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]

-4/35*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/7*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/a/d+152/105*a^2*tan(d*x+c)/d/(
a+a*sec(d*x+c))^(1/2)+38/105*a*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 116, normalized size of antiderivative = 1.00, 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 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 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 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) (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 = 0.18, 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])])

________________________________________________________________________________________

fricas [A]  time = 0.68, size = 87, normalized size = 0.75 \[ \frac {2 \, {\left (104 \, a \cos \left (d x + c\right )^{3} + 52 \, a \cos \left (d x + c\right )^{2} + 39 \, a \cos \left (d x + c\right ) + 15 \, a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]

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)

________________________________________________________________________________________

giac [A]  time = 9.72, size = 151, normalized size = 1.30 \[ -\frac {4 \, {\left (105 \, \sqrt {2} a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (140 \, \sqrt {2} a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 19 \, {\left (2 \, \sqrt {2} a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, \sqrt {2} a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.84, size = 83, normalized size = 0.72 \[ -\frac {2 \left (104 \left (\cos ^{4}\left (d x +c \right )\right )-52 \left (\cos ^{3}\left (d x +c \right )\right )-13 \left (\cos ^{2}\left (d x +c \right )\right )-24 \cos \left (d x +c \right )-15\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{105 d \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105/d*(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*(1+cos(d*x+c))/cos(d*x+c))^(1/
2)/cos(d*x+c)^3/sin(d*x+c)*a

________________________________________________________________________________________

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))^(3/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [B]  time = 5.08, size = 346, normalized size = 2.98 \[ -\frac {\left (\frac {a\,16{}\mathrm {i}}{7\,d}+\frac {a\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,16{}\mathrm {i}}{7\,d}\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\left (\frac {a\,8{}\mathrm {i}}{3\,d}-\frac {a\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,104{}\mathrm {i}}{105\,d}\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}+\frac {\left (\frac {a\,8{}\mathrm {i}}{5\,d}+\frac {a\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,184{}\mathrm {i}}{35\,d}\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {a\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,208{}\mathrm {i}}{105\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((a*8i)/(3*d) - (a*exp(c*1i + d*x*1i)*104i)/(105*d))*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^
(1/2))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) - (((a*16i)/(7*d) + (a*exp(c*1i + d*x*1i)*16i)/(7*d
))*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i
) + 1)^3) + (((a*8i)/(5*d) + (a*exp(c*1i + d*x*1i)*184i)/(35*d))*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d
*x*1i)/2))^(1/2))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) - (a*exp(c*1i + d*x*1i)*(a + a/(exp(-
c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*208i)/(105*d*(exp(c*1i + d*x*1i) + 1))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \sec ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________