3.108 \(\int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=146 \[ \frac {832 a^3 \tan (c+d x)}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {208 a^2 \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}-\frac {4 \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d}+\frac {26 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d} \]

[Out]

26/105*a*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d-4/63*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/d+2/9*(a+a*sec(d*x+c))^(7/
2)*tan(d*x+c)/a/d+832/315*a^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+208/315*a^2*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c
)/d

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3800, 4001, 3793, 3792} \[ \frac {208 a^2 \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {832 a^3 \tan (c+d x)}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}-\frac {4 \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d}+\frac {26 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(832*a^3*Tan[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (208*a^2*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(315
*d) + (26*a*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(105*d) - (4*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(63
*d) + (2*(a + a*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*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))^{5/2} \, dx &=\frac {2 (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {2 \int \sec (c+d x) \left (\frac {7 a}{2}-a \sec (c+d x)\right ) (a+a \sec (c+d x))^{5/2} \, dx}{9 a}\\ &=-\frac {4 (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {13}{21} \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx\\ &=\frac {26 a (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}-\frac {4 (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {1}{105} (104 a) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac {208 a^2 \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {26 a (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}-\frac {4 (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {1}{315} \left (416 a^2\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {832 a^3 \tan (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {208 a^2 \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {26 a (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}-\frac {4 (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.53, size = 70, normalized size = 0.48 \[ \frac {2 a^3 \tan (c+d x) \left (35 \sec ^4(c+d x)+130 \sec ^3(c+d x)+219 \sec ^2(c+d x)+292 \sec (c+d x)+584\right )}{315 d \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(2*a^3*(584 + 292*Sec[c + d*x] + 219*Sec[c + d*x]^2 + 130*Sec[c + d*x]^3 + 35*Sec[c + d*x]^4)*Tan[c + d*x])/(3
15*d*Sqrt[a*(1 + Sec[c + d*x])])

________________________________________________________________________________________

fricas [A]  time = 0.69, size = 108, normalized size = 0.74 \[ \frac {2 \, {\left (584 \, a^{2} \cos \left (d x + c\right )^{4} + 292 \, a^{2} \cos \left (d x + c\right )^{3} + 219 \, a^{2} \cos \left (d x + c\right )^{2} + 130 \, a^{2} \cos \left (d x + c\right ) + 35 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

2/315*(584*a^2*cos(d*x + c)^4 + 292*a^2*cos(d*x + c)^3 + 219*a^2*cos(d*x + c)^2 + 130*a^2*cos(d*x + c) + 35*a^
2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)

________________________________________________________________________________________

giac [A]  time = 5.91, size = 180, normalized size = 1.23 \[ \frac {8 \, {\left (315 \, \sqrt {2} a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (630 \, \sqrt {2} a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 13 \, {\left (63 \, \sqrt {2} a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 4 \, {\left (2 \, \sqrt {2} a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \sqrt {2} a^{7} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{315 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

8/315*(315*sqrt(2)*a^7*sgn(cos(d*x + c)) - (630*sqrt(2)*a^7*sgn(cos(d*x + c)) - 13*(63*sqrt(2)*a^7*sgn(cos(d*x
+ c)) + 4*(2*sqrt(2)*a^7*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)^2 - 9*sqrt(2)*a^7*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)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^
2 - a)^4*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)

________________________________________________________________________________________

maple [A]  time = 0.87, size = 95, normalized size = 0.65 \[ -\frac {2 \left (584 \left (\cos ^{5}\left (d x +c \right )\right )-292 \left (\cos ^{4}\left (d x +c \right )\right )-73 \left (\cos ^{3}\left (d x +c \right )\right )-89 \left (\cos ^{2}\left (d x +c \right )\right )-95 \cos \left (d x +c \right )-35\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{315 d \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2),x)

[Out]

-2/315/d*(584*cos(d*x+c)^5-292*cos(d*x+c)^4-73*cos(d*x+c)^3-89*cos(d*x+c)^2-95*cos(d*x+c)-35)*(a*(1+cos(d*x+c)
)/cos(d*x+c))^(1/2)/cos(d*x+c)^4/sin(d*x+c)*a^2

________________________________________________________________________________________

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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [B]  time = 8.18, size = 456, normalized size = 3.12 \[ \frac {\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 (\frac {a^2\,32{}\mathrm {i}}{9\,d}-\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,32{}\mathrm {i}}{9\,d}\right )}{\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 )}^4}-\frac {\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 (\frac {a^2\,96{}\mathrm {i}}{7\,d}-\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,32{}\mathrm {i}}{63\,d}\right )}{\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 {\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 (\frac {a^2\,8{}\mathrm {i}}{3\,d}-\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,584{}\mathrm {i}}{315\,d}\right )}{\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 {\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 (\frac {a^2\,56{}\mathrm {i}}{5\,d}+\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,904{}\mathrm {i}}{105\,d}\right )}{\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^2\,{\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}}}\,1168{}\mathrm {i}}{315\,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))^(5/2)/cos(c + d*x)^3,x)

[Out]

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((a^2*32i)/(9*d) - (a^2*exp(c*1i + d*x*1i)*32i)
/(9*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i +
d*x*1i)/2))^(1/2)*((a^2*96i)/(7*d) - (a^2*exp(c*1i + d*x*1i)*32i)/(63*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i
+ d*x*2i) + 1)^3) + ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((a^2*8i)/(3*d) - (a^2*exp
(c*1i + d*x*1i)*584i)/(315*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) + ((a + a/(exp(- c*1i - d*
x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((a^2*56i)/(5*d) + (a^2*exp(c*1i + d*x*1i)*904i)/(105*d)))/((exp(c*1i +
d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) - (a^2*exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i
+ d*x*1i)/2))^(1/2)*1168i)/(315*d*(exp(c*1i + d*x*1i) + 1))

________________________________________________________________________________________

sympy [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))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________