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

Optimal. Leaf size=172 \[ \frac{2 a^2 (12 A+20 B+15 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^{3/2} C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 a (3 A+5 B) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]

[Out]

(2*a^(3/2)*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^2*(12*A + 20*B + 15*C)*Sqrt[Se
c[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(3*A + 5*B)*Sqrt[a + a*Sec[c + d*x]]*Sin[c +
d*x])/(15*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2))

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Rubi [A]  time = 0.51202, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {4086, 4017, 4015, 3801, 215} \[ \frac{2 a^2 (12 A+20 B+15 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^{3/2} C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 a (3 A+5 B) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(3/2)*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^2*(12*A + 20*B + 15*C)*Sqrt[Se
c[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(3*A + 5*B)*Sqrt[a + a*Sec[c + d*x]]*Sin[c +
d*x])/(15*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2))

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 4017

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[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))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx &=\frac{2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \sec (c+d x))^{3/2} \left (\frac{1}{2} a (3 A+5 B)+\frac{5}{2} a C \sec (c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{5 a}\\ &=\frac{2 a (3 A+5 B) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 \int \frac{\sqrt{a+a \sec (c+d x)} \left (\frac{1}{4} a^2 (12 A+20 B+15 C)+\frac{15}{4} a^2 C \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{15 a}\\ &=\frac{2 a^2 (12 A+20 B+15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (3 A+5 B) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+(a C) \int \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a^2 (12 A+20 B+15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (3 A+5 B) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{(2 a C) \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 )}{d}\\ &=\frac{2 a^{3/2} C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}+\frac{2 a^2 (12 A+20 B+15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (3 A+5 B) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 1.84905, size = 162, normalized size = 0.94 \[ \frac{\sec ^3\left (\frac{1}{2} (c+d x)\right ) (a (\sec (c+d x)+1))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right ) (2 (9 A+5 B) \cos (c+d x)+3 A \cos (2 (c+d x))+39 A+50 B+30 C)+15 \sqrt{2} C \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{15 d \sec ^{\frac{7}{2}}(c+d x) (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[(c + d*x)/2]^3*(a*(1 + Sec[c + d*x]))^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(15*Sqrt[2]*C*ArcTanh
[Sqrt[2]*Sin[(c + d*x)/2]] + (39*A + 50*B + 30*C + 2*(9*A + 5*B)*Cos[c + d*x] + 3*A*Cos[2*(c + d*x)])*Sin[(c +
d*x)/2]))/(15*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sec[c + d*x]^(7/2))

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Maple [A]  time = 0.411, size = 245, normalized size = 1.4 \begin{align*} -{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{30\,d\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( -15\,C\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{2}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) \right ) \sin \left ( dx+c \right ) +15\,C\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{2}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) \right ) \sin \left ( dx+c \right ) +12\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+24\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+20\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+36\,A\cos \left ( dx+c \right ) +80\,B\cos \left ( dx+c \right ) +60\,C\cos \left ( dx+c \right ) -72\,A-100\,B-60\,C \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30/d*a*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-15*C*(-2/(cos(d*x+c)+1))^(1/2)*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)))*sin(d*x+c)+15*C*(-2/(cos(d*x+c)+1))^(1/2)*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)))*sin(d*x+c)+12*A*cos(d*x+c)^3+24*A*cos(d*x+c)^2+20
*B*cos(d*x+c)^2+36*A*cos(d*x+c)+80*B*cos(d*x+c)+60*C*cos(d*x+c)-72*A-100*B-60*C)*cos(d*x+c)^3*(1/cos(d*x+c))^(
5/2)/sin(d*x+c)

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Maxima [B]  time = 2.31892, size = 705, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(3*sqrt(2)*(20*a*cos(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) + 5*a*
cos(2/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) - 20*a*cos(5/2*d*x + 5/2*c)*
sin(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) - 5*a*cos(5/2*d*x + 5/2*c)*sin(2/5*arctan2(sin(5/
2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 2*a*sin(5/2*d*x + 5/2*c) + 5*a*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c),
cos(5/2*d*x + 5/2*c))) + 20*a*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*A*sqrt(a) + 15*sqr
t(2)*(sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*s
qrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(
2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin
(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a*log(2*c
os(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/
2*c) + 2) + 8*a*sin(1/2*d*x + 1/2*c))*C*sqrt(a) + 20*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x
+ 1/2*c))*B*sqrt(a))/d

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Fricas [A]  time = 0.627288, size = 1089, normalized size = 6.33 \begin{align*} \left [\frac{15 \,{\left (C a \cos \left (d x + c\right ) + C a\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 (3 \, A a \cos \left (d x + c\right )^{3} +{\left (9 \, A + 5 \, B\right )} a \cos \left (d x + c\right )^{2} +{\left (18 \, A + 25 \, B + 15 \, C\right )} a \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 )}}}{30 \,{\left (d \cos \left (d x + c\right ) + d\right )}}, \frac{15 \,{\left (C a \cos \left (d x + c\right ) + C a\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 (3 \, A a \cos \left (d x + c\right )^{3} +{\left (9 \, A + 5 \, B\right )} a \cos \left (d x + c\right )^{2} +{\left (18 \, A + 25 \, B + 15 \, C\right )} a \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 )}}}{15 \,{\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(15*(C*a*cos(d*x + c) + C*a)*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*(3*A*a*cos(d*x + c)^3 + (9*A + 5*B)*a*cos(d*x + c)^2 + (18*A + 25*B + 15*C)*a*co
s(d*x + c))*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) + d), 1/1
5*(15*(C*a*cos(d*x + c) + C*a)*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*A*a*cos(d*x + c)^3 + (9*A + 5*B)*a*cos(d
*x + c)^2 + (18*A + 25*B + 15*C)*a*cos(d*x + c))*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) + 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((a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/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 )}{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}{\sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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