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

Optimal. Leaf size=97 \[ -\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}+\frac {2 a^2 \sqrt {a \sec (c+d x)+a}}{d}+\frac {2 a (a \sec (c+d x)+a)^{3/2}}{3 d}+\frac {2 (a \sec (c+d x)+a)^{5/2}}{5 d} \]

[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*a^2*(a+a*sec(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.07, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3880, 50, 63, 207} \[ \frac {2 a^2 \sqrt {a \sec (c+d x)+a}}{d}-\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}+\frac {2 a (a \sec (c+d x)+a)^{3/2}}{3 d}+\frac {2 (a \sec (c+d x)+a)^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x],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*Se
c[c + d*x])^(3/2))/(3*d) + (2*(a + a*Sec[c + d*x])^(5/2))/(5*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 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 (c+d x) \, dx &=\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 {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 {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 {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 {\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}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 82, normalized size = 0.85 \[ \frac {2 (a (\sec (c+d x)+1))^{5/2} \left (\sqrt {\sec (c+d x)+1} \left (3 \sec ^2(c+d x)+11 \sec (c+d x)+23\right )-15 \tanh ^{-1}\left (\sqrt {\sec (c+d x)+1}\right )\right )}{15 d (\sec (c+d x)+1)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a*(1 + Sec[c + d*x]))^(5/2)*(-15*ArcTanh[Sqrt[1 + Sec[c + d*x]]] + Sqrt[1 + Sec[c + d*x]]*(23 + 11*Sec[c +
d*x] + 3*Sec[c + d*x]^2)))/(15*d*(1 + Sec[c + d*x])^(5/2))

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fricas [A]  time = 1.26, size = 282, normalized size = 2.91 \[ \left [\frac {15 \, a^{\frac {5}{2}} \cos \left (d x + c\right )^{2} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 4 \, {\left (23 \, a^{2} \cos \left (d x + c\right )^{2} + 11 \, a^{2} \cos \left (d x + c\right ) + 3 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{30 \, d \cos \left (d x + c\right )^{2}}, \frac {15 \, \sqrt {-a} a^{2} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{2} + 2 \, {\left (23 \, a^{2} \cos \left (d x + c\right )^{2} + 11 \, a^{2} \cos \left (d x + c\right ) + 3 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(15*a^(5/2)*cos(d*x + c)^2*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*(23*a^2*cos(d*x + c)^2 + 11*a^2*cos(d*x + c) + 3*
a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x + c)^2), 1/15*(15*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)^2 + 2*(23*a^2*cos(d*x +
c)^2 + 11*a^2*cos(d*x + c) + 3*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x + c)^2)]

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giac [A]  time = 1.73, size = 148, normalized size = 1.53 \[ \frac {\sqrt {2} {\left (\frac {15 \, \sqrt {2} a \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (15 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} a - 10 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a^{2} + 12 \, a^{3}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*sqrt(2)*(15*sqrt(2)*a*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/sqrt(-a) + 2*(15*(
a*tan(1/2*d*x + 1/2*c)^2 - a)^2*a - 10*(a*tan(1/2*d*x + 1/2*c)^2 - a)*a^2 + 12*a^3)/((a*tan(1/2*d*x + 1/2*c)^2
- a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))*a^2*sgn(cos(d*x + c))/d

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maple [A]  time = 0.19, size = 74, normalized size = 0.76 \[ \frac {\frac {2 \left (a +a \sec \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +a \sec \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 a^{2} \sqrt {a +a \sec \left (d x +c \right )}-2 a^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {a +a \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.64, size = 105, normalized size = 1.08 \[ \frac {15 \, a^{\frac {5}{2}} \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 6 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}} + 10 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a + 30 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}} a^{2}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*a^(5/2)*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/cos(d*x + c)) + sqrt(a))) + 6*(a + a/cos
(d*x + c))^(5/2) + 10*(a + a/cos(d*x + c))^(3/2)*a + 30*sqrt(a + a/cos(d*x + c))*a^2)/d

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mupad [B]  time = 1.82, size = 92, normalized size = 0.95 \[ \frac {2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{5\,d}+\frac {2\,a\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{3\,d}+\frac {2\,a^2\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}}{d}+\frac {a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(a + a/cos(c + d*x))^(5/2))/(5*d) + (a^(5/2)*atan(((a + a/cos(c + d*x))^(1/2)*1i)/a^(1/2))*2i)/d + (2*a*(a
+ a/cos(c + d*x))^(3/2))/(3*d) + (2*a^2*(a + a/cos(c + d*x))^(1/2))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}} \tan {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sec(c + d*x) + 1))**(5/2)*tan(c + d*x), x)

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