3.203 \(\int \frac {\sec (c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=130 \[ \frac {(3 A+19 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(7 A-9 C) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

[Out]

1/32*(3*A+19*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)-1/4*(A+C)*sec(
d*x+c)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)+1/16*(7*A-9*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)

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Rubi [A]  time = 0.26, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4079, 4000, 3795, 203} \[ \frac {(3 A+19 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(7 A-9 C) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*A + 19*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A
+ C)*Sec[c + d*x]*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) + ((7*A - 9*C)*Tan[c + d*x])/(16*a*d*(a + a*S
ec[c + d*x])^(3/2))

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rule 4000

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

Rule 4079

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

Rubi steps

\begin {align*} \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx &=-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {\int \frac {\sec (c+d x) \left (a (3 A-C)-\frac {1}{2} a (A-7 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(7 A-9 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(3 A+19 C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(7 A-9 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(3 A+19 C) \operatorname {Subst}\left (\int \frac {1}{2 a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{16 a^2 d}\\ &=\frac {(3 A+19 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(7 A-9 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 2.59, size = 120, normalized size = 0.92 \[ \frac {\tan ^3\left (\frac {1}{2} (c+d x)\right ) ((9 C-7 A) \cos (c+d x)-3 A+13 C)-\frac {(3 A+19 C) \sin (c+d x) \sqrt {\sec (c+d x)-1} \tan ^{-1}\left (\frac {\sqrt {\sec (c+d x)-1}}{\sqrt {2}}\right )}{\sqrt {2}}}{16 a^2 d (\cos (c+d x)-1) \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(((3*A + 19*C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]/Sqrt[2]]*Sqrt[-1 + Sec[c + d*x]]*Sin[c + d*x])/Sqrt[2]) + (-3*
A + 13*C + (-7*A + 9*C)*Cos[c + d*x])*Tan[(c + d*x)/2]^3)/(16*a^2*d*(-1 + Cos[c + d*x])*Sqrt[a*(1 + Sec[c + d*
x])])

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fricas [A]  time = 0.49, size = 479, normalized size = 3.68 \[ \left [-\frac {\sqrt {2} {\left ({\left (3 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 19 \, C\right )} \cos \left (d x + c\right ) + 3 \, A + 19 \, C\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (7 \, A - 9 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A - 13 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac {\sqrt {2} {\left ({\left (3 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 19 \, C\right )} \cos \left (d x + c\right ) + 3 \, A + 19 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, {\left ({\left (7 \, A - 9 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A - 13 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[-1/64*(sqrt(2)*((3*A + 19*C)*cos(d*x + c)^3 + 3*(3*A + 19*C)*cos(d*x + c)^2 + 3*(3*A + 19*C)*cos(d*x + c) + 3
*A + 19*C)*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c)
+ 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*((7*A - 9*C)*cos(d*x +
c)^2 + (3*A - 13*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3
+ 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), -1/32*(sqrt(2)*((3*A + 19*C)*cos(d*x + c)^3 + 3*(3*
A + 19*C)*cos(d*x + c)^2 + 3*(3*A + 19*C)*cos(d*x + c) + 3*A + 19*C)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 2*((7*A - 9*C)*cos(d*x + c)^2 + (3*A - 13*C)*cos(
d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2
+ 3*a^3*d*cos(d*x + c) + a^3*d)]

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giac [A]  time = 2.39, size = 190, normalized size = 1.46 \[ \frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left (\frac {2 \, \sqrt {2} {\left (A a^{5} + C a^{5}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{8} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {\sqrt {2} {\left (5 \, A a^{5} - 11 \, C a^{5}\right )}}{a^{8} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {\sqrt {2} {\left (3 \, A + 19 \, C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/32*(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*(A*a^5 + C*a^5)*tan(1/2*d*x + 1/2*c)^2/(a^8*sgn(tan(1/2*d
*x + 1/2*c)^2 - 1)) - sqrt(2)*(5*A*a^5 - 11*C*a^5)/(a^8*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c)
+ sqrt(2)*(3*A + 19*C)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-
a)*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d

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maple [B]  time = 1.83, size = 602, normalized size = 4.63 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right ) \left (3 A \left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+19 C \left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+6 A \sin \left (d x +c \right ) \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+38 C \sin \left (d x +c \right ) \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+3 A \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-14 A \left (\cos ^{3}\left (d x +c \right )\right )+19 C \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+18 C \left (\cos ^{3}\left (d x +c \right )\right )+8 A \left (\cos ^{2}\left (d x +c \right )\right )+8 C \left (\cos ^{2}\left (d x +c \right )\right )+6 A \cos \left (d x +c \right )-26 C \cos \left (d x +c \right )\right )}{32 d \left (1+\cos \left (d x +c \right )\right ) \sin \left (d x +c \right )^{3} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x)

[Out]

-1/32/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(3*A*cos(d*x+c)^2*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+19*C*cos(d*x+c)^
2*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*sin(d*x+c)+6*A*sin(d*x+c)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))
*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+38*C*sin(d*x+c)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(
d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+3*A*ln(((-2*cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-14*A*cos(
d*x+c)^3+19*C*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+
cos(d*x+c)))^(1/2)*sin(d*x+c)+18*C*cos(d*x+c)^3+8*A*cos(d*x+c)^2+8*C*cos(d*x+c)^2+6*A*cos(d*x+c)-26*C*cos(d*x+
c))/(1+cos(d*x+c))/sin(d*x+c)^3/a^3

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{\cos \left (c+d\,x\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)*(a + a/cos(c + d*x))^(5/2)),x)

[Out]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)*(a + a/cos(c + d*x))^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)/(a*(sec(c + d*x) + 1))**(5/2), x)

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