3.1.53 \(\int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx\) [53]

Optimal. Leaf size=176 \[ \frac {6 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {10 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 b^5 d}+\frac {2 B \sin (c+d x)}{7 b^2 d (b \sec (c+d x))^{5/2}}+\frac {2 C \sin (c+d x)}{5 b^3 d (b \sec (c+d x))^{3/2}}+\frac {10 B \sin (c+d x)}{21 b^4 d \sqrt {b \sec (c+d x)}} \]

[Out]

2/7*B*sin(d*x+c)/b^2/d/(b*sec(d*x+c))^(5/2)+2/5*C*sin(d*x+c)/b^3/d/(b*sec(d*x+c))^(3/2)+10/21*B*sin(d*x+c)/b^4
/d/(b*sec(d*x+c))^(1/2)+6/5*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^4/d/cos(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2)+10/21*B*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elli
pticF(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^5/d

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Rubi [A]
time = 0.10, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {4132, 3854, 3856, 2720, 12, 16, 2719} \begin {gather*} \frac {10 B \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^5 d}+\frac {10 B \sin (c+d x)}{21 b^4 d \sqrt {b \sec (c+d x)}}+\frac {6 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 C \sin (c+d x)}{5 b^3 d (b \sec (c+d x))^{3/2}}+\frac {2 B \sin (c+d x)}{7 b^2 d (b \sec (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*C*EllipticE[(c + d*x)/2, 2])/(5*b^4*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (10*B*Sqrt[Cos[c + d*x]]*E
llipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(21*b^5*d) + (2*B*Sin[c + d*x])/(7*b^2*d*(b*Sec[c + d*x])^(5/2)
) + (2*C*Sin[c + d*x])/(5*b^3*d*(b*Sec[c + d*x])^(3/2)) + (10*B*Sin[c + d*x])/(21*b^4*d*Sqrt[b*Sec[c + d*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 2719

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

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 4132

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

Rubi steps

\begin {align*} \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx &=\frac {B \int \frac {1}{(b \sec (c+d x))^{7/2}} \, dx}{b}+\int \frac {C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx\\ &=\frac {2 B \sin (c+d x)}{7 b^2 d (b \sec (c+d x))^{5/2}}+\frac {(5 B) \int \frac {1}{(b \sec (c+d x))^{3/2}} \, dx}{7 b^3}+C \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx\\ &=\frac {2 B \sin (c+d x)}{7 b^2 d (b \sec (c+d x))^{5/2}}+\frac {10 B \sin (c+d x)}{21 b^4 d \sqrt {b \sec (c+d x)}}+\frac {(5 B) \int \sqrt {b \sec (c+d x)} \, dx}{21 b^5}+\frac {C \int \frac {1}{(b \sec (c+d x))^{5/2}} \, dx}{b^2}\\ &=\frac {2 B \sin (c+d x)}{7 b^2 d (b \sec (c+d x))^{5/2}}+\frac {2 C \sin (c+d x)}{5 b^3 d (b \sec (c+d x))^{3/2}}+\frac {10 B \sin (c+d x)}{21 b^4 d \sqrt {b \sec (c+d x)}}+\frac {(3 C) \int \frac {1}{\sqrt {b \sec (c+d x)}} \, dx}{5 b^4}+\frac {\left (5 B \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 b^5}\\ &=\frac {10 B \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^5 d}+\frac {2 B \sin (c+d x)}{7 b^2 d (b \sec (c+d x))^{5/2}}+\frac {2 C \sin (c+d x)}{5 b^3 d (b \sec (c+d x))^{3/2}}+\frac {10 B \sin (c+d x)}{21 b^4 d \sqrt {b \sec (c+d x)}}+\frac {(3 C) \int \sqrt {\cos (c+d x)} \, dx}{5 b^4 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\\ &=\frac {6 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {10 B \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^5 d}+\frac {2 B \sin (c+d x)}{7 b^2 d (b \sec (c+d x))^{5/2}}+\frac {2 C \sin (c+d x)}{5 b^3 d (b \sec (c+d x))^{3/2}}+\frac {10 B \sin (c+d x)}{21 b^4 d \sqrt {b \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.44, size = 104, normalized size = 0.59 \begin {gather*} \frac {\sqrt {b \sec (c+d x)} \left (252 C \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+100 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+(65 B+42 C \cos (c+d x)+15 B \cos (2 (c+d x))) \sin (2 (c+d x))\right )}{210 b^5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*Sec[c + d*x]]*(252*C*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 100*B*Sqrt[Cos[c + d*x]]*EllipticF
[(c + d*x)/2, 2] + (65*B + 42*C*Cos[c + d*x] + 15*B*Cos[2*(c + d*x)])*Sin[2*(c + d*x)]))/(210*b^5*d)

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Maple [C] Result contains complex when optimal does not.
time = 23.06, size = 493, normalized size = 2.80

method	result	size
default	\(-\frac {2 \left (-25 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 )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-63 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 )+63 i C \cos \left (d x +c \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 )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+15 B \left (\cos ^{5}\left (d x +c \right )\right )-25 i B \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 )}}-63 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 )}}+63 i C \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 ) \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 )}}+21 C \left (\cos ^{4}\left (d x +c \right )\right )+10 B \left (\cos ^{3}\left (d x +c \right )\right )+42 C \left (\cos ^{2}\left (d x +c \right )\right )-25 B \cos \left (d x +c \right )-63 C \cos \left (d x +c \right )\right )}{105 d \cos \left (d x +c \right )^{5} \sin \left (d x +c \right ) \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {9}{2}}}\)	\(493\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(9/2),x,method=_RETURNVERBOSE)

[Out]

-2/105/d*(-25*I*B*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)-63*I*C*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)*(1/(1+co
s(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+63*I*C*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)
*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+15*B*cos(d*x+c)^5-25*I*B*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)-63*I*C*Ellip
ticF(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)+63*
I*C*EllipticE(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)+21*C*cos(d*x+c)^4+10*B*cos(d*x+c)^3+42*C*cos(d*x+c)^2-25*B*cos(d*x+c)-63*C*cos(d*x+c))/cos(d*x+c)^5/sin(
d*x+c)/(b/cos(d*x+c))^(9/2)

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.45, size = 175, normalized size = 0.99 \begin {gather*} \frac {-25 i \, \sqrt {2} B \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 25 i \, \sqrt {2} B \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 63 i \, \sqrt {2} C \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 63 i \, \sqrt {2} C \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (15 \, B \cos \left (d x + c\right )^{3} + 21 \, C \cos \left (d x + c\right )^{2} + 25 \, B \cos \left (d x + c\right )\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, b^{5} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(-25*I*sqrt(2)*B*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 25*I*sqrt(2)*B*sqrt
(b)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 63*I*sqrt(2)*C*sqrt(b)*weierstrassZeta(-4, 0,
weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 63*I*sqrt(2)*C*sqrt(b)*weierstrassZeta(-4, 0, wei
erstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(15*B*cos(d*x + c)^3 + 21*C*cos(d*x + c)^2 + 25*B*
cos(d*x + c))*sqrt(b/cos(d*x + c))*sin(d*x + c))/(b^5*d)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3879 deep

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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