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

Optimal. Leaf size=86 \[ \frac{8 a^2 \tan (c+d x)}{5 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{5 d}+\frac{2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.119957, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3798, 3793, 3792} \[ \frac{8 a^2 \tan (c+d x)}{5 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{5 d}+\frac{2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3798

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

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 ^2(c+d x) (a+a \sec (c+d x))^{3/2} \, dx &=\frac{2 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{3}{5} \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac{2 a \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{5 d}+\frac{2 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{1}{5} (4 a) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{8 a^2 \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{5 d}+\frac{2 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.129918, size = 48, normalized size = 0.56 \[ \frac{2 a^2 \tan (c+d x) \left (\sec ^2(c+d x)+3 \sec (c+d x)+6\right )}{5 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(6 + 3*Sec[c + d*x] + Sec[c + d*x]^2)*Tan[c + d*x])/(5*d*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [A]  time = 0.135, size = 73, normalized size = 0.9 \begin{align*} -{\frac{2\,a \left ( 6\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}-3\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-2\,\cos \left ( dx+c \right ) -1 \right ) }{5\,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)^2*(a+a*sec(d*x+c))^(3/2),x)

[Out]

-2/5/d*a*(6*cos(d*x+c)^3-3*cos(d*x+c)^2-2*cos(d*x+c)-1)*(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(-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)^2*(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.71473, size = 189, normalized size = 2.2 \begin{align*} \frac{2 \,{\left (6 \, a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + a\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{5 \,{\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)^2*(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

2/5*(6*a*cos(d*x + c)^2 + 3*a*cos(d*x + c) + a)*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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}} \sec ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sec(c + d*x) + 1))**(3/2)*sec(c + d*x)**2, x)

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Giac [A]  time = 4.93236, size = 163, normalized size = 1.9 \begin{align*} \frac{4 \,{\left (5 \, \sqrt{2} a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) +{\left (2 \, \sqrt{2} a^{4} \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^{4} \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 )}{5 \,{\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)^2*(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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