3.6 \(\int \frac {A+B \sec (c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=147 \[ \frac {6 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}+\frac {2 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 b^3 d}+\frac {2 B \sin (c+d x)}{3 b^2 d \sqrt {b \sec (c+d x)}} \]

[Out]

2/5*A*sin(d*x+c)/b/d/(b*sec(d*x+c))^(3/2)+2/3*B*sin(d*x+c)/b^2/d/(b*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))/b^2/d/cos(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2
)+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)
*(b*sec(d*x+c))^(1/2)/b^3/d

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3787, 3769, 3771, 2639, 2641} \[ \frac {6 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}+\frac {2 B \sin (c+d x)}{3 b^2 d \sqrt {b \sec (c+d x)}}+\frac {2 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*A*EllipticE[(c + d*x)/2, 2])/(5*b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*B*Sqrt[Cos[c + d*x]]*El
lipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*b^3*d) + (2*A*Sin[c + d*x])/(5*b*d*(b*Sec[c + d*x])^(3/2)) +
(2*B*Sin[c + d*x])/(3*b^2*d*Sqrt[b*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)}{(b \sec (c+d x))^{5/2}} \, dx &=A \int \frac {1}{(b \sec (c+d x))^{5/2}} \, dx+\frac {B \int \frac {1}{(b \sec (c+d x))^{3/2}} \, dx}{b}\\ &=\frac {2 A \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}+\frac {2 B \sin (c+d x)}{3 b^2 d \sqrt {b \sec (c+d x)}}+\frac {(3 A) \int \frac {1}{\sqrt {b \sec (c+d x)}} \, dx}{5 b^2}+\frac {B \int \sqrt {b \sec (c+d x)} \, dx}{3 b^3}\\ &=\frac {2 A \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}+\frac {2 B \sin (c+d x)}{3 b^2 d \sqrt {b \sec (c+d x)}}+\frac {(3 A) \int \sqrt {\cos (c+d x)} \, dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {\left (B \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 b^3}\\ &=\frac {6 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 b^3 d}+\frac {2 A \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}+\frac {2 B \sin (c+d x)}{3 b^2 d \sqrt {b \sec (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.51, size = 88, normalized size = 0.60 \[ \frac {2 \left (\sin (c+d x) \sqrt {\cos (c+d x)} (3 A \cos (c+d x)+5 B)+9 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 B F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{15 d \cos ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(9*A*EllipticE[(c + d*x)/2, 2] + 5*B*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(5*B + 3*A*Cos[c + d*x]
)*Sin[c + d*x]))/(15*d*Cos[c + d*x]^(5/2)*(b*Sec[c + d*x])^(5/2))

________________________________________________________________________________________

fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {b \sec \left (d x + c\right )}}{b^{3} \sec \left (d x + c\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c))/(b^3*sec(d*x + c)^3), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 1.33, size = 482, normalized size = 3.28 \[ \frac {\frac {6 i A \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )}{5}-\frac {6 i A \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )}{5}+\frac {2 i B \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )}{3}+\frac {6 i A \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{5}-\frac {6 i A \sin \left (d x +c \right ) \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{5}+\frac {2 i B \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )}{3}-\frac {2 A \left (\cos ^{4}\left (d x +c \right )\right )}{5}-\frac {2 B \left (\cos ^{3}\left (d x +c \right )\right )}{3}-\frac {4 A \left (\cos ^{2}\left (d x +c \right )\right )}{5}+\frac {6 A \cos \left (d x +c \right )}{5}+\frac {2 B \cos \left (d x +c \right )}{3}}{d \cos \left (d x +c \right )^{3} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15/d*(9*I*A*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*EllipticF(I*(-1+cos(d*x+c)
)/sin(d*x+c),I)*cos(d*x+c)-9*I*A*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c
)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*sin(d*x+c)+5*I*B*cos(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x
+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)+9*I*A*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),
I)*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-9*I*A*EllipticE(I*(-1+cos(d*x+c))/sin
(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+5*I*B*sin(d*x+c)*EllipticF(I*
(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-3*A*cos(d*x+c)^4-5*B*
cos(d*x+c)^3-6*A*cos(d*x+c)^2+9*A*cos(d*x+c)+5*B*cos(d*x+c))/cos(d*x+c)^3/(b/cos(d*x+c))^(5/2)/sin(d*x+c)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________