3.231 \(\int \frac{(a+a \sec (c+d x))^{3/2}}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=161 \[ \frac{26 a^2 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{208 a^2 \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{104 a^2 \sin (c+d x)}{105 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

[Out]

(2*a^2*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (26*a^2*Sin[c + d*x])/(35*d*Sec[c + d
*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (104*a^2*Sin[c + d*x])/(105*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]
]) + (208*a^2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.232306, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3813, 21, 3805, 3804} \[ \frac{26 a^2 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{208 a^2 \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{104 a^2 \sin (c+d x)}{105 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(3/2)/Sec[c + d*x]^(7/2),x]

[Out]

(2*a^2*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (26*a^2*Sin[c + d*x])/(35*d*Sec[c + d
*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (104*a^2*Sin[c + d*x])/(105*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]
]) + (208*a^2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]])

Rule 3813

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[a/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 2)*(d*Csc[e + f*x])^(n + 1)*(b*(m - 2*n - 2) - a*(m + 2*n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d,
e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && (LtQ[n, -1] || (EqQ[m, 3/2] && EqQ[n, -2^(-1)])) && IntegerQ[2
*m]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3805

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

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))^{3/2}}{\sec ^{\frac{7}{2}}(c+d x)} \, dx &=\frac{2 a^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{1}{7} (2 a) \int \frac{\frac{13 a}{2}+\frac{13}{2} a \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{2 a^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{1}{7} (13 a) \int \frac{\sqrt{a+a \sec (c+d x)}}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{26 a^2 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{1}{35} (52 a) \int \frac{\sqrt{a+a \sec (c+d x)}}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{26 a^2 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{104 a^2 \sin (c+d x)}{105 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{1}{105} (104 a) \int \frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 a^2 \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{26 a^2 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{104 a^2 \sin (c+d x)}{105 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{208 a^2 \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.337498, size = 72, normalized size = 0.45 \[ \frac{a (253 \cos (c+d x)+78 \cos (2 (c+d x))+15 \cos (3 (c+d x))+494) \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)}}{210 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^(3/2)/Sec[c + d*x]^(7/2),x]

[Out]

(a*(494 + 253*Cos[c + d*x] + 78*Cos[2*(c + d*x)] + 15*Cos[3*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*
x)/2])/(210*d*Sqrt[Sec[c + d*x]])

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Maple [A]  time = 0.205, size = 93, normalized size = 0.6 \begin{align*} -{\frac{2\,a \left ( 15\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+24\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+13\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+52\,\cos \left ( dx+c \right ) -104 \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{105\,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))^(3/2)/sec(d*x+c)^(7/2),x)

[Out]

-2/105/d*a*(15*cos(d*x+c)^4+24*cos(d*x+c)^3+13*cos(d*x+c)^2+52*cos(d*x+c)-104)*(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 = 2.82408, size = 409, normalized size = 2.54 \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)/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/840*sqrt(2)*(735*a*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) + 175*a
*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) + 63*a*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) - 735*a*cos(7/2*d*x + 7/2*c)*sin(6/7*arctan2(sin
(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 175*a*cos(7/2*d*x + 7/2*c)*sin(4/7*arctan2(sin(7/2*d*x + 7/2*c), c
os(7/2*d*x + 7/2*c))) - 63*a*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)))
+ 30*a*sin(7/2*d*x + 7/2*c) + 63*a*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 175*a*sin(3
/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 735*a*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 = 1.65348, size = 255, normalized size = 1.58 \begin{align*} \frac{2 \,{\left (15 \, a \cos \left (d x + c\right )^{4} + 39 \, a \cos \left (d x + c\right )^{3} + 52 \, a \cos \left (d x + c\right )^{2} + 104 \, 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 )}{105 \,{\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))^(3/2)/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

2/105*(15*a*cos(d*x + c)^4 + 39*a*cos(d*x + c)^3 + 52*a*cos(d*x + c)^2 + 104*a*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))**(3/2)/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{3}{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))^(3/2)/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^(3/2)/sec(d*x + c)^(7/2), x)