3.238 \(\int \frac{(a+a \sec (c+d x))^{5/2}}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=119 \[ \frac{64 a^3 \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d \sqrt{\sec (c+d x)}}+\frac{2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]

[Out]

(64*a^3*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*Sqrt[a + a*Sec[c + d*x]]*Si
n[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 = 0.169419, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {3809, 3804} \[ \frac{64 a^3 \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d \sqrt{\sec (c+d x)}}+\frac{2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a^3*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*Sqrt[a + a*Sec[c + d*x]]*Si
n[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 3809

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(a*Co
t[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*m), x] + Dist[(b*(2*m - 1))/(d*m), Int[(a + b*C
sc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& EqQ[m + n, 0] && GtQ[m, 1/2] && IntegerQ[2*m]

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{(a+a \sec (c+d x))^{5/2}}{\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{1}{5} (8 a) \int \frac{(a+a \sec (c+d x))^{3/2}}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{16 a^2 \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{1}{15} \left (32 a^2\right ) \int \frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{64 a^3 \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{16 a^2 \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 = 0.28466, size = 64, normalized size = 0.54 \[ \frac{a^2 (28 \cos (c+d x)+3 \cos (2 (c+d x))+89) \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)}}{15 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(89 + 28*Cos[c + d*x] + 3*Cos[2*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(15*d*Sqrt[Sec[c
+ d*x]])

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Maple [A]  time = 0.183, size = 85, normalized size = 0.7 \begin{align*} -{\frac{2\,{a}^{2} \left ( 3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+11\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+29\,\cos \left ( dx+c \right ) -43 \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+a*sec(d*x+c))^(5/2)/sec(d*x+c)^(5/2),x)

[Out]

-2/15/d*a^2*(3*cos(d*x+c)^3+11*cos(d*x+c)^2+29*cos(d*x+c)-43)*(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 [A]  time = 2.25436, size = 81, normalized size = 0.68 \begin{align*} \frac{{\left (3 \, \sqrt{2} a^{2} \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 25 \, \sqrt{2} a^{2} \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 150 \, \sqrt{2} a^{2} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} \sqrt{a}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(3*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 25*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 150*sqrt(2)*a^2*sin(1/2*d*x +
1/2*c))*sqrt(a)/d

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Fricas [A]  time = 1.65321, size = 230, normalized size = 1.93 \begin{align*} \frac{2 \,{\left (3 \, a^{2} \cos \left (d x + c\right )^{3} + 14 \, a^{2} \cos \left (d x + c\right )^{2} + 43 \, a^{2} \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+a*sec(d*x+c))^(5/2)/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*a^2*cos(d*x + c)^3 + 14*a^2*cos(d*x + c)^2 + 43*a^2*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+a*sec(d*x+c))**(5/2)/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 (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{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))^(5/2)/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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