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

Optimal. Leaf size=116 \[ \frac{152 a^2 \tan (c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}-\frac{4 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac{38 a \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 d} \]

[Out]

(152*a^2*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (38*a*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(105*d)
- (4*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*d) + (2*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*a*d)

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Rubi [A]  time = 0.193941, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3800, 4001, 3793, 3792} \[ \frac{152 a^2 \tan (c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}-\frac{4 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac{38 a \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(152*a^2*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (38*a*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(105*d)
- (4*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*d) + (2*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*a*d)

Rule 3800

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 4001

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B,
0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 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 ^3(c+d x) (a+a \sec (c+d x))^{3/2} \, dx &=\frac{2 (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 a d}+\frac{2 \int \sec (c+d x) \left (\frac{5 a}{2}-a \sec (c+d x)\right ) (a+a \sec (c+d x))^{3/2} \, dx}{7 a}\\ &=-\frac{4 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 a d}+\frac{19}{35} \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac{38 a \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 d}-\frac{4 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 a d}+\frac{1}{105} (76 a) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{152 a^2 \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{38 a \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 d}-\frac{4 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac{2 (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 a d}\\ \end{align*}

Mathematica [A]  time = 0.178844, size = 60, normalized size = 0.52 \[ \frac{2 a^2 \tan (c+d x) \left (15 \sec ^3(c+d x)+39 \sec ^2(c+d x)+52 \sec (c+d x)+104\right )}{105 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(104 + 52*Sec[c + d*x] + 39*Sec[c + d*x]^2 + 15*Sec[c + d*x]^3)*Tan[c + d*x])/(105*d*Sqrt[a*(1 + Sec[c
+ d*x])])

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Maple [A]  time = 0.148, size = 83, normalized size = 0.7 \begin{align*} -{\frac{2\,a \left ( 104\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}-52\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}-13\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-24\,\cos \left ( dx+c \right ) -15 \right ) }{105\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}\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)^3*(a+a*sec(d*x+c))^(3/2),x)

[Out]

-2/105/d*a*(104*cos(d*x+c)^4-52*cos(d*x+c)^3-13*cos(d*x+c)^2-24*cos(d*x+c)-15)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(
1/2)/cos(d*x+c)^3/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)^3*(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.69004, size = 230, normalized size = 1.98 \begin{align*} \frac{2 \,{\left (104 \, a \cos \left (d x + c\right )^{3} + 52 \, a \cos \left (d x + c\right )^{2} + 39 \, a \cos \left (d x + c\right ) + 15 \, a\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 )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

2/105*(104*a*cos(d*x + c)^3 + 52*a*cos(d*x + c)^2 + 39*a*cos(d*x + c) + 15*a)*sqrt((a*cos(d*x + c) + a)/cos(d*
x + c))*sin(d*x + c)/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)

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

[Out]

Timed out

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Giac [A]  time = 4.83551, size = 204, normalized size = 1.76 \begin{align*} -\frac{4 \,{\left (105 \, \sqrt{2} a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (140 \, \sqrt{2} a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 19 \,{\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} - 7 \, \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 )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{105 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{3} \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)^3*(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-4/105*(105*sqrt(2)*a^5*sgn(cos(d*x + c)) - (140*sqrt(2)*a^5*sgn(cos(d*x + c)) + 19*(2*sqrt(2)*a^5*sgn(cos(d*x
+ c))*tan(1/2*d*x + 1/2*c)^2 - 7*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)^
2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)