3.1.22 \(\int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx\) [22]

Optimal. Leaf size=112 \[ \frac {2 (5 A+7 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 b^4 d}+\frac {2 (5 A+7 C) \sin (c+d x)}{21 b^3 d \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}} \]

[Out]

2/21*(5*A+7*C)*sin(d*x+c)/b^3/d/(b*sec(d*x+c))^(1/2)+2/21*(5*A+7*C)*(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^4/d+2/7*A*tan(d*x+c)/d/(b*
sec(d*x+c))^(7/2)

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Rubi [A]
time = 0.06, 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 = {4130, 3854, 3856, 2720} \begin {gather*} \frac {2 (5 A+7 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 b^4 d}+\frac {2 (5 A+7 C) \sin (c+d x)}{21 b^3 d \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(7/2),x]

[Out]

(2*(5*A + 7*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(21*b^4*d) + (2*(5*A + 7*C)*
Sin[c + d*x])/(21*b^3*d*Sqrt[b*Sec[c + d*x]]) + (2*A*Tan[c + d*x])/(7*d*(b*Sec[c + d*x])^(7/2))

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3854

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 3856

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 4130

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))^{7/2}} \, dx &=\frac {2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}}+\frac {(5 A+7 C) \int \frac {1}{(b \sec (c+d x))^{3/2}} \, dx}{7 b^2}\\ &=\frac {2 (5 A+7 C) \sin (c+d x)}{21 b^3 d \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}}+\frac {(5 A+7 C) \int \sqrt {b \sec (c+d x)} \, dx}{21 b^4}\\ &=\frac {2 (5 A+7 C) \sin (c+d x)}{21 b^3 d \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}}+\frac {\left ((5 A+7 C) \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 b^4}\\ &=\frac {2 (5 A+7 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 b^4 d}+\frac {2 (5 A+7 C) \sin (c+d x)}{21 b^3 d \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.47, size = 79, normalized size = 0.71 \begin {gather*} \frac {\frac {4 (5 A+7 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{\sqrt {\cos (c+d x)}}+2 (13 A+14 C+3 A \cos (2 (c+d x))) \sin (c+d x)}{42 b^3 d \sqrt {b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(7/2),x]

[Out]

((4*(5*A + 7*C)*EllipticF[(c + d*x)/2, 2])/Sqrt[Cos[c + d*x]] + 2*(13*A + 14*C + 3*A*Cos[2*(c + d*x)])*Sin[c +
d*x])/(42*b^3*d*Sqrt[b*Sec[c + d*x]])

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Maple [C] Result contains complex when optimal does not.
time = 21.00, size = 241, normalized size = 2.15

method	result	size
default	\(-\frac {2 \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (5 i A \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+7 i C \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-3 A \left (\cos ^{4}\left (d x +c \right )\right )+3 A \left (\cos ^{3}\left (d x +c \right )\right )-5 A \left (\cos ^{2}\left (d x +c \right )\right )-7 C \left (\cos ^{2}\left (d x +c \right )\right )+5 A \cos \left (d x +c \right )+7 C \cos \left (d x +c \right )\right )}{21 d \sin \left (d x +c \right )^{3} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \cos \left (d x +c \right )^{4}}\)	\(241\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

-2/21/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))*(5*I*A*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*Ell
ipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)+7*I*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)-3*A*cos(d*x+c)^4+3*A*cos(d*x+c)^3-5*A*cos(d*x+c)^2-7
*C*cos(d*x+c)^2+5*A*cos(d*x+c)+7*C*cos(d*x+c))/sin(d*x+c)^3/(b/cos(d*x+c))^(7/2)/cos(d*x+c)^4

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(7/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.00, size = 119, normalized size = 1.06 \begin {gather*} \frac {\sqrt {2} {\left (-5 i \, A - 7 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (5 i \, A + 7 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (3 \, A \cos \left (d x + c\right )^{3} + {\left (5 \, A + 7 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{21 \, b^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/21*(sqrt(2)*(-5*I*A - 7*I*C)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(5*
I*A + 7*I*C)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(3*A*cos(d*x + c)^3 + (5*A
+ 7*C)*cos(d*x + c))*sqrt(b/cos(d*x + c))*sin(d*x + c))/(b^4*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + C \sec ^{2}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)**2)/(b*sec(d*x+c))**(7/2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)/(b*sec(c + d*x))**(7/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(7/2),x)

[Out]

int((A + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(7/2), x)

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