3.1.47 \(\int (b \sec (c+d x))^{3/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [47]

Optimal. Leaf size=169 \[ -\frac {6 b^2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 b B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {6 b C \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 B (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 C (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b d} \]

[Out]

2/3*B*(b*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/5*C*(b*sec(d*x+c))^(5/2)*sin(d*x+c)/b/d-6/5*b^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))/d/cos(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2)+6/
5*b*C*sin(d*x+c)*(b*sec(d*x+c))^(1/2)/d+2/3*b*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)/d

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Rubi [A]
time = 0.10, antiderivative size = 169, 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, 3853, 3856, 2720, 12, 16, 2719} \begin {gather*} -\frac {6 b^2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 B \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}+\frac {2 b 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 d}+\frac {2 C \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 b d}+\frac {6 b C \sin (c+d x) \sqrt {b \sec (c+d x)}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Sec[c + d*x])^(3/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(-6*b^2*C*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*b*B*Sqrt[Cos[c + d*x]]
*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*d) + (6*b*C*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*
B*(b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + (2*C*(b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*b*d)

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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[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 (b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {B \int (b \sec (c+d x))^{5/2} \, dx}{b}+\int C \sec ^2(c+d x) (b \sec (c+d x))^{3/2} \, dx\\ &=\frac {2 B (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {1}{3} (b B) \int \sqrt {b \sec (c+d x)} \, dx+C \int \sec ^2(c+d x) (b \sec (c+d x))^{3/2} \, dx\\ &=\frac {2 B (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {C \int (b \sec (c+d x))^{7/2} \, dx}{b^2}+\frac {1}{3} \left (b B \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 b 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 d}+\frac {2 B (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 C (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b d}+\frac {1}{5} (3 C) \int (b \sec (c+d x))^{3/2} \, dx\\ &=\frac {2 b 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 d}+\frac {6 b C \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 B (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 C (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b d}-\frac {1}{5} \left (3 b^2 C\right ) \int \frac {1}{\sqrt {b \sec (c+d x)}} \, dx\\ &=\frac {2 b 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 d}+\frac {6 b C \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 B (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 C (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b d}-\frac {\left (3 b^2 C\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\\ &=-\frac {6 b^2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 b 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 d}+\frac {6 b C \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 B (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 C (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b d}\\ \end {align*}

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Mathematica [A]
time = 0.32, size = 102, normalized size = 0.60 \begin {gather*} \frac {(b \sec (c+d x))^{5/2} \left (-36 C \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 B \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+21 C \sin (c+d x)+10 B \sin (2 (c+d x))+9 C \sin (3 (c+d x))\right )}{30 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[c + d*x])^(3/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((b*Sec[c + d*x])^(5/2)*(-36*C*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 20*B*Cos[c + d*x]^(5/2)*Elliptic
F[(c + d*x)/2, 2] + 21*C*Sin[c + d*x] + 10*B*Sin[2*(c + d*x)] + 9*C*Sin[3*(c + d*x)]))/(30*b*d)

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Maple [C] Result contains complex when optimal does not.
time = 33.61, size = 526, normalized size = 3.11

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 )^{2} \left (5 i B \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\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 )-9 i C \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\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 )+9 i C \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\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 i B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\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 )-9 i C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\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 )+9 i C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\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 B \left (\cos ^{3}\left (d x +c \right )\right )-9 C \left (\cos ^{3}\left (d x +c \right )\right )+6 C \left (\cos ^{2}\left (d x +c \right )\right )+5 B \cos \left (d x +c \right )+3 C \right ) \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}}{15 d \sin \left (d x +c \right )^{5} \cos \left (d x +c \right )}\)	\(526\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

2/15/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(5*I*B*sin(d*x+c)*cos(d*x+c)^3*(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)-9*I*C*sin(d*x+c)*cos(d*x+c)^3*(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)+9*I*C*sin(d*x+c)*cos(d*x+
c)^3*(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*
B*sin(d*x+c)*cos(d*x+c)^2*(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)-9*I*C*sin(d*x+c)*cos(d*x+c)^2*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*Elli
pticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+9*I*C*sin(d*x+c)*cos(d*x+c)^2*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)-5*B*cos(d*x+c)^3-9*C*cos(d*x+c)^3+6*C*cos(d*x+c)^2+
5*B*cos(d*x+c)+3*C)*(b/cos(d*x+c))^(3/2)/sin(d*x+c)^5/cos(d*x+c)

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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))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*(b*sec(d*x + c))^(3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.66, size = 207, normalized size = 1.22 \begin {gather*} \frac {-5 i \, \sqrt {2} B b^{\frac {3}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} B b^{\frac {3}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 i \, \sqrt {2} C b^{\frac {3}{2}} \cos \left (d x + c\right )^{2} {\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 ) + 9 i \, \sqrt {2} C b^{\frac {3}{2}} \cos \left (d x + c\right )^{2} {\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 (9 \, C b \cos \left (d x + c\right )^{2} + 5 \, B b \cos \left (d x + c\right ) + 3 \, C b\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/15*(-5*I*sqrt(2)*B*b^(3/2)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*I*sq
rt(2)*B*b^(3/2)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 9*I*sqrt(2)*C*b^(3/
2)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 9*I*sqrt
(2)*C*b^(3/2)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)))
+ 2*(9*C*b*cos(d*x + c)^2 + 5*B*b*cos(d*x + c) + 3*C*b)*sqrt(b/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \left (B + C \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))**(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((b*sec(c + d*x))**(3/2)*(B + C*sec(c + d*x))*sec(c + d*x), 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))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*(b*sec(d*x + c))^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,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))^(3/2),x)

[Out]

int((B/cos(c + d*x) + C/cos(c + d*x)^2)*(b/cos(c + d*x))^(3/2), x)

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