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

Optimal. Leaf size=89 \[ \frac{64 a^3 \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d}+\frac{2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

[Out]

(64*a^3*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(15*d)
+ (2*a*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)

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Rubi [A]  time = 0.0949433, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3793, 3792} \[ \frac{64 a^3 \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d}+\frac{2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a^3*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(15*d)
+ (2*a*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)

Rule 3793

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

Rule 3792

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

Rubi steps

\begin{align*} \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx &=\frac{2 a (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{1}{5} (8 a) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac{16 a^2 \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 a (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{1}{15} \left (32 a^2\right ) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{64 a^3 \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{16 a^2 \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 a (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.0907647, size = 50, normalized size = 0.56 \[ \frac{2 a^3 \tan (c+d x) \left (3 \sec ^2(c+d x)+14 \sec (c+d x)+43\right )}{15 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(43 + 14*Sec[c + d*x] + 3*Sec[c + d*x]^2)*Tan[c + d*x])/(15*d*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [A]  time = 0.128, size = 75, normalized size = 0.8 \begin{align*} -{\frac{2\,{a}^{2} \left ( 43\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}-29\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-11\,\cos \left ( dx+c \right ) -3 \right ) }{15\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15/d*a^2*(43*cos(d*x+c)^3-29*cos(d*x+c)^2-11*cos(d*x+c)-3)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)^2
/sin(d*x+c)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.96229, size = 204, normalized size = 2.29 \begin{align*} \frac{2 \,{\left (43 \, a^{2} \cos \left (d x + c\right )^{2} + 14 \, a^{2} \cos \left (d x + c\right ) + 3 \, a^{2}\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 )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(43*a^2*cos(d*x + c)^2 + 14*a^2*cos(d*x + c) + 3*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c
)/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)

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

[Out]

Timed out

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Giac [A]  time = 4.91895, size = 165, normalized size = 1.85 \begin{align*} \frac{8 \,{\left (15 \, \sqrt{2} a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 4 \,{\left (2 \, \sqrt{2} a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 5 \, \sqrt{2} a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{15 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/15*(15*sqrt(2)*a^5*sgn(cos(d*x + c)) + 4*(2*sqrt(2)*a^5*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)^2 - 5*sqrt(2)
*a^5*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a
*tan(1/2*d*x + 1/2*c)^2 + a)*d)