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

Optimal. Leaf size=136 \[ -\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}+\frac {3 \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}-\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d} \]

[Out]

-sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))+3*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d-3*(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-(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 = 0.11, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3818, 3787, 3771, 2641, 3768, 2639} \[ -\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}+\frac {3 \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}-\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*d) - (Sqrt[Cos[c + d*x]]*EllipticF[(c
+ d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*d) + (3*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(a*d) - (Sec[c + d*x]^(3/2)*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 3818

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

Rubi steps

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

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Mathematica [C]  time = 1.78, size = 262, normalized size = 1.93 \[ \frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (\frac {\sqrt {\sec (c+d x)} \left (6 \csc (c) \cos (d x)-2 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {2 i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (3 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i (c+d x)}\right )-\left (-1+e^{2 i c}\right ) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-e^{2 i (c+d x)}\right )+3 \left (1+e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) d}\right )}{a (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^2*Sec[c + d*x]*(((-2*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*(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))] - E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2,
5/4, -E^((2*I)*(c + d*x))]))/(d*E^(I*(c + d*x))*(-1 + E^((2*I)*c))) + (Sqrt[Sec[c + d*x]]*(6*Cos[d*x]*Csc[c]
- 2*Tan[(c + d*x)/2]))/d))/(a*(1 + Sec[c + d*x]))

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fricas [F]  time = 0.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sec \left (d x + c\right )^{\frac {5}{2}}}{a \sec \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sec(d*x + c)^(5/2)/(a*sec(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 4.25, size = 253, normalized size = 1.86 \[ -\frac {-\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (\EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+6 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-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)*(-2*sin(1/2*d*x+1/2*c)^4+s
in(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+6*(
-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)^4-5*(-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)/a/(-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*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 \[ \int \frac {\sec \left (d x + c\right )^{\frac {5}{2}}}{a \sec \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(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 = -0.01 \[ \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(c + d*x)**(5/2)/(sec(c + d*x) + 1), x)/a

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