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

Optimal. Leaf size=145 \[ \frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}+\frac {2 (a \sec (c+d x)+a)^{9/2}}{9 a^2 d}-\frac {2 a^2 \sqrt {a \sec (c+d x)+a}}{d}-\frac {2 (a \sec (c+d x)+a)^{7/2}}{7 a d}-\frac {2 (a \sec (c+d x)+a)^{5/2}}{5 d}-\frac {2 a (a \sec (c+d x)+a)^{3/2}}{3 d} \]

[Out]

2*a^(5/2)*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/d-2/3*a*(a+a*sec(d*x+c))^(3/2)/d-2/5*(a+a*sec(d*x+c))^(5/2)/
d-2/7*(a+a*sec(d*x+c))^(7/2)/a/d+2/9*(a+a*sec(d*x+c))^(9/2)/a^2/d-2*a^2*(a+a*sec(d*x+c))^(1/2)/d

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3880, 80, 50, 63, 207} \[ \frac {2 (a \sec (c+d x)+a)^{9/2}}{9 a^2 d}-\frac {2 a^2 \sqrt {a \sec (c+d x)+a}}{d}+\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}-\frac {2 (a \sec (c+d x)+a)^{7/2}}{7 a d}-\frac {2 (a \sec (c+d x)+a)^{5/2}}{5 d}-\frac {2 a (a \sec (c+d x)+a)^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(5/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/d - (2*a^2*Sqrt[a + a*Sec[c + d*x]])/d - (2*a*(a + a*Sec
[c + d*x])^(3/2))/(3*d) - (2*(a + a*Sec[c + d*x])^(5/2))/(5*d) - (2*(a + a*Sec[c + d*x])^(7/2))/(7*a*d) + (2*(
a + a*Sec[c + d*x])^(9/2))/(9*a^2*d)

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 3880

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(d*b^(m - 1)
)^(-1), Subst[Int[((-a + b*x)^((m - 1)/2)*(a + b*x)^((m - 1)/2 + n))/x, x], x, Csc[c + d*x]], x] /; FreeQ[{a,
b, c, d, n}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^{5/2} \tan ^3(c+d x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(-a+a x) (a+a x)^{7/2}}{x} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac {2 (a+a \sec (c+d x))^{9/2}}{9 a^2 d}-\frac {\operatorname {Subst}\left (\int \frac {(a+a x)^{7/2}}{x} \, dx,x,\sec (c+d x)\right )}{a d}\\ &=-\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a d}+\frac {2 (a+a \sec (c+d x))^{9/2}}{9 a^2 d}-\frac {\operatorname {Subst}\left (\int \frac {(a+a x)^{5/2}}{x} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac {2 (a+a \sec (c+d x))^{5/2}}{5 d}-\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a d}+\frac {2 (a+a \sec (c+d x))^{9/2}}{9 a^2 d}-\frac {a \operatorname {Subst}\left (\int \frac {(a+a x)^{3/2}}{x} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac {2 a (a+a \sec (c+d x))^{3/2}}{3 d}-\frac {2 (a+a \sec (c+d x))^{5/2}}{5 d}-\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a d}+\frac {2 (a+a \sec (c+d x))^{9/2}}{9 a^2 d}-\frac {a^2 \operatorname {Subst}\left (\int \frac {\sqrt {a+a x}}{x} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac {2 a^2 \sqrt {a+a \sec (c+d x)}}{d}-\frac {2 a (a+a \sec (c+d x))^{3/2}}{3 d}-\frac {2 (a+a \sec (c+d x))^{5/2}}{5 d}-\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a d}+\frac {2 (a+a \sec (c+d x))^{9/2}}{9 a^2 d}-\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac {2 a^2 \sqrt {a+a \sec (c+d x)}}{d}-\frac {2 a (a+a \sec (c+d x))^{3/2}}{3 d}-\frac {2 (a+a \sec (c+d x))^{5/2}}{5 d}-\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a d}+\frac {2 (a+a \sec (c+d x))^{9/2}}{9 a^2 d}-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1+\frac {x^2}{a}} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{d}\\ &=\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {2 a^2 \sqrt {a+a \sec (c+d x)}}{d}-\frac {2 a (a+a \sec (c+d x))^{3/2}}{3 d}-\frac {2 (a+a \sec (c+d x))^{5/2}}{5 d}-\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a d}+\frac {2 (a+a \sec (c+d x))^{9/2}}{9 a^2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.61, size = 102, normalized size = 0.70 \[ \frac {2 (a (\sec (c+d x)+1))^{5/2} \left (\sqrt {\sec (c+d x)+1} \left (35 \sec ^4(c+d x)+95 \sec ^3(c+d x)+12 \sec ^2(c+d x)-226 \sec (c+d x)-493\right )+315 \tanh ^{-1}\left (\sqrt {\sec (c+d x)+1}\right )\right )}{315 d (\sec (c+d x)+1)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a*(1 + Sec[c + d*x]))^(5/2)*(315*ArcTanh[Sqrt[1 + Sec[c + d*x]]] + Sqrt[1 + Sec[c + d*x]]*(-493 - 226*Sec[
c + d*x] + 12*Sec[c + d*x]^2 + 95*Sec[c + d*x]^3 + 35*Sec[c + d*x]^4)))/(315*d*(1 + Sec[c + d*x])^(5/2))

________________________________________________________________________________________

fricas [A]  time = 0.79, size = 334, normalized size = 2.30 \[ \left [\frac {315 \, a^{\frac {5}{2}} \cos \left (d x + c\right )^{4} \log \left (-8 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) - 4 \, {\left (493 \, a^{2} \cos \left (d x + c\right )^{4} + 226 \, a^{2} \cos \left (d x + c\right )^{3} - 12 \, a^{2} \cos \left (d x + c\right )^{2} - 95 \, a^{2} \cos \left (d x + c\right ) - 35 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{630 \, d \cos \left (d x + c\right )^{4}}, -\frac {315 \, \sqrt {-a} a^{2} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{4} + 2 \, {\left (493 \, a^{2} \cos \left (d x + c\right )^{4} + 226 \, a^{2} \cos \left (d x + c\right )^{3} - 12 \, a^{2} \cos \left (d x + c\right )^{2} - 95 \, a^{2} \cos \left (d x + c\right ) - 35 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{315 \, d \cos \left (d x + c\right )^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/630*(315*a^(5/2)*cos(d*x + c)^4*log(-8*a*cos(d*x + c)^2 - 4*(2*cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)*sqrt(
(a*cos(d*x + c) + a)/cos(d*x + c)) - 8*a*cos(d*x + c) - a) - 4*(493*a^2*cos(d*x + c)^4 + 226*a^2*cos(d*x + c)^
3 - 12*a^2*cos(d*x + c)^2 - 95*a^2*cos(d*x + c) - 35*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x
+ c)^4), -1/315*(315*sqrt(-a)*a^2*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(2*a*
cos(d*x + c) + a))*cos(d*x + c)^4 + 2*(493*a^2*cos(d*x + c)^4 + 226*a^2*cos(d*x + c)^3 - 12*a^2*cos(d*x + c)^2
- 95*a^2*cos(d*x + c) - 35*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x + c)^4)]

________________________________________________________________________________________

giac [A]  time = 3.92, size = 198, normalized size = 1.37 \[ -\frac {\sqrt {2} {\left (\frac {315 \, \sqrt {2} a^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (315 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} a^{2} - 210 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} a^{3} + 252 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} a^{4} - 360 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a^{5} - 560 \, a^{6}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{315 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/315*sqrt(2)*(315*sqrt(2)*a^2*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/sqrt(-a) + 2*
(315*(a*tan(1/2*d*x + 1/2*c)^2 - a)^4*a^2 - 210*(a*tan(1/2*d*x + 1/2*c)^2 - a)^3*a^3 + 252*(a*tan(1/2*d*x + 1/
2*c)^2 - a)^2*a^4 - 360*(a*tan(1/2*d*x + 1/2*c)^2 - a)*a^5 - 560*a^6)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^4*sqrt(-
a*tan(1/2*d*x + 1/2*c)^2 + a)))*a*sgn(cos(d*x + c))/d

________________________________________________________________________________________

maple [B]  time = 1.24, size = 362, normalized size = 2.50 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (315 \sqrt {2}\, \left (\cos ^{4}\left (d x +c \right )\right ) \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {9}{2}}+1260 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}\, \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {9}{2}} \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )+1890 \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {9}{2}}+1260 \sqrt {2}\, \cos \left (d x +c \right ) \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {9}{2}}+315 \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {9}{2}}+15776 \left (\cos ^{4}\left (d x +c \right )\right )+7232 \left (\cos ^{3}\left (d x +c \right )\right )-384 \left (\cos ^{2}\left (d x +c \right )\right )-3040 \cos \left (d x +c \right )-1120\right ) a^{2}}{5040 d \cos \left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5040/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(315*2^(1/2)*cos(d*x+c)^4*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)+1260*2^(1/2)*cos(d*x+c)^3*arctan(1/2*(-2*cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)+1890*2^(1/2)*cos(d*x+c)^2*arctan(1/2*(-2*cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)+1260*2^(1/2)*cos(d*x+c)*arctan(1/2
*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)+315*2^(1/2)*arctan(1/2*(-2
*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)+15776*cos(d*x+c)^4+7232*cos(d*
x+c)^3-384*cos(d*x+c)^2-3040*cos(d*x+c)-1120)/cos(d*x+c)^4*a^2

________________________________________________________________________________________

maxima [A]  time = 0.61, size = 143, normalized size = 0.99 \[ -\frac {315 \, a^{\frac {5}{2}} \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 126 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}} - \frac {70 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {9}{2}}}{a^{2}} + \frac {90 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {7}{2}}}{a} + 210 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a + 630 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}} a^{2}}{315 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/315*(315*a^(5/2)*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/cos(d*x + c)) + sqrt(a))) + 126*(a +
a/cos(d*x + c))^(5/2) - 70*(a + a/cos(d*x + c))^(9/2)/a^2 + 90*(a + a/cos(d*x + c))^(7/2)/a + 210*(a + a/cos(d
*x + c))^(3/2)*a + 630*sqrt(a + a/cos(d*x + c))*a^2)/d

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^3\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________