3.239 \(\int \frac{(a+a \sec (c+d x))^{5/2}}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=156 \[ \frac{64 a^3 \sin (c+d x) \sqrt{\sec (c+d x)}}{21 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{21 d \sqrt{\sec (c+d x)}}+\frac{2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

[Out]

(64*a^3*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(21*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*Sqrt[a + a*Sec[c + d*x]]*Si
n[c + d*x])/(21*d*Sqrt[Sec[c + d*x]]) + (2*a*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(7*d*Sec[c + d*x]^(3/2))
+ (2*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2))

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Rubi [A]  time = 0.236141, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3812, 3809, 3804} \[ \frac{64 a^3 \sin (c+d x) \sqrt{\sec (c+d x)}}{21 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{21 d \sqrt{\sec (c+d x)}}+\frac{2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(5/2)/Sec[c + d*x]^(7/2),x]

[Out]

(64*a^3*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(21*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*Sqrt[a + a*Sec[c + d*x]]*Si
n[c + d*x])/(21*d*Sqrt[Sec[c + d*x]]) + (2*a*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(7*d*Sec[c + d*x]^(3/2))
+ (2*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2))

Rule 3812

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[
e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*(m + 1)), x] + Dist[(a*m)/(b*d*(m + 1)), Int[(a + b*Csc
[e + f*x])^m*(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 + 1, 0] &&  !LtQ[m, -2^(-1)]

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{7}{2}}(c+d x)} \, dx &=\frac{2 (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{5}{7} \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)}{7 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{7} (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)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 a (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{21} \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)}{21 d \sqrt{a+a \sec (c+d x)}}+\frac{16 a^2 \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 a (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 0.336332, size = 74, normalized size = 0.47 \[ \frac{a^2 (101 \cos (c+d x)+24 \cos (2 (c+d x))+3 \cos (3 (c+d x))+208) \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)}}{42 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^(5/2)/Sec[c + d*x]^(7/2),x]

[Out]

(a^2*(208 + 101*Cos[c + d*x] + 24*Cos[2*(c + d*x)] + 3*Cos[3*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d
*x)/2])/(42*d*Sqrt[Sec[c + d*x]])

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Maple [A]  time = 0.187, size = 95, normalized size = 0.6 \begin{align*} -{\frac{2\,{a}^{2} \left ( 3\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+9\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+11\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+23\,\cos \left ( dx+c \right ) -46 \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{21\,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{7}{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)^(7/2),x)

[Out]

-2/21/d*a^2*(3*cos(d*x+c)^4+9*cos(d*x+c)^3+11*cos(d*x+c)^2+23*cos(d*x+c)-46)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/
2)*cos(d*x+c)^4*(1/cos(d*x+c))^(7/2)/sin(d*x+c)

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Maxima [B]  time = 2.16188, size = 436, normalized size = 2.79 \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))^(5/2)/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/168*sqrt(2)*(315*a^2*cos(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 77*
a^2*cos(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 21*a^2*cos(2/7*arctan2
(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 315*a^2*cos(7/2*d*x + 7/2*c)*sin(6/7*arct
an2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 77*a^2*cos(7/2*d*x + 7/2*c)*sin(4/7*arctan2(sin(7/2*d*x + 7
/2*c), cos(7/2*d*x + 7/2*c))) - 21*a^2*cos(7/2*d*x + 7/2*c)*sin(2/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x
+ 7/2*c))) + 6*a^2*sin(7/2*d*x + 7/2*c) + 21*a^2*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))
+ 77*a^2*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 315*a^2*sin(1/7*arctan2(sin(7/2*d*x +
7/2*c), cos(7/2*d*x + 7/2*c))))*sqrt(a)/d

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Fricas [A]  time = 1.70664, size = 262, normalized size = 1.68 \begin{align*} \frac{2 \,{\left (3 \, a^{2} \cos \left (d x + c\right )^{4} + 12 \, a^{2} \cos \left (d x + c\right )^{3} + 23 \, a^{2} \cos \left (d x + c\right )^{2} + 46 \, 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 )}{21 \,{\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)^(7/2),x, algorithm="fricas")

[Out]

2/21*(3*a^2*cos(d*x + c)^4 + 12*a^2*cos(d*x + c)^3 + 23*a^2*cos(d*x + c)^2 + 46*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)**(7/2),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{7}{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)^(7/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^(5/2)/sec(d*x + c)^(7/2), x)