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

Optimal. Leaf size=77 \[ \frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}} \]

[Out]

2/5*(3*A+5*C)*(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)+2/5*A*tan(d*x+c)/d/(b*sec(d*x+c))^(5/2)

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Rubi [A]  time = 0.06, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {4045, 3771, 2639} \[ \frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*A + 5*C)*EllipticE[(c + d*x)/2, 2])/(5*b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*A*Tan[c + d*x
])/(5*d*(b*Sec[c + d*x])^(5/2))

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 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 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx &=\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}+\frac {(3 A+5 C) \int \frac {1}{\sqrt {b \sec (c+d x)}} \, dx}{5 b^2}\\ &=\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}+\frac {(3 A+5 C) \int \sqrt {\cos (c+d x)} \, dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\\ &=\frac {2 (3 A+5 C) 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 \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 1.31, size = 133, normalized size = 1.73 \[ \frac {e^{-i d x} \sec ^2(c+d x) (\cos (d x)+i \sin (d x)) \left (-\frac {8 i (3 A+5 C) e^{2 i (c+d x)} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+6 A \sin (2 (c+d x))+12 i (3 A+5 C)\right )}{30 d (b \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])*((12*I)*(3*A + 5*C) - ((8*I)*(3*A + 5*C)*E^((2*I)*(c + d*x))*Hypergeom
etric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + 6*A*Sin[2*(c + d*x)]))/(30*d*E^
(I*d*x)*(b*Sec[c + d*x])^(5/2))

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {b \sec \left (d x + c\right )}}{b^{3} \sec \left (d x + c\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*sec(d*x + c)^2 + A)*sqrt(b*sec(d*x + c))/(b^3*sec(d*x + c)^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.43, size = 614, normalized size = 7.97 \[ \frac {\frac {6 i A \sin \left (d x +c \right ) \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 )}{5}-\frac {6 i A \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 ) \cos \left (d x +c \right )}{5}+2 i C \sin \left (d x +c \right ) \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 )-2 i C \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 ) \cos \left (d x +c \right )+\frac {6 i A \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 )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )}{5}-\frac {6 i A \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 )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )}{5}+2 i C \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 )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-2 i C \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 )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-\frac {2 A \left (\cos ^{4}\left (d x +c \right )\right )}{5}-\frac {4 A \left (\cos ^{2}\left (d x +c \right )\right )}{5}-2 C \left (\cos ^{2}\left (d x +c \right )\right )+\frac {6 A \cos \left (d x +c \right )}{5}+2 C \cos \left (d x +c \right )}{d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5/d*(3*I*A*sin(d*x+c)*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)-3*I*A*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos
(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)+5*I*C*sin(d*x+c)*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)-5*I*C*sin(d*x+c)*cos(d*x+c)*EllipticE(I*(-1+co
s(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*I*A*sin(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)-3*I*A*sin(d*x+c)*(1
/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+5*I*C*sin(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)-5*I
*C*sin(d*x+c)*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)-A*cos(d*x+c)^4-2*A*cos(d*x+c)^2-5*C*cos(d*x+c)^2+3*A*cos(d*x+c)+5*C*cos(d*x+c))/sin(d*x+c)/cos(d*x+c)^3/(
b/cos(d*x+c))^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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