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

Optimal. Leaf size=101 \[ \frac{8 a^2 (5 A+3 B) \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a (5 A+3 B) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d}+\frac{2 B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

[Out]

(8*a^2*(5*A + 3*B)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(5*A + 3*B)*Sqrt[a + a*Sec[c + d*x]]*T
an[c + d*x])/(15*d) + (2*B*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)

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Rubi [A]  time = 0.140036, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {4001, 3793, 3792} \[ \frac{8 a^2 (5 A+3 B) \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a (5 A+3 B) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d}+\frac{2 B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*(5*A + 3*B)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(5*A + 3*B)*Sqrt[a + a*Sec[c + d*x]]*T
an[c + d*x])/(15*d) + (2*B*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)

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 (c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx &=\frac{2 B (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{1}{5} (5 A+3 B) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac{2 a (5 A+3 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 B (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{1}{15} (4 a (5 A+3 B)) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{8 a^2 (5 A+3 B) \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (5 A+3 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 B (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.284577, size = 70, normalized size = 0.69 \[ \frac{2 a \sqrt{a (\sec (c+d x)+1)} ((25 A+18 B) \sin (c+d x)+\tan (c+d x) (5 A+3 B \sec (c+d x)+9 B))}{15 d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Sqrt[a*(1 + Sec[c + d*x])]*((25*A + 18*B)*Sin[c + d*x] + (5*A + 9*B + 3*B*Sec[c + d*x])*Tan[c + d*x]))/(1
5*d*(1 + Cos[c + d*x]))

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Maple [A]  time = 0.227, size = 95, normalized size = 0.9 \begin{align*} -{\frac{2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 25\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+18\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+5\,A\cos \left ( dx+c \right ) +9\,B\cos \left ( dx+c \right ) +3\,B \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))^(3/2)*(A+B*sec(d*x+c)),x)

[Out]

-2/15/d*a*(-1+cos(d*x+c))*(25*A*cos(d*x+c)^2+18*B*cos(d*x+c)^2+5*A*cos(d*x+c)+9*B*cos(d*x+c)+3*B)*(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)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.468321, size = 225, normalized size = 2.23 \begin{align*} \frac{2 \,{\left ({\left (25 \, A + 18 \, B\right )} a \cos \left (d x + c\right )^{2} +{\left (5 \, A + 9 \, B\right )} a \cos \left (d x + c\right ) + 3 \, B a\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))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

2/15*((25*A + 18*B)*a*cos(d*x + c)^2 + (5*A + 9*B)*a*cos(d*x + c) + 3*B*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}} \left (A + B \sec{\left (c + d x \right )}\right ) \sec{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 5.31794, size = 238, normalized size = 2.36 \begin{align*} \frac{4 \,{\left (15 \, \sqrt{2} A a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 15 \, \sqrt{2} B a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (25 \, \sqrt{2} A a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 15 \, \sqrt{2} B a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 2 \,{\left (5 \, \sqrt{2} A a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, \sqrt{2} B 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 )^{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))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

4/15*(15*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 15*sqrt(2)*B*a^4*sgn(cos(d*x + c)) - (25*sqrt(2)*A*a^4*sgn(cos(d*x
+ c)) + 15*sqrt(2)*B*a^4*sgn(cos(d*x + c)) - 2*(5*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 3*sqrt(2)*B*a^4*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)^
2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)