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

Optimal. Leaf size=233 \[ \frac{a^3 (24 A-54 B-49 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{24 d \sqrt{a \sec (c+d x)+a}}+\frac{a^2 (24 A+42 B+31 C) \sin (c+d x) \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}{24 d}+\frac{a^{5/2} (40 A+38 B+25 C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{8 d}+\frac{a (6 B+5 C) \sin (c+d x) \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{12 d}+\frac{C \sin (c+d x) \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d} \]

[Out]

(a^(5/2)*(40*A + 38*B + 25*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*d) + (a^3*(24*A - 5
4*B - 49*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(24*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(24*A + 42*B + 31*C)*Sqrt[
Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(24*d) + (a*(6*B + 5*C)*Sqrt[Sec[c + d*x]]*(a + a*Sec[c +
d*x])^(3/2)*Sin[c + d*x])/(12*d) + (C*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.725627, 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 = {4088, 4018, 4015, 3801, 215} \[ \frac{a^3 (24 A-54 B-49 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{24 d \sqrt{a \sec (c+d x)+a}}+\frac{a^2 (24 A+42 B+31 C) \sin (c+d x) \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}{24 d}+\frac{a^{5/2} (40 A+38 B+25 C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{8 d}+\frac{a (6 B+5 C) \sin (c+d x) \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{12 d}+\frac{C \sin (c+d x) \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}{3 d} \]

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))/Sqrt[Sec[c + d*x]],x]

[Out]

(a^(5/2)*(40*A + 38*B + 25*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*d) + (a^3*(24*A - 5
4*B - 49*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(24*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(24*A + 42*B + 31*C)*Sqrt[
Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(24*d) + (a*(6*B + 5*C)*Sqrt[Sec[c + d*x]]*(a + a*Sec[c +
d*x])^(3/2)*Sin[c + d*x])/(12*d) + (C*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(3*d)

Rule 4088

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[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d
*Csc[e + f*x])^n)/(f*(m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n*
Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b*B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A,
B, C, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] &&  !LtQ[n, -2^(-1)] && NeQ[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 )}{\sqrt{\sec (c+d x)}} \, dx &=\frac{C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac{\int \frac{(a+a \sec (c+d x))^{5/2} \left (\frac{1}{2} a (6 A-C)+\frac{1}{2} a (6 B+5 C) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{3 a}\\ &=\frac{a (6 B+5 C) \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac{C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac{\int \frac{(a+a \sec (c+d x))^{3/2} \left (\frac{3}{4} a^2 (8 A-2 B-3 C)+\frac{1}{4} a^2 (24 A+42 B+31 C) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{6 a}\\ &=\frac{a^2 (24 A+42 B+31 C) \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{a (6 B+5 C) \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac{C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac{\int \frac{\sqrt{a+a \sec (c+d x)} \left (\frac{1}{8} a^3 (24 A-54 B-49 C)+\frac{3}{8} a^3 (40 A+38 B+25 C) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{6 a}\\ &=\frac{a^3 (24 A-54 B-49 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (24 A+42 B+31 C) \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{a (6 B+5 C) \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac{C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac{1}{16} \left (a^2 (40 A+38 B+25 C)\right ) \int \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^3 (24 A-54 B-49 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (24 A+42 B+31 C) \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{a (6 B+5 C) \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac{C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}-\frac{\left (a^2 (40 A+38 B+25 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 )}{8 d}\\ &=\frac{a^{5/2} (40 A+38 B+25 C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{8 d}+\frac{a^3 (24 A-54 B-49 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (24 A+42 B+31 C) \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{a (6 B+5 C) \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac{C \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 1.75494, size = 158, normalized size = 0.68 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{5}{2}}(c+d x) \sqrt{a (\sec (c+d x)+1)} \left (4 \sin \left (\frac{1}{2} (c+d x)\right ) (4 (18 A+6 B+17 C) \cos (c+d x)+3 (8 A+22 B+25 C) \cos (2 (c+d x))+24 A \cos (3 (c+d x))+24 A+66 B+91 C)+12 \sqrt{2} (40 A+38 B+25 C) \cos ^3(c+d x) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{192 d} \]

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))/Sqrt[Sec[c + d*x]],x]

[Out]

(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]^(5/2)*Sqrt[a*(1 + Sec[c + d*x])]*(12*Sqrt[2]*(40*A + 38*B + 25*C)*ArcTanh[S
qrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^3 + 4*(24*A + 66*B + 91*C + 4*(18*A + 6*B + 17*C)*Cos[c + d*x] + 3*(8*A
+ 22*B + 25*C)*Cos[2*(c + d*x)] + 24*A*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(192*d)

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Maple [B]  time = 0.399, size = 568, 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)^(1/2),x)

[Out]

1/96/d*a^2*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(120*A*(-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)))*sin(d*x+c)*2^(1/2)*cos(d*x+c)^3-120*A*(-2/(cos(d*x+c)+1))^(1/2)*arct
an(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*sin(d*x+c)*2^(1/2)*cos(d*x+c)^3+114*B*2^(1
/2)*cos(d*x+c)^3*(-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)))*sin(d*x+c)-114*B*2^(1/2)*cos(d*x+c)^3*(-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)))*sin(d*x+c)+75*C*(-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)))*sin(d*x+c)*2^(1/2)*cos(d*x+c)^3-75*C*(-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)))*sin(d*x+c)*2^(1/2)*cos(d*x+c)^3-192*A*cos(d*x+c)^4
+96*A*cos(d*x+c)^3-264*B*cos(d*x+c)^3-300*C*cos(d*x+c)^3+96*A*cos(d*x+c)^2+216*B*cos(d*x+c)^2+164*C*cos(d*x+c)
^2+48*B*cos(d*x+c)+104*C*cos(d*x+c)+32*C)*(1/cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*x+c)^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((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.53903, size = 1349, normalized size = 5.79 \begin{align*} \left [\frac{3 \,{\left ({\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{2}\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 (48 \, A a^{2} \cos \left (d x + c\right )^{3} + 3 \,{\left (8 \, A + 22 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \,{\left (6 \, B + 17 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, 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 )}}}{96 \,{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, \frac{3 \,{\left ({\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{2}\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 (48 \, A a^{2} \cos \left (d x + c\right )^{3} + 3 \,{\left (8 \, A + 22 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \,{\left (6 \, B + 17 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, 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 )^{3} + d \cos \left (d x + c\right )^{2}\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)^(1/2),x, algorithm="fricas")

[Out]

[1/96*(3*((40*A + 38*B + 25*C)*a^2*cos(d*x + c)^3 + (40*A + 38*B + 25*C)*a^2*cos(d*x + c)^2)*sqrt(a)*log((a*co
s(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*(48*A*a^2*cos(d*x + c)
^3 + 3*(8*A + 22*B + 25*C)*a^2*cos(d*x + c)^2 + 2*(6*B + 17*C)*a^2*cos(d*x + c) + 8*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)^3 + d*cos(d*x + c)^2), 1/48*(3*((40*A +
38*B + 25*C)*a^2*cos(d*x + c)^3 + (40*A + 38*B + 25*C)*a^2*cos(d*x + c)^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*
cos(d*x + c) + a)/cos(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
*(48*A*a^2*cos(d*x + c)^3 + 3*(8*A + 22*B + 25*C)*a^2*cos(d*x + c)^2 + 2*(6*B + 17*C)*a^2*cos(d*x + c) + 8*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)^3 + d*cos(d*x + c
)^2)]

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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)**(1/2),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}}}{\sqrt{\sec \left (d x + c\right )}}\,{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)^(1/2),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)/sqrt(sec(d*x + c)), x)