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

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

[Out]

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

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Rubi [A]  time = 0.11219, antiderivative size = 73, normalized size of antiderivative = 1., 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 = {2675, 2667, 63, 206} \[ \frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} d}+\frac{\sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2675

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p + 1)), x] + Dist[(a*(m + p + 1))/(g^2*(p + 1)), Int[(
g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ[m + 1/2, 2*p]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

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 206

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

Rubi steps

\begin{align*} \int \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=\frac{\sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{2 d}+\frac{1}{4} a \int \sec (c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=\frac{\sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{2 d}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,a \sin (c+d x)\right )}{4 d}\\ &=\frac{\sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{2 d}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+a \sin (c+d x)}\right )}{2 d}\\ &=\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+a \sin (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} d}+\frac{\sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.249573, size = 72, normalized size = 0.99 \[ \frac{a \left (\sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a (\sin (c+d x)+1)}}{\sqrt{2} \sqrt{a}}\right )-\frac{2 \sqrt{a (\sin (c+d x)+1)}}{\sin (c+d x)-1}\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.102, size = 70, normalized size = 1. \begin{align*} 2\,{\frac{{a}^{3}}{d} \left ( -1/4\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }}{a \left ( a\sin \left ( dx+c \right ) -a \right ) }}+1/8\,{\frac{\sqrt{2}}{{a}^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a^3*(-1/4*(a+a*sin(d*x+c))^(1/2)/a/(a*sin(d*x+c)-a)+1/8/a^(3/2)*2^(1/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2
^(1/2)/a^(1/2)))/d

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.69733, size = 266, normalized size = 3.64 \begin{align*} \frac{{\left (\sqrt{2} a \sin \left (d x + c\right ) - \sqrt{2} a\right )} \sqrt{a} \log \left (-\frac{a \sin \left (d x + c\right ) + 2 \, \sqrt{2} \sqrt{a \sin \left (d x + c\right ) + a} \sqrt{a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 4 \, \sqrt{a \sin \left (d x + c\right ) + a} a}{8 \,{\left (d \sin \left (d x + c\right ) - d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((sqrt(2)*a*sin(d*x + c) - sqrt(2)*a)*sqrt(a)*log(-(a*sin(d*x + c) + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sq
rt(a) + 3*a)/(sin(d*x + c) - 1)) - 4*sqrt(a*sin(d*x + c) + a)*a)/(d*sin(d*x + c) - d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out