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

Optimal. Leaf size=163 \[ \frac{(19 A+5 B+3 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(9 A-B-7 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(A-B+C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

[Out]

((19*A + 5*B + 3*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16
*Sqrt[2]*a^(5/2)*d) - ((A - B + C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((9*A -
B - 7*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2))

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Rubi [A]  time = 0.408229, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.089, Rules used = {4084, 4012, 3808, 206} \[ \frac{(19 A+5 B+3 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(9 A-B-7 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(A-B+C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((19*A + 5*B + 3*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16
*Sqrt[2]*a^(5/2)*d) - ((A - B + C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((9*A -
B - 7*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2))

Rule 4084

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((a*A - b*B + a*C)*Cot[e + f*x]*(a + b*Cs
c[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[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(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, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 4012

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

Rule 3808

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

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 1.9, size = 119, normalized size = 0.73 \[ \frac{\sec \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) \left (8 (19 A+5 B+3 C) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-4 \sin \left (\frac{1}{2} (c+d x)\right ) ((13 A-5 B-3 C) \cos (c+d x)+9 A-B-7 C)\right )}{64 a d (a (\sec (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[(c + d*x)/2]*Sec[c + d*x]^(3/2)*(8*(19*A + 5*B + 3*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^4 - 4*(9
*A - B - 7*C + (13*A - 5*B - 3*C)*Cos[c + d*x])*Sin[(c + d*x)/2]))/(64*a*d*(a*(1 + Sec[c + d*x]))^(3/2))

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Maple [B]  time = 0.374, size = 482, normalized size = 3. \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)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x)

[Out]

1/16/d/a^3*(1/cos(d*x+c))^(1/2)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*cos(d*x+c)*(-1+cos(d*x+c))^2*(13*A*cos(d*x
+c)^2*(-2/(cos(d*x+c)+1))^(1/2)+19*A*sin(d*x+c)*cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))-5*
B*(-2/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^2+5*B*sin(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*cos(d
*x+c)-3*C*cos(d*x+c)^2*(-2/(cos(d*x+c)+1))^(1/2)+3*C*sin(d*x+c)*cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+
c)+1))^(1/2))-4*A*cos(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)+19*A*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*s
in(d*x+c)+4*B*(-2/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+5*B*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*sin(d*
x+c)-4*C*cos(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)+3*C*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)-
9*A*(-2/(cos(d*x+c)+1))^(1/2)+B*(-2/(cos(d*x+c)+1))^(1/2)+7*C*(-2/(cos(d*x+c)+1))^(1/2))/sin(d*x+c)^5/(-2/(cos
(d*x+c)+1))^(1/2)

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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)^(1/2)*(A+B*sec(d*x+c)+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 = 0.54422, size = 1400, normalized size = 8.59 \begin{align*} \left [\frac{\sqrt{2}{\left ({\left (19 \, A + 5 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (19 \, A + 5 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (19 \, A + 5 \, B + 3 \, C\right )} \cos \left (d x + c\right ) + 19 \, A + 5 \, B + 3 \, C\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{2} \sqrt{a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - \frac{4 \,{\left ({\left (13 \, A - 5 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (9 \, A - B - 7 \, 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 )}{\sqrt{\cos \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 (19 \, A + 5 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (19 \, A + 5 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (19 \, A + 5 \, B + 3 \, C\right )} \cos \left (d x + c\right ) + 19 \, A + 5 \, B + 3 \, C\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{2} \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) + \frac{2 \,{\left ({\left (13 \, A - 5 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (9 \, A - B - 7 \, 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 )}{\sqrt{\cos \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)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/64*(sqrt(2)*((19*A + 5*B + 3*C)*cos(d*x + c)^3 + 3*(19*A + 5*B + 3*C)*cos(d*x + c)^2 + 3*(19*A + 5*B + 3*C)
*cos(d*x + c) + 19*A + 5*B + 3*C)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)
/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)
) - 4*((13*A - 5*B - 3*C)*cos(d*x + c)^2 + (9*A - B - 7*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c
))*sin(d*x + c)/sqrt(cos(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)*((19*A + 5*B + 3*C)*cos(d*x + c)^3 + 3*(19*A + 5*B + 3*C)*cos(d*x + c)^2 + 3*(19*A + 5*B
+ 3*C)*cos(d*x + c) + 19*A + 5*B + 3*C)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c
))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) + 2*((13*A - 5*B - 3*C)*cos(d*x + c)^2 + (9*A - B - 7*C)*cos(d*x + c))
*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(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(-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)**(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt{\sec \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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