3.502 \(\int \frac{\cos ^{\frac{5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=204 \[ -\frac{5 (3 A-2 B) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a^2 d}+\frac{7 (8 A-5 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac{(3 A-2 B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{a^2 d (\cos (c+d x)+1)}+\frac{7 (8 A-5 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{15 a^2 d}-\frac{5 (3 A-2 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 a^2 d}-\frac{(A-B) \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

(7*(8*A - 5*B)*EllipticE[(c + d*x)/2, 2])/(5*a^2*d) - (5*(3*A - 2*B)*EllipticF[(c + d*x)/2, 2])/(3*a^2*d) - (5
*(3*A - 2*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a^2*d) + (7*(8*A - 5*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(15*
a^2*d) - ((3*A - 2*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(a^2*d*(1 + Cos[c + d*x])) - ((A - B)*Cos[c + d*x]^(7/2
)*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

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Rubi [A]  time = 0.418915, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2954, 2977, 2748, 2635, 2641, 2639} \[ -\frac{5 (3 A-2 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{7 (8 A-5 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac{(3 A-2 B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{a^2 d (\cos (c+d x)+1)}+\frac{7 (8 A-5 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{15 a^2 d}-\frac{5 (3 A-2 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 a^2 d}-\frac{(A-B) \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*(8*A - 5*B)*EllipticE[(c + d*x)/2, 2])/(5*a^2*d) - (5*(3*A - 2*B)*EllipticF[(c + d*x)/2, 2])/(3*a^2*d) - (5
*(3*A - 2*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a^2*d) + (7*(8*A - 5*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(15*
a^2*d) - ((3*A - 2*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(a^2*d*(1 + Cos[c + d*x])) - ((A - B)*Cos[c + d*x]^(7/2
)*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

Rule 2954

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2635

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

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{\cos ^{\frac{5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx &=\int \frac{\cos ^{\frac{7}{2}}(c+d x) (B+A \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx\\ &=-\frac{(A-B) \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{\cos ^{\frac{5}{2}}(c+d x) \left (-\frac{7}{2} a (A-B)+\frac{1}{2} a (11 A-5 B) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(3 A-2 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac{(A-B) \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \cos ^{\frac{3}{2}}(c+d x) \left (-\frac{15}{2} a^2 (3 A-2 B)+\frac{7}{2} a^2 (8 A-5 B) \cos (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{(3 A-2 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac{(A-B) \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{(7 (8 A-5 B)) \int \cos ^{\frac{5}{2}}(c+d x) \, dx}{6 a^2}-\frac{(5 (3 A-2 B)) \int \cos ^{\frac{3}{2}}(c+d x) \, dx}{2 a^2}\\ &=-\frac{5 (3 A-2 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac{7 (8 A-5 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a^2 d}-\frac{(3 A-2 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac{(A-B) \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{(7 (8 A-5 B)) \int \sqrt{\cos (c+d x)} \, dx}{10 a^2}-\frac{(5 (3 A-2 B)) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2}\\ &=\frac{7 (8 A-5 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac{5 (3 A-2 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{5 (3 A-2 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac{7 (8 A-5 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a^2 d}-\frac{(3 A-2 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac{(A-B) \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 6.83331, size = 1396, normalized size = 6.84 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((28*I)/5)*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x])*((2*E^((2*I)*d*x)*Hyper
geometric2F1[1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)
*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d
*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*
I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d
*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 +
E^((2*I)*d*x))*Sin[c])))/((B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2) - (((7*I)/2)*B*Cos[c/2 + (d*x)/2]^4*Cs
c[c/2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x])*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, -(E^((2*I
)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*
x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-1
+ E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[
(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c]
+ I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])))/((B + A*Cos
[c + d*x])*(a + a*Sec[c + d*x])^2) + (10*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4},
Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x])*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin
[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan
[Cot[c]]]])/(d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^2) - (20*B*Cos[c/2 + (d*x)/2]^4*Cs
c[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c +
d*x])*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x -
ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Se
c[c + d*x])^2) + (Cos[c/2 + (d*x)/2]^4*(A + B*Sec[c + d*x])*((4*(-20*A + 15*B - 36*A*Cos[c] + 20*B*Cos[c])*Csc
[c])/(5*d) + (8*(-2*A + B)*Cos[d*x]*Sin[c])/(3*d) + (4*A*Cos[2*d*x]*Sin[2*c])/(5*d) - (2*Sec[c/2]*Sec[c/2 + (d
*x)/2]^3*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/(3*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(-4*A*Sin[(d*x)/2] + 3*B
*Sin[(d*x)/2]))/d + (8*(-2*A + B)*Cos[c]*Sin[d*x])/(3*d) + (4*A*Cos[2*c]*Sin[2*d*x])/(5*d) - (2*(-A + B)*Sec[c
/2 + (d*x)/2]^2*Tan[c/2])/(3*d)))/(Sqrt[Cos[c + d*x]]*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2)

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Maple [A]  time = 2.219, size = 465, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(96*A*cos(1/2*d*x+1/2*c)^10-352*A*cos(1/2*d*x+1/
2*c)^8+80*B*cos(1/2*d*x+1/2*c)^8+120*A*cos(1/2*d*x+1/2*c)^6-150*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x
+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3-336*A*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+60*B*cos(1/2*d
*x+1/2*c)^6+100*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),
2^(1/2))*cos(1/2*d*x+1/2*c)^3+210*B*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+266*A*cos(1/2*d*x+1/2*c)^4-240*B*cos(1/2*d*x+1/2*c)^4-135*A*co
s(1/2*d*x+1/2*c)^2+105*B*cos(1/2*d*x+1/2*c)^2+5*A-5*B)/a^2/cos(1/2*d*x+1/2*c)^3/(-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*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + A \cos \left (d x + c\right )^{2}\right )} \sqrt{\cos \left (d x + c\right )}}{a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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