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

Optimal. Leaf size=165 \[ \frac{5 (A-15 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(A+9 C) \tan (c+d x)}{4 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{(A+C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac{(3 A-13 C) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]

[Out]

(5*(A - 15*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]^2*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((3*A - 13*C)*Tan[c + d*x])/(16*a*d*(a +
a*Sec[c + d*x])^(3/2)) + ((A + 9*C)*Tan[c + d*x])/(4*a^2*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.456373, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4085, 4008, 4001, 3795, 203} \[ \frac{5 (A-15 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(A+9 C) \tan (c+d x)}{4 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{(A+C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac{(3 A-13 C) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(A - 15*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]^2*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((3*A - 13*C)*Tan[c + d*x])/(16*a*d*(a +
a*Sec[c + d*x])^(3/2)) + ((A + 9*C)*Tan[c + d*x])/(4*a^2*d*Sqrt[a + a*Sec[c + d*x]])

Rule 4085

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

Rule 4008

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

Rule 4001

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

Rubi steps

\begin{align*} \int \frac{\sec ^2(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 ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{\int \frac{\sec ^2(c+d x) \left (-2 a (A-C)-\frac{1}{2} a (A+9 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(3 A-13 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{\sec (c+d x) \left (\frac{3}{4} a^2 (3 A-13 C)+a^2 (A+9 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(3 A-13 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(A+9 C) \tan (c+d x)}{4 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(5 (A-15 C)) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(3 A-13 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(A+9 C) \tan (c+d x)}{4 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{(5 (A-15 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{5 (A-15 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 ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(3 A-13 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(A+9 C) \tan (c+d x)}{4 a^2 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 3.2032, size = 136, normalized size = 0.82 \[ \frac{\tan ^3\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) (-(10 (A+17 C) \cos (c+d x)+(A+49 C) \cos (2 (c+d x))+A+113 C))-5 \sqrt{2} (A-15 C) \sin (c+d x) \sqrt{\sec (c+d x)-1} \tan ^{-1}\left (\frac{\sqrt{\sec (c+d x)-1}}{\sqrt{2}}\right )}{32 a^2 d (\cos (c+d x)-1) \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*Sqrt[2]*(A - 15*C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]/Sqrt[2]]*Sqrt[-1 + Sec[c + d*x]]*Sin[c + d*x] - (A + 113
*C + 10*(A + 17*C)*Cos[c + d*x] + (A + 49*C)*Cos[2*(c + d*x)])*Sec[c + d*x]*Tan[(c + d*x)/2]^3)/(32*a^2*d*(-1
+ Cos[c + d*x])*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [B]  time = 0.301, size = 597, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/d/a^3*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(5*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(
-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^2-75*C*(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c
))*sin(d*x+c)*cos(d*x+c)^2+10*A*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(-(-(-2*cos(d*x+c
)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-150*C*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d
*x+c)+1))^(1/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))+5*A*(-2*cos(d*
x+c)/(cos(d*x+c)+1))^(1/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*sin
(d*x+c)-2*A*cos(d*x+c)^3-75*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)-98*C*cos(d*x+c)^3-8*A*cos(d*x+c)^2-72*C*cos(d*x+c)^2+10*A*cos(
d*x+c)+106*C*cos(d*x+c)+64*C)/sin(d*x+c)^5

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(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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Fricas [A]  time = 0.602017, size = 1238, normalized size = 7.5 \begin{align*} \left [\frac{5 \, \sqrt{2}{\left ({\left (A - 15 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (A - 15 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (A - 15 \, C\right )} \cos \left (d x + c\right ) + A - 15 \, 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 (A + 49 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \,{\left (A + 17 \, C\right )} \cos \left (d x + c\right ) + 32 \, C\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{5 \, \sqrt{2}{\left ({\left (A - 15 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (A - 15 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (A - 15 \, C\right )} \cos \left (d x + c\right ) + A - 15 \, 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 (A + 49 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \,{\left (A + 17 \, C\right )} \cos \left (d x + c\right ) + 32 \, C\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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/64*(5*sqrt(2)*((A - 15*C)*cos(d*x + c)^3 + 3*(A - 15*C)*cos(d*x + c)^2 + 3*(A - 15*C)*cos(d*x + c) + A - 15
*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*((A + 49*C)*cos(d*x + c)^2 +
5*(A + 17*C)*cos(d*x + c) + 32*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*(5*sqrt(2)*((A - 15*C)*cos(d*x + c)^3 + 3*(A
- 15*C)*cos(d*x + c)^2 + 3*(A - 15*C)*cos(d*x + c) + A - 15*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*((A + 49*C)*cos(d*x + c)^2 + 5*(A + 17*C)*cos(d*x + c
) + 32*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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(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)**2/(a*(sec(c + d*x) + 1))**(5/2), x)

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Giac [B]  time = 9.0881, size = 386, normalized size = 2.34 \begin{align*} \frac{\frac{{\left ({\left (\frac{2 \,{\left (\sqrt{2} A a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) + \sqrt{2} C a^{6} \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 )^{2}}{a^{8}} + \frac{\sqrt{2} A a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) + 17 \, \sqrt{2} C a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}{a^{8}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{3 \, \sqrt{2} A a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) + 83 \, \sqrt{2} C a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}{a^{8}}\right )} \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}} + \frac{5 \,{\left (\sqrt{2} A - 15 \, \sqrt{2} 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(((2*(sqrt(2)*A*a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + sqrt(2)*C*a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(
1/2*d*x + 1/2*c)^2/a^8 + (sqrt(2)*A*a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 17*sqrt(2)*C*a^6*sgn(tan(1/2*d*x + 1
/2*c)^2 - 1))/a^8)*tan(1/2*d*x + 1/2*c)^2 - (3*sqrt(2)*A*a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 83*sqrt(2)*C*a^
6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))/a^8)*tan(1/2*d*x + 1/2*c)/sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a) + 5*(sqrt(2)*
A - 15*sqrt(2)*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