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]

2/5*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+14/15*a*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-4/15*(a+a*sec(d*x+c))^(1
/2)*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 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]

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)]

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 = 0.11, 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])])

________________________________________________________________________________________

fricas [A]  time = 0.69, size = 72, normalized size = 0.84 \[ \frac {2 \, {\left (8 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 3\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]

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)

________________________________________________________________________________________

giac [A]  time = 3.67, size = 101, normalized size = 1.17 \[ \frac {2 \, \sqrt {2} {\left (15 \, a^{3} + {\left (7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 10 \, a^{3}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{15 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

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)

________________________________________________________________________________________

maple [A]  time = 1.01, size = 72, normalized size = 0.84 \[ -\frac {2 \left (8 \left (\cos ^{3}\left (d x +c \right )\right )-4 \left (\cos ^{2}\left (d x +c \right )\right )-\cos \left (d x +c \right )-3\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{15 d \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )} \]

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*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^2/sin(d*x+
c)

________________________________________________________________________________________

maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

________________________________________________________________________________________

mupad [B]  time = 4.46, size = 115, normalized size = 1.34 \[ \frac {8\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,5{}\mathrm {i}-{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,5{}\mathrm {i}-{\mathrm {e}}^{c\,5{}\mathrm {i}+d\,x\,5{}\mathrm {i}}\,2{}\mathrm {i}+2{}\mathrm {i}\right )}{15\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^(1/2)/cos(c + d*x)^3,x)

[Out]

(8*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*2i + d*x*2i)*5i - exp(c*3i + d*x*3i)*5
i - exp(c*5i + d*x*5i)*2i + 2i))/(15*d*(exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \sec ^{3}{\left (c + d x \right )}\, dx \]

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)

________________________________________________________________________________________