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

Optimal. Leaf size=233 \[ \frac{a^3 (432 A+392 B+299 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{192 d \sqrt{a \sec (c+d x)+a}}+\frac{a^2 (16 A+24 B+17 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{32 d}+\frac{a^{5/2} (304 A+200 B+163 C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{64 d}+\frac{a (8 B+5 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{24 d}+\frac{C \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d} \]

[Out]

(a^(5/2)*(304*A + 200*B + 163*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(64*d) + (a^3*(432*
A + 392*B + 299*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(192*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(16*A + 24*B + 17*
C)*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(32*d) + (a*(8*B + 5*C)*Sec[c + d*x]^(3/2)*(a + a
*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(24*d) + (C*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(4*
d)

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Rubi [A]  time = 0.752273, 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, 4016, 3801, 215} \[ \frac{a^3 (432 A+392 B+299 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{192 d \sqrt{a \sec (c+d x)+a}}+\frac{a^2 (16 A+24 B+17 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{32 d}+\frac{a^{5/2} (304 A+200 B+163 C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{64 d}+\frac{a (8 B+5 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{24 d}+\frac{C \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(5/2)*(304*A + 200*B + 163*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(64*d) + (a^3*(432*
A + 392*B + 299*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(192*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(16*A + 24*B + 17*
C)*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(32*d) + (a*(8*B + 5*C)*Sec[c + d*x]^(3/2)*(a + a
*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(24*d) + (C*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(4*
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 4016

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[(-2*b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]
), x] + Dist[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)), Int[Sqrt[a + b*Csc[e + f*x]]*(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] && 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 \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{C \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac{\int \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{5/2} \left (\frac{1}{2} a (8 A+C)+\frac{1}{2} a (8 B+5 C) \sec (c+d x)\right ) \, dx}{4 a}\\ &=\frac{a (8 B+5 C) \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac{C \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac{\int \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \left (\frac{1}{4} a^2 (48 A+8 B+11 C)+\frac{3}{4} a^2 (16 A+24 B+17 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac{a^2 (16 A+24 B+17 C) \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac{a (8 B+5 C) \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac{C \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac{\int \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \left (\frac{1}{8} a^3 (240 A+104 B+95 C)+\frac{1}{8} a^3 (432 A+392 B+299 C) \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac{a^3 (432 A+392 B+299 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{192 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (16 A+24 B+17 C) \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac{a (8 B+5 C) \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac{C \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac{1}{128} \left (a^2 (304 A+200 B+163 C)\right ) \int \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^3 (432 A+392 B+299 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{192 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (16 A+24 B+17 C) \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac{a (8 B+5 C) \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac{C \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}-\frac{\left (a^2 (304 A+200 B+163 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 )}{64 d}\\ &=\frac{a^{5/2} (304 A+200 B+163 C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{64 d}+\frac{a^3 (432 A+392 B+299 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{192 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (16 A+24 B+17 C) \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac{a (8 B+5 C) \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac{C \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 2.15909, size = 179, normalized size = 0.77 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a (\sec (c+d x)+1)} \left (4 \sin \left (\frac{1}{2} (c+d x)\right ) ((1584 A+2056 B+2203 C) \cos (c+d x)+4 (48 A+136 B+163 C) \cos (2 (c+d x))+528 A \cos (3 (c+d x))+192 A+600 B \cos (3 (c+d x))+544 B+489 C \cos (3 (c+d x))+844 C)+24 \sqrt{2} (304 A+200 B+163 C) \cos ^4(c+d x) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{3072 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]^(7/2)*Sqrt[a*(1 + Sec[c + d*x])]*(24*Sqrt[2]*(304*A + 200*B + 163*C)*ArcTan
h[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^4 + 4*(192*A + 544*B + 844*C + (1584*A + 2056*B + 2203*C)*Cos[c + d*x
] + 4*(48*A + 136*B + 163*C)*Cos[2*(c + d*x)] + 528*A*Cos[3*(c + d*x)] + 600*B*Cos[3*(c + d*x)] + 489*C*Cos[3*
(c + d*x)])*Sin[(c + d*x)/2]))/(3072*d)

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Maple [B]  time = 0.375, size = 641, normalized size = 2.8 \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+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/768/d*a^2*(912*A*cos(d*x+c)^4*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*2^(1/2
)-912*A*cos(d*x+c)^4*2^(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)))+600*B*cos
(d*x+c)^4*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*2^(1/2)-600*B*cos(d*x+c)^4*2
^(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)))+489*C*arctan(1/4*2^(1/2)*(-2/(c
os(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*2^(1/2)*cos(d*x+c)^4-489*C*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1
))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*2^(1/2)*cos(d*x+c)^4+1056*A*sin(d*x+c)*cos(d*x+c)^3*(-2/(cos(d*x+c)+1))^(1
/2)+1200*B*sin(d*x+c)*cos(d*x+c)^3*(-2/(cos(d*x+c)+1))^(1/2)+978*C*sin(d*x+c)*cos(d*x+c)^3*(-2/(cos(d*x+c)+1))
^(1/2)+192*A*cos(d*x+c)^2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)+544*B*cos(d*x+c)^2*sin(d*x+c)*(-2/(cos(d*x+c)+1
))^(1/2)+652*C*sin(d*x+c)*cos(d*x+c)^2*(-2/(cos(d*x+c)+1))^(1/2)+128*B*cos(d*x+c)*sin(d*x+c)*(-2/(cos(d*x+c)+1
))^(1/2)+368*C*sin(d*x+c)*cos(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)+96*C*(-2/(cos(d*x+c)+1))^(1/2)*sin(d*x+c))*(a*(
cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(1/cos(d*x+c))^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)/sin(d*x+c)^2/cos(d*x+c)^3*(cos(
d*x+c)^2-1)

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

[Out]

Timed out

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Fricas [A]  time = 2.37755, size = 1435, normalized size = 6.16 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/768*(3*((304*A + 200*B + 163*C)*a^2*cos(d*x + c)^4 + (304*A + 200*B + 163*C)*a^2*cos(d*x + c)^3)*sqrt(a)*lo
g((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*(3*(176*A + 200
*B + 163*C)*a^2*cos(d*x + c)^3 + 2*(48*A + 136*B + 163*C)*a^2*cos(d*x + c)^2 + 8*(8*B + 23*C)*a^2*cos(d*x + c)
+ 48*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)^4 + d*co
s(d*x + c)^3), 1/384*(3*((304*A + 200*B + 163*C)*a^2*cos(d*x + c)^4 + (304*A + 200*B + 163*C)*a^2*cos(d*x + c)
^3)*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*(3*(176*A + 200*B + 163*C)*a^2*cos(d*x + c)^3 + 2*(48*A + 136*B + 163*
C)*a^2*cos(d*x + c)^2 + 8*(8*B + 23*C)*a^2*cos(d*x + c) + 48*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*si
n(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\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(sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^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)