3.394 \(\int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=155 \[ \frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {\sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}-\frac {\sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3}-\frac {4 \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a \sec (c+d x)+a)^2} \]

[Out]

1/10*(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))/a^3/d+1/6*(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))/a^3/d-1/5*sin(d*x+c)/d/cos(d*x+c)
^(3/2)/(a+a*sec(d*x+c))^3-4/15*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^2/cos(d*x+c)^(1/2)+1/6*sin(d*x+c)/d/(a^3+a^3*se
c(d*x+c))/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.38, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {4264, 3816, 4019, 4020, 3787, 3771, 2639, 2641} \[ \frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {\sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}-\frac {\sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3}-\frac {4 \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^3),x]

[Out]

EllipticE[(c + d*x)/2, 2]/(10*a^3*d) + EllipticF[(c + d*x)/2, 2]/(6*a^3*d) - Sin[c + d*x]/(5*d*Cos[c + d*x]^(3
/2)*(a + a*Sec[c + d*x])^3) - (4*Sin[c + d*x])/(15*a*d*Sqrt[Cos[c + d*x]]*(a + a*Sec[c + d*x])^2) + Sin[c + d*
x]/(6*d*Sqrt[Cos[c + d*x]]*(a^3 + a^3*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 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 3816

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

Rule 4019

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

Rule 4020

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

Rule 4264

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps

\begin {align*} \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^3} \, dx\\ &=-\frac {\sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)} \left (\frac {a}{2}-\frac {7}{2} a \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {4 \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-2 a^2-\frac {9}{2} a^2 \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))} \, dx}{15 a^4}\\ &=-\frac {\sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {4 \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2}+\frac {\sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {3 a^3}{4}-\frac {5}{4} a^3 \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{15 a^6}\\ &=-\frac {\sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {4 \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2}+\frac {\sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{20 a^3}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx}{12 a^3}\\ &=-\frac {\sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {4 \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2}+\frac {\sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\int \sqrt {\cos (c+d x)} \, dx}{20 a^3}+\frac {\int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3}\\ &=\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac {\sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {4 \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2}+\frac {\sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [C]  time = 1.60, size = 342, normalized size = 2.21 \[ \frac {\cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {\csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (4 \cos \left (\frac {1}{2} (c-d x)\right )+26 \cos \left (\frac {1}{2} (3 c+d x)\right )+10 \cos \left (\frac {1}{2} (c+3 d x)\right )+5 \cos \left (\frac {1}{2} (5 c+3 d x)\right )+3 \cos \left (\frac {1}{2} (3 c+5 d x)\right )\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right )}{8 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4 i \sqrt {2} e^{-i (c+d x)} \left (3 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i (c+d x)}\right )-5 \left (-1+e^{2 i c}\right ) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-e^{2 i (c+d x)}\right )+3 \left (1+e^{2 i (c+d x)}\right )\right ) \sec ^3(c+d x)}{\left (-1+e^{2 i c}\right ) d \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}\right )}{15 a^3 (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^3),x]

[Out]

(Cos[(c + d*x)/2]^6*(-1/8*((4*Cos[(c - d*x)/2] + 26*Cos[(3*c + d*x)/2] + 10*Cos[(c + 3*d*x)/2] + 5*Cos[(5*c +
3*d*x)/2] + 3*Cos[(3*c + 5*d*x)/2])*Csc[c/2]*Sec[c/2]*Sec[(c + d*x)/2]^5)/(d*Cos[c + d*x]^(5/2)) + ((4*I)*Sqrt
[2]*(3*(1 + E^((2*I)*(c + d*x))) + 3*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4,
1/2, 3/4, -E^((2*I)*(c + d*x))] - 5*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeom
etric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))])*Sec[c + d*x]^3)/(d*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[(1 +
E^((2*I)*(c + d*x)))/E^(I*(c + d*x))])))/(15*a^3*(1 + Sec[c + d*x])^3)

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\cos \left (d x + c\right )}}{a^{3} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right ) + a^{3} \cos \left (d x + c\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

integral(sqrt(cos(d*x + c))/(a^3*cos(d*x + c)^3*sec(d*x + c)^3 + 3*a^3*cos(d*x + c)^3*sec(d*x + c)^2 + 3*a^3*c
os(d*x + c)^3*sec(d*x + c) + a^3*cos(d*x + c)^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 3.56, size = 270, normalized size = 1.74 \[ \frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (12 \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-22 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3\right )}{60 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*cos(1/2*d*x+1/2*c)^8-10*(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)^5+6*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*cos(1/2*d*x+1/2*c)^5*EllipticE(cos(1/2*d*x+1/2*c),2^(1
/2))-22*cos(1/2*d*x+1/2*c)^6+6*cos(1/2*d*x+1/2*c)^4+7*cos(1/2*d*x+1/2*c)^2-3)/a^3/cos(1/2*d*x+1/2*c)^5/(-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(-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(1/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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