3.91 \(\int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=86 \[ \frac{2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}-\frac{4 \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d}+\frac{14 a \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(14*a*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) - (4*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(15*d) + (2*(a
 + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*a*d)

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Rubi [A]  time = 0.152013, 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 = {3800, 4001, 3792} \[ \frac{2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}-\frac{4 \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d}+\frac{14 a \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(14*a*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) - (4*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(15*d) + (2*(a
 + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*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 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) \sqrt{a+a \sec (c+d x)} \, dx &=\frac{2 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 a d}+\frac{2 \int \sec (c+d x) \left (\frac{3 a}{2}-a \sec (c+d x)\right ) \sqrt{a+a \sec (c+d x)} \, dx}{5 a}\\ &=-\frac{4 \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 a d}+\frac{7}{15} \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{14 a \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}-\frac{4 \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 a d}\\ \end{align*}

Mathematica [A]  time = 0.102349, size = 48, normalized size = 0.56 \[ \frac{2 a \tan (c+d x) \left (3 \sec ^2(c+d x)+4 \sec (c+d x)+8\right )}{15 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(2*a*(8 + 4*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 = 0.169, size = 72, normalized size = 0.8 \begin{align*} -{\frac{16\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}-8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-2\,\cos \left ( dx+c \right ) -6}{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)^3*(a+a*sec(d*x+c))^(1/2),x)

[Out]

-2/15/d*(8*cos(d*x+c)^3-4*cos(d*x+c)^2-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(-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))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.67688, size = 185, normalized size = 2.15 \begin{align*} \frac{2 \,{\left (8 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 3\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)^3*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/15*(8*cos(d*x + c)^2 + 4*cos(d*x + c) + 3)*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 \sqrt{a \left (\sec{\left (c + d x \right )} + 1\right )} \sec ^{3}{\left (c + d x \right )}\, dx \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))**(1/2),x)

[Out]

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

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Giac [A]  time = 4.85842, size = 136, normalized size = 1.58 \begin{align*} \frac{2 \, \sqrt{2}{\left (15 \, a^{3} +{\left (7 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 10 \, a^{3}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \mathrm{sgn}\left (\cos \left (d x + c\right )\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)^3*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

2/15*sqrt(2)*(15*a^3 + (7*a^3*tan(1/2*d*x + 1/2*c)^2 - 10*a^3)*tan(1/2*d*x + 1/2*c)^2)*sgn(cos(d*x + c))*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)