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+40 (3 A+5 C) \cos \left (\frac{1}{2} (c+d x)\right ) \cos ^{\frac{5}{2}}(c+d x) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-i \sin \left (\frac{1}{2} (c+d x)\right )\right )+2 i (2 (45 A+56 C) \cos (c+d x)+6 (5 A+7 C) \cos (2 (c+d x))+15 i A \sin (c+d x)+15 i A \sin (3 (c+d x))+30 A \cos (3 (c+d x))+30 A+31 i C \sin (c+d x)-4 i C \sin (2 (c+d x))+19 i C \sin (3 (c+d x))+44 C \cos (3 (c+d x))+54 C)\right )}{60 a d (\sec (c+d x)+1)} \]

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]

-(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 + 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)]))*(Cos[(c + 3*d*x)/2] + I*Sin[(c + 3*d*x)/2]))/(60*a*d*E^(I*d*x)*(1 + Sec[c + d*x]))

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Maple [B]  time = 7.168, size = 803, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

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*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*(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)))+(-A-C)*(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)*(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)-2/5*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*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x
+1/2*c)-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*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(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)*(-(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*sin(1/2*d*x+1/2*c)^4+sin(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*si
n(1/2*d*x+1/2*c)^2-1))/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., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)