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

Optimal. Leaf size=127 \[ \frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {6 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d} \]

[Out]

2/5*a*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/3*b*sin(d*x+c)/d/sec(d*x+c)^(1/2)+6/5*a*(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)/d+2/3*b*(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)/d

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Rubi [A]  time = 0.09, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3787, 3769, 3771, 2639, 2641} \[ \frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {6 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*b*Sqrt[Cos[c + d*x]]*Elliptic
F[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*a*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (2*b*Sin[c + d*x])
/(3*d*Sqrt[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 3769

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

Rubi steps

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

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Mathematica [A]  time = 0.40, size = 88, normalized size = 0.69 \[ \frac {\sqrt {\sec (c+d x)} \left (\sin (2 (c+d x)) (3 a \cos (c+d x)+5 b)+18 a \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 b \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*(18*a*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 10*b*Sqrt[Cos[c + d*x]]*EllipticF[(c
+ d*x)/2, 2] + (5*b + 3*a*Cos[c + d*x])*Sin[2*(c + d*x)]))/(15*d)

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \sec \left (d x + c\right ) + a}{\sec \left (d x + c\right )^{\frac {5}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 3.79, size = 262, normalized size = 2.06 \[ -\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-24 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (24 a +20 b \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-6 a -10 b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, b -9 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, a \right )}{15 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*a*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(
24*a+20*b)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-6*a-10*b)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*b-9*(sin(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)^2-1)^(1/2)*a)/(-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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \sec \left (d x + c\right ) + a}{\sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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