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

Optimal. Leaf size=73 \[ -\frac{2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{d}+\frac{2 a \sqrt{a \sec (c+d x)+a}}{d}+\frac{2 (a \sec (c+d x)+a)^{3/2}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0552833, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3880, 50, 63, 207} \[ -\frac{2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{d}+\frac{2 a \sqrt{a \sec (c+d x)+a}}{d}+\frac{2 (a \sec (c+d x)+a)^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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])

Rubi steps

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

Mathematica [A]  time = 0.137176, size = 70, normalized size = 0.96 \[ \frac{2 (a (\sec (c+d x)+1))^{3/2} \left (\sqrt{\sec (c+d x)+1} (\sec (c+d x)+4)-3 \tanh ^{-1}\left (\sqrt{\sec (c+d x)+1}\right )\right )}{3 d (\sec (c+d x)+1)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a*(1 + Sec[c + d*x]))^(3/2)*(-3*ArcTanh[Sqrt[1 + Sec[c + d*x]]] + Sqrt[1 + Sec[c + d*x]]*(4 + Sec[c + d*x]
)))/(3*d*(1 + Sec[c + d*x])^(3/2))

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Maple [A]  time = 0.031, size = 57, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({\frac{2}{3} \left ( a+a\sec \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}+2\,a\sqrt{a+a\sec \left ( dx+c \right ) }-2\,{a}^{3/2}{\it Artanh} \left ({\frac{\sqrt{a+a\sec \left ( dx+c \right ) }}{\sqrt{a}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.9428, size = 626, normalized size = 8.58 \begin{align*} \left [\frac{3 \, a^{\frac{3}{2}} \cos \left (d x + c\right ) \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 (4 \, a \cos \left (d x + c\right ) + a\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{6 \, d \cos \left (d x + c\right )}, \frac{3 \, \sqrt{-a} a \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 \,{\left (4 \, a \cos \left (d x + c\right ) + a\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{3 \, d \cos \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*a^(3/2)*cos(d*x + c)*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*(4*a*cos(d*x + c) + a)*sqrt((a*cos(d*x + c) + a)/cos(
d*x + c)))/(d*cos(d*x + c)), 1/3*(3*sqrt(-a)*a*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*(4*a*cos(d*x + c) + a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)
))/(d*cos(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}} \tan{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 4.69551, size = 163, normalized size = 2.23 \begin{align*} \frac{\sqrt{2} a^{4}{\left (\frac{3 \, \sqrt{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} a^{2}} + \frac{2 \,{\left (3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 5 \, a\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} a^{2}}\right )} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(2)*a^4*(3*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*a^2) + 2
*(3*a*tan(1/2*d*x + 1/2*c)^2 - 5*a)/((a*tan(1/2*d*x + 1/2*c)^2 - a)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a^2))*
sgn(cos(d*x + c))/d