3.90 \(\int \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=122 \[ \frac{2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt{a \sec (c+d x)+a}}+\frac{12 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 a d}-\frac{8 \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{35 d}+\frac{4 a \tan (c+d x)}{5 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(4*a*Tan[c + d*x])/(5*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c +
d*x]]) - (8*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(35*d) + (12*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*a
*d)

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Rubi [A]  time = 0.206857, antiderivative size = 122, 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 = {3803, 3800, 4001, 3792} \[ \frac{2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt{a \sec (c+d x)+a}}+\frac{12 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 a d}-\frac{8 \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{35 d}+\frac{4 a \tan (c+d x)}{5 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(4*a*Tan[c + d*x])/(5*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c +
d*x]]) - (8*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(35*d) + (12*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*a
*d)

Rule 3803

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

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 ^4(c+d x) \sqrt{a+a \sec (c+d x)} \, dx &=\frac{2 a \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{6}{7} \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{12 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}+\frac{12 \int \sec (c+d x) \left (\frac{3 a}{2}-a \sec (c+d x)\right ) \sqrt{a+a \sec (c+d x)} \, dx}{35 a}\\ &=\frac{2 a \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}-\frac{8 \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{35 d}+\frac{12 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}+\frac{2}{5} \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{4 a \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}-\frac{8 \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{35 d}+\frac{12 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}\\ \end{align*}

Mathematica [A]  time = 0.139788, size = 58, normalized size = 0.48 \[ \frac{2 a \tan (c+d x) \left (5 \sec ^3(c+d x)+6 \sec ^2(c+d x)+8 \sec (c+d x)+16\right )}{35 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(2*a*(16 + 8*Sec[c + d*x] + 6*Sec[c + d*x]^2 + 5*Sec[c + d*x]^3)*Tan[c + d*x])/(35*d*Sqrt[a*(1 + Sec[c + d*x])
])

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Maple [A]  time = 0.211, size = 82, normalized size = 0.7 \begin{align*} -{\frac{32\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}-16\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}-4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-2\,\cos \left ( dx+c \right ) -10}{35\,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)^4*(a+a*sec(d*x+c))^(1/2),x)

[Out]

-2/35/d*(16*cos(d*x+c)^4-8*cos(d*x+c)^3-2*cos(d*x+c)^2-cos(d*x+c)-5)*(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)^4*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.66307, size = 212, normalized size = 1.74 \begin{align*} \frac{2 \,{\left (16 \, \cos \left (d x + c\right )^{3} + 8 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) + 5\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{35 \,{\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)^4*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/35*(16*cos(d*x + c)^3 + 8*cos(d*x + c)^2 + 6*cos(d*x + c) + 5)*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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (\sec{\left (c + d x \right )} + 1\right )} \sec ^{4}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sec(c + d*x) + 1))*sec(c + d*x)**4, x)

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Giac [A]  time = 4.85081, size = 162, normalized size = 1.33 \begin{align*} -\frac{2 \, \sqrt{2}{\left (35 \, a^{4} -{\left (35 \, a^{4} +{\left (9 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 49 \, a^{4}\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 )} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{35 \,{\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)^4*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-2/35*sqrt(2)*(35*a^4 - (35*a^4 + (9*a^4*tan(1/2*d*x + 1/2*c)^2 - 49*a^4)*tan(1/2*d*x + 1/2*c)^2)*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)^3*sqrt(-a*tan(1/2*d*x + 1/2
*c)^2 + a)*d)