3.23 \(\int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx\)

Optimal. Leaf size=112 \[ \frac {2 (7 A+9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 b^4 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 (7 A+9 C) \sin (c+d x)}{45 b^3 d (b \sec (c+d x))^{3/2}}+\frac {2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}} \]

[Out]

2/45*(7*A+9*C)*sin(d*x+c)/b^3/d/(b*sec(d*x+c))^(3/2)+2/15*(7*A+9*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^4/d/cos(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2)+2/9*A*tan(d*x+c)/d/(b*
sec(d*x+c))^(9/2)

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Rubi [A]  time = 0.09, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4045, 3769, 3771, 2639} \[ \frac {2 (7 A+9 C) \sin (c+d x)}{45 b^3 d (b \sec (c+d x))^{3/2}}+\frac {2 (7 A+9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 b^4 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(9/2),x]

[Out]

(2*(7*A + 9*C)*EllipticE[(c + d*x)/2, 2])/(15*b^4*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*(7*A + 9*C)*
Sin[c + d*x])/(45*b^3*d*(b*Sec[c + d*x])^(3/2)) + (2*A*Tan[c + d*x])/(9*d*(b*Sec[c + d*x])^(9/2))

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 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 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))^{9/2}} \, dx &=\frac {2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}+\frac {(7 A+9 C) \int \frac {1}{(b \sec (c+d x))^{5/2}} \, dx}{9 b^2}\\ &=\frac {2 (7 A+9 C) \sin (c+d x)}{45 b^3 d (b \sec (c+d x))^{3/2}}+\frac {2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}+\frac {(7 A+9 C) \int \frac {1}{\sqrt {b \sec (c+d x)}} \, dx}{15 b^4}\\ &=\frac {2 (7 A+9 C) \sin (c+d x)}{45 b^3 d (b \sec (c+d x))^{3/2}}+\frac {2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}+\frac {(7 A+9 C) \int \sqrt {\cos (c+d x)} \, dx}{15 b^4 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\\ &=\frac {2 (7 A+9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 b^4 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 (7 A+9 C) \sin (c+d x)}{45 b^3 d (b \sec (c+d x))^{3/2}}+\frac {2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 1.56, size = 143, normalized size = 1.28 \[ \frac {e^{-i d x} (\cos (d x)+i \sin (d x)) \left (-\frac {32 i (7 A+9 C) e^{2 i (c+d x)} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+(76 A+72 C) \sin (2 (c+d x))+10 A \sin (4 (c+d x))+336 i A+432 i C\right )}{360 b^4 d \sqrt {b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(9/2),x]

[Out]

((Cos[d*x] + I*Sin[d*x])*((336*I)*A + (432*I)*C - ((32*I)*(7*A + 9*C)*E^((2*I)*(c + d*x))*Hypergeometric2F1[1/
2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + (76*A + 72*C)*Sin[2*(c + d*x)] + 10*A*Sin[
4*(c + d*x)]))/(360*b^4*d*E^(I*d*x)*Sqrt[b*Sec[c + d*x]])

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {b \sec \left (d x + c\right )}}{b^{5} \sec \left (d x + c\right )^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(9/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \sec \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(9/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(9/2), x)

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maple [C]  time = 1.57, size = 636, normalized size = 5.68 \[ -\frac {2 \left (5 A \left (\cos ^{6}\left (d x +c \right )\right )-21 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 )+21 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 )-27 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 )+27 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 )-21 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 )+21 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 )-27 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 )+27 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 )+2 A \left (\cos ^{4}\left (d x +c \right )\right )+9 C \left (\cos ^{4}\left (d x +c \right )\right )+14 A \left (\cos ^{2}\left (d x +c \right )\right )+18 C \left (\cos ^{2}\left (d x +c \right )\right )-21 A \cos \left (d x +c \right )-27 C \cos \left (d x +c \right )\right )}{45 d \cos \left (d x +c \right )^{5} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {9}{2}} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(9/2),x)

[Out]

-2/45/d*(5*A*cos(d*x+c)^6-21*I*A*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(
1+cos(d*x+c)))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)+21*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)-27*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)+27*I*C*EllipticE(I*(-1+c
os(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)-21*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)+21*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)-27*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)+27*I*C*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*(1/(1+cos(d*x+c)))^(1/2)+2*A*cos(d*x+c)^4+9*C*cos(d*x+c)^4+14*A*cos(d*x+c)^2+18*C*cos(d*x+c)^2-21*A*cos
(d*x+c)-27*C*cos(d*x+c))/cos(d*x+c)^5/(b/cos(d*x+c))^(9/2)/sin(d*x+c)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \sec \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(9/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(9/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(9/2),x)

[Out]

int((A + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(9/2), 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((A+C*sec(d*x+c)**2)/(b*sec(d*x+c))**(9/2),x)

[Out]

Timed out

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