3.2 \(\int (b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=136 \[ -\frac {2 A b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}+\frac {2 B \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}+\frac {2 b B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3787, 3768, 3771, 2639, 2641} \[ -\frac {2 A b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}+\frac {2 B \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}+\frac {2 b B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*b^2*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*b*B*Sqrt[Cos[c + d*x]]*E
llipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*d) + (2*A*b*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/d + (2*B*(b*S
ec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d)

Rule 2639

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

Rule 2641

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

Rule 3768

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 3771

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 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.27, size = 87, normalized size = 0.64 \[ \frac {(b \sec (c+d x))^{3/2} \left (2 \sin (c+d x) (3 A \cos (c+d x)+B)-6 A \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 B \cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B b \sec \left (d x + c\right )^{2} + A b \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right )}, x\right ) \]

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)),x, algorithm="fricas")

[Out]

integral((B*b*sec(d*x + c)^2 + A*b*sec(d*x + c))*sqrt(b*sec(d*x + c)), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}\,{d x} \]

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)),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [C]  time = 1.41, size = 499, normalized size = 3.67 \[ \frac {2 \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (3 i A \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 ) \sin \left (d x +c \right )-3 i A \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 ) \sin \left (d x +c \right )+i B \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 ) \sin \left (d x +c \right )+3 i A \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 )}}\, \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 )-3 i A \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 ) \sin \left (d x +c \right )+i B \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 ) \sin \left (d x +c \right )-3 A \left (\cos ^{2}\left (d x +c \right )\right )-B \left (\cos ^{2}\left (d x +c \right )\right )+3 A \cos \left (d x +c \right )+B \right ) \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}}{3 d \sin \left (d x +c \right )^{5}} \]

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)),x)

[Out]

2/3/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(3*I*A*cos(d*x+c)^2*(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)*sin(d*x+c)-3*I*A*cos(d*x+c)^2*(1/(1+cos(d*x+c)))^(1/2)*(co
s(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)+I*B*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)*sin(d*x+c)+3*I*A*c
os(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)
*sin(d*x+c)-3*I*A*cos(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)*sin(d*x+c)+I*B*cos(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*Ellip
ticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)-3*A*cos(d*x+c)^2-B*cos(d*x+c)^2+3*A*cos(d*x+c)+B)*(b/cos(d*x+c
))^(3/2)/sin(d*x+c)^5

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}\,{d x} \]

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)),x, algorithm="maxima")

[Out]

integrate((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 \[ \int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \left (A + B \sec {\left (c + d x \right )}\right )\, dx \]

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)),x)

[Out]

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

________________________________________________________________________________________