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

Optimal. Leaf size=122 \[ \frac{2 a (7 A+15 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d \sqrt{\sec (c+d x)}} \]

[Out]

(2*a*(7*A + 15*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*A*Sqrt[a + a*Sec[c + d
*x]]*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (2*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d
*x]])

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Rubi [A]  time = 0.314592, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.081, Rules used = {4087, 4013, 3804} \[ \frac{2 a (7 A+15 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(7*A + 15*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*A*Sqrt[a + a*Sec[c + d
*x]]*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (2*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d
*x]])

Rule 4087

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

Rule 4013

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

Rule 3804

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Simp[(-2*a*Co
t[e + f*x])/(f*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]

Rubi steps

\begin{align*} \int \frac{\sqrt{a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx &=\frac{2 A \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \int \frac{\sqrt{a+a \sec (c+d x)} \left (\frac{a A}{2}+\frac{1}{2} a (2 A+5 C) \sec (c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{5 a}\\ &=\frac{2 A \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{1}{15} (7 A+15 C) \int \frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 a (7 A+15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 A \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.531613, size = 68, normalized size = 0.56 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} (8 A \cos (c+d x)+3 A \cos (2 (c+d x))+19 A+30 C)}{15 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((19*A + 30*C + 8*A*Cos[c + d*x] + 3*A*Cos[2*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(15*d*Sq
rt[Sec[c + d*x]])

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Maple [A]  time = 0.371, size = 87, normalized size = 0.7 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( 3\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+4\,A\cos \left ( dx+c \right ) +8\,A+15\,C \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15\,d\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+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+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2),x)

[Out]

-2/15/d*(-1+cos(d*x+c))*(3*A*cos(d*x+c)^2+4*A*cos(d*x+c)+8*A+15*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*cos(d*x
+c)^3*(1/cos(d*x+c))^(5/2)/sin(d*x+c)

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Maxima [B]  time = 1.94444, size = 302, normalized size = 2.48 \begin{align*} \frac{\sqrt{2}{\left (30 \, \cos \left (\frac{4}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 5 \, \cos \left (\frac{2}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) - 30 \, \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) \sin \left (\frac{4}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) - 5 \, \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) \sin \left (\frac{2}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 6 \, \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 5 \, \sin \left (\frac{3}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 30 \, \sin \left (\frac{1}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right )\right )} A \sqrt{a} + 120 \, \sqrt{2} C \sqrt{a} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(sqrt(2)*(30*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*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) - 30*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*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))) + 6*sin(5/2*d*x + 5/2*c) + 5*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x +
5/2*c))) + 30*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*A*sqrt(a) + 120*sqrt(2)*C*sqrt(a)*
sin(1/2*d*x + 1/2*c))/d

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Fricas [A]  time = 0.485701, size = 231, normalized size = 1.89 \begin{align*} \frac{2 \,{\left (3 \, A \cos \left (d x + c\right )^{3} + 4 \, A \cos \left (d x + c\right )^{2} +{\left (8 \, A + 15 \, 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 )}{15 \,{\left (d \cos \left (d x + c\right ) + d\right )} \sqrt{\cos \left (d x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*A*cos(d*x + c)^3 + 4*A*cos(d*x + c)^2 + (8*A + 15*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*sin(d*x + c)/((d*cos(d*x + c) + d)*sqrt(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+C*sec(d*x+c)**2)*(a+a*sec(d*x+c))**(1/2)/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} + A\right )} \sqrt{a \sec \left (d x + c\right ) + a}}{\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+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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