3.1.51 \(\int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\) [51]

Optimal. Leaf size=116 \[ \frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 b^3 d}+\frac {2 B \sin (c+d x)}{3 b^2 d \sqrt {b \sec (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 8, 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 {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)}}+\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.12, size = 81, normalized size = 0.70 \begin {gather*} \frac {6 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 B F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\frac {B \sin (2 (c+d x))}{\sqrt {\cos (c+d x)}}}{3 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*C*EllipticE[(c + d*x)/2, 2] + 2*B*EllipticF[(c + d*x)/2, 2] + (B*Sin[2*(c + d*x)])/Sqrt[Cos[c + d*x]])/(3*b
^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]])

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Maple [C] Result contains complex when optimal does not.
time = 23.18, size = 470, normalized size = 4.05

method	result	size
default	\(-\frac {2 \left (-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 )+3 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 )-3 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 )-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 )}}+3 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 )}}-3 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 )}}+B \left (\cos ^{3}\left (d x +c \right )\right )+3 C \left (\cos ^{2}\left (d x +c \right )\right )-B \cos \left (d x +c \right )-3 C \cos \left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3} \sin \left (d x +c \right ) \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}\)	\(470\)

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))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/3/d*(-I*B*cos(d*x+c)*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*I*C*cos(d*x+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)-3*I*C*cos(d*x+c)*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)-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*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)-3*I*C*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)+B*cos(d*x+c)^3+3*C*cos(d
*x+c)^2-B*cos(d*x+c)-3*C*cos(d*x+c))/cos(d*x+c)^3/sin(d*x+c)/(b/cos(d*x+c))^(5/2)

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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))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.38, size = 150, normalized size = 1.29 \begin {gather*} \frac {2 \, B \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 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 ) + 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 ) + 3 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 ) - 3 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 )}{3 \, b^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

1/3*(2*B*sqrt(b/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) - I*sqrt(2)*B*sqrt(b)*weierstrassPInverse(-4, 0, cos(d
*x + c) + I*sin(d*x + c)) + I*sqrt(2)*B*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*
I*sqrt(2)*C*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sq
rt(2)*C*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(b^3*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \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))**(5/2),x)

[Out]

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

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))/(b*sec(d*x + c))^(5/2), 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 )}^{5/2}} \,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))^(5/2),x)

[Out]

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

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