3.1.65 \(\int (b \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [65]

Optimal. Leaf size=178 \[ -\frac {2 b^2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 b B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b (5 A+3 C) \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 B (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 C (b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]

[Out]

2/3*B*(b*sec(d*x+c))^(3/2)*sin(d*x+c)/d-2/5*b^2*(5*A+3*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elli
pticE(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)+2/5*b*(5*A+3*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))*co
s(d*x+c)^(1/2)*(b*sec(d*x+c))^(1/2)/d+2/5*C*(b*sec(d*x+c))^(3/2)*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4132, 3853, 3856, 2720, 4131, 2719} \begin {gather*} -\frac {2 b^2 (5 A+3 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 (5 A+3 C) \sin (c+d x) \sqrt {b \sec (c+d x)}}{5 d}+\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 \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^2*(5*A + 3*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) + (2*b*(5*A + 3*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])^(3/2)*Tan[c + d*x])/(
5*d)

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 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

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

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 4.54, size = 308, normalized size = 1.73 \begin {gather*} \frac {2 e^{-i c} \left (-1+e^{2 i c}\right ) \cos ^3(c+d x) \csc (c) \left (5 B-15 A e^{i (c+d x)}-3 C e^{i (c+d x)}-30 A e^{3 i (c+d x)}-24 C e^{3 i (c+d x)}-5 B e^{4 i (c+d x)}-15 A e^{5 i (c+d x)}-9 C e^{5 i (c+d x)}-5 i B \left (1+e^{2 i (c+d x)}\right )^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+(5 A+3 C) e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )\right ) (b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{15 d \left (1+e^{2 i (c+d x)}\right )^2 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-1 + E^((2*I)*c))*Cos[c + d*x]^3*Csc[c]*(5*B - 15*A*E^(I*(c + d*x)) - 3*C*E^(I*(c + d*x)) - 30*A*E^((3*I)*
(c + d*x)) - 24*C*E^((3*I)*(c + d*x)) - 5*B*E^((4*I)*(c + d*x)) - 15*A*E^((5*I)*(c + d*x)) - 9*C*E^((5*I)*(c +
d*x)) - (5*I)*B*(1 + E^((2*I)*(c + d*x)))^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (5*A + 3*C)*E^(I*(
c + d*x))*(1 + E^((2*I)*(c + d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*(b*Sec[c + d
*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(15*d*E^(I*c)*(1 + E^((2*I)*(c + d*x)))^2*(A + 2*C + 2*B*C
os[c + d*x] + A*Cos[2*(c + d*x)]))

________________________________________________________________________________________

Maple [C] Result contains complex when optimal does not.
time = 37.26, size = 832, normalized size = 4.67

method	result	size
default	\(\text {Expression too large to display}\)	\(832\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(d*x+c))^(3/2)*(A+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*(15*I*A*cos(d*x+c)^2*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)+15*I*A*cos(d*x+c)^3*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)+9*I*C*cos(d*x+c)^2*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)-1
5*I*A*cos(d*x+c)^3*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)-15*I*A*cos(d*x+c)^2*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)-9*I*C*cos(d*x+c)^3*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)+9*I*C*cos(d*x+c)^3*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)-5*I*B*cos(d*x+c)^3*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)-9*I*
C*cos(d*x+c)^2*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)-5*I*B*cos(d*x+c)^2*sin(d*x+c)*(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)+15*A*cos(d*x+c)^3+5*B*cos(d*x+c)^3+9*C*cos(d*x+c)^3-15*A*cos(d*x+c)^2-6*
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)

________________________________________________________________________________________

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)*(A+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) + A)*(b*sec(d*x + c))^(3/2), x)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.95, size = 225, normalized size = 1.26 \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 ) - 3 i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} 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 ) + 3 i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} 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 (3 \, {\left (5 \, A + 3 \, C\right )} 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)*(A+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)) - 3*I*sqrt(2)*(5*A +
3*C)*b^(3/2)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)))
+ 3*I*sqrt(2)*(5*A + 3*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*(3*(5*A + 3*C)*b*cos(d*x + c)^2 + 5*B*b*cos(d*x + c) + 3*C*b)*sqrt(b/cos(d*x + c))*si
n(d*x + c))/(d*cos(d*x + c)^2)

________________________________________________________________________________________

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 (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)*(A+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) + A)*(b*sec(d*x + c))^(3/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________