3.599 \(\int \frac{(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=233 \[ \frac{a^3 (56 A+12 B-27 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{12 d \sqrt{a \sec (c+d x)+a}}-\frac{a^2 (8 A-12 B-21 C) \sin (c+d x) \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}{12 d}+\frac{a^{5/2} (8 A+20 B+19 C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 d}-\frac{a (4 A-3 C) \sin (c+d x) \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{6 d}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d \sqrt{\sec (c+d x)}} \]

[Out]

(a^(5/2)*(8*A + 20*B + 19*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (a^3*(56*A + 12
*B - 27*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(12*d*Sqrt[a + a*Sec[c + d*x]]) - (a^2*(8*A - 12*B - 21*C)*Sqrt[Se
c[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(12*d) - (a*(4*A - 3*C)*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d
*x])^(3/2)*Sin[c + d*x])/(6*d) + (2*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.741483, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {4086, 4018, 4015, 3801, 215} \[ \frac{a^3 (56 A+12 B-27 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{12 d \sqrt{a \sec (c+d x)+a}}-\frac{a^2 (8 A-12 B-21 C) \sin (c+d x) \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}{12 d}+\frac{a^{5/2} (8 A+20 B+19 C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 d}-\frac{a (4 A-3 C) \sin (c+d x) \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{6 d}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(3/2),x]

[Out]

(a^(5/2)*(8*A + 20*B + 19*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (a^3*(56*A + 12
*B - 27*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(12*d*Sqrt[a + a*Sec[c + d*x]]) - (a^2*(8*A - 12*B - 21*C)*Sqrt[Se
c[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(12*d) - (a*(4*A - 3*C)*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d
*x])^(3/2)*Sin[c + d*x])/(6*d) + (2*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])

Rule 4086

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*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])

Rule 4018

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

Rule 4015

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

Rule 3801

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

Rule 215

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

Rubi steps

\begin{align*} \int \frac{(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx &=\frac{2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 \int \frac{(a+a \sec (c+d x))^{5/2} \left (\frac{1}{2} a (5 A+3 B)-\frac{1}{2} a (4 A-3 C) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{3 a}\\ &=-\frac{a (4 A-3 C) \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac{2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{\int \frac{(a+a \sec (c+d x))^{3/2} \left (\frac{3}{4} a^2 (8 A+4 B-C)-\frac{1}{4} a^2 (8 A-12 B-21 C) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{3 a}\\ &=-\frac{a^2 (8 A-12 B-21 C) \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{12 d}-\frac{a (4 A-3 C) \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac{2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{\int \frac{\sqrt{a+a \sec (c+d x)} \left (\frac{1}{8} a^3 (56 A+12 B-27 C)+\frac{3}{8} a^3 (8 A+20 B+19 C) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{3 a}\\ &=\frac{a^3 (56 A+12 B-27 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2 (8 A-12 B-21 C) \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{12 d}-\frac{a (4 A-3 C) \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac{2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{1}{8} \left (a^2 (8 A+20 B+19 C)\right ) \int \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^3 (56 A+12 B-27 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2 (8 A-12 B-21 C) \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{12 d}-\frac{a (4 A-3 C) \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac{2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\frac{\left (a^2 (8 A+20 B+19 C)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{a}}} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{4 d}\\ &=\frac{a^{5/2} (8 A+20 B+19 C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{4 d}+\frac{a^3 (56 A+12 B-27 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2 (8 A-12 B-21 C) \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{12 d}-\frac{a (4 A-3 C) \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac{2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.27296, size = 155, normalized size = 0.67 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) \sqrt{a (\sec (c+d x)+1)} \left (4 \sin \left (\frac{1}{2} (c+d x)\right ) (3 (2 A+4 B+11 C) \cos (c+d x)+4 (8 A+3 B) \cos (2 (c+d x))+2 A \cos (3 (c+d x))+32 A+12 B+6 C)+6 \sqrt{2} (8 A+20 B+19 C) \cos ^2(c+d x) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{48 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(3/2),x]

[Out]

(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]^(3/2)*Sqrt[a*(1 + Sec[c + d*x])]*(6*Sqrt[2]*(8*A + 20*B + 19*C)*ArcTanh[Sqr
t[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^2 + 4*(32*A + 12*B + 6*C + 3*(2*A + 4*B + 11*C)*Cos[c + d*x] + 4*(8*A + 3*
B)*Cos[2*(c + d*x)] + 2*A*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)

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Maple [B]  time = 0.351, size = 549, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x)

[Out]

-1/48/d*a^2*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-24*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2*(-2/(cos(d*x+c)+1))^(1/
2)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))+24*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)^
2*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*(-2/(cos(d*x+c)+1))^(1/2)-60*B*arcta
n(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*(-2/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^2*2^(1
/2)*sin(d*x+c)+60*B*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d
*x+c)))*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)-57*C*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2*(-2/(cos(d*x+c)+1))^(1/2)*arctan(
1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))+57*C*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2*arctan(1
/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*(-2/(cos(d*x+c)+1))^(1/2)+32*A*cos(d*x+c)^4+22
4*A*cos(d*x+c)^3+96*B*cos(d*x+c)^3-256*A*cos(d*x+c)^2-48*B*cos(d*x+c)^2+132*C*cos(d*x+c)^2-48*B*cos(d*x+c)-108
*C*cos(d*x+c)-24*C)*(1/cos(d*x+c))^(3/2)/sin(d*x+c)

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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((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.50063, size = 1308, normalized size = 5.61 \begin{align*} \left [\frac{3 \,{\left ({\left (8 \, A + 20 \, B + 19 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} +{\left (8 \, A + 20 \, B + 19 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac{4 \,{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt{a} \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 )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac{4 \,{\left (8 \, A a^{2} \cos \left (d x + c\right )^{3} + 8 \,{\left (8 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (4 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right ) + 6 \, C a^{2}\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 )}}}{48 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac{3 \,{\left ({\left (8 \, A + 20 \, B + 19 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} +{\left (8 \, A + 20 \, B + 19 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt{-a} \arctan \left (\frac{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 )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac{2 \,{\left (8 \, A a^{2} \cos \left (d x + c\right )^{3} + 8 \,{\left (8 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (4 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right ) + 6 \, C a^{2}\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 )}}}{24 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[1/48*(3*((8*A + 20*B + 19*C)*a^2*cos(d*x + c)^2 + (8*A + 20*B + 19*C)*a^2*cos(d*x + c))*sqrt(a)*log((a*cos(d*
x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x
+ c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(8*A*a^2*cos(d*x + c)^3 +
8*(8*A + 3*B)*a^2*cos(d*x + c)^2 + 3*(4*B + 11*C)*a^2*cos(d*x + c) + 6*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*
x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2 + d*cos(d*x + c)), 1/24*(3*((8*A + 20*B + 19*C)*a^2
*cos(d*x + c)^2 + (8*A + 20*B + 19*C)*a^2*cos(d*x + c))*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/c
os(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)) + 2*(8*A*a^2*cos(d*x +
c)^3 + 8*(8*A + 3*B)*a^2*cos(d*x + c)^2 + 3*(4*B + 11*C)*a^2*cos(d*x + c) + 6*C*a^2)*sqrt((a*cos(d*x + c) + a
)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2 + d*cos(d*x + c))]

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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((a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(3/2),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 )}{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\sec \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(5/2)/sec(d*x + c)^(3/2), x)