3.110 \(\int \sec ^3(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=95 \[ -\frac{3 a}{4 d \sqrt{a \sin (c+d x)+a}}+\frac{3 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} d}+\frac{\sec ^2(c+d x) \sqrt{a \sin (c+d x)+a}}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.123447, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2675, 2667, 51, 63, 206} \[ -\frac{3 a}{4 d \sqrt{a \sin (c+d x)+a}}+\frac{3 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} d}+\frac{\sec ^2(c+d x) \sqrt{a \sin (c+d x)+a}}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(3*Sqrt[a]*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(4*Sqrt[2]*d) - (3*a)/(4*d*Sqrt[a + a*Sin[c +
d*x]]) + (Sec[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]])/(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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 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) \sqrt{a+a \sin (c+d x)} \, dx &=\frac{\sec ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{2 d}+\frac{1}{4} (3 a) \int \frac{\sec (c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=\frac{\sec ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{2 d}+\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{3/2}} \, dx,x,a \sin (c+d x)\right )}{4 d}\\ &=-\frac{3 a}{4 d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{2 d}+\frac{(3 a) \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,a \sin (c+d x)\right )}{8 d}\\ &=-\frac{3 a}{4 d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{2 d}+\frac{(3 a) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+a \sin (c+d x)}\right )}{4 d}\\ &=\frac{3 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+a \sin (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} d}-\frac{3 a}{4 d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^2(c+d x) \sqrt{a+a \sin (c+d x)}}{2 d}\\ \end{align*}

Mathematica [C]  time = 0.348932, size = 271, normalized size = 2.85 \[ \frac{\sqrt{a (\sin (c+d x)+1)} \left (\frac{\left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{2 \sin \left (\frac{d x}{2}\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+(-3+3 i) \sqrt [4]{-1} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \sec \left (\frac{d x}{4}\right ) \left (\sin \left (\frac{1}{4} (2 c+d x)\right )+\cos \left (\frac{1}{4} (2 c+d x)\right )\right )\right )-2\right )}{4 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

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

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Maple [A]  time = 0.145, size = 90, normalized size = 1. \begin{align*} 2\,{\frac{{a}^{3}}{d} \left ( -1/4\,{\frac{1}{{a}^{2}} \left ( 1/2\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }}{a\sin \left ( dx+c \right ) -a}}-3/4\,{\frac{\sqrt{2}}{\sqrt{a}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) } \right ) }-1/4\,{\frac{1}{{a}^{2}\sqrt{a+a\sin \left ( dx+c \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))^(1/2),x)

[Out]

2*a^3*(-1/4/a^2*(1/2*(a+a*sin(d*x+c))^(1/2)/(a*sin(d*x+c)-a)-3/4*2^(1/2)/a^(1/2)*arctanh(1/2*(a+a*sin(d*x+c))^
(1/2)*2^(1/2)/a^(1/2)))-1/4/a^2/(a+a*sin(d*x+c))^(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))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.71906, size = 274, normalized size = 2.88 \begin{align*} \frac{3 \, \sqrt{2} \sqrt{a} \cos \left (d x + c\right )^{2} \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}{\left (3 \, \sin \left (d x + c\right ) - 1\right )}}{16 \, d \cos \left (d x + c\right )^{2}} \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))^(1/2),x, algorithm="fricas")

[Out]

1/16*(3*sqrt(2)*sqrt(a)*cos(d*x + c)^2*log(-(a*sin(d*x + c) + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a) + 3*a
)/(sin(d*x + c) - 1)) + 4*sqrt(a*sin(d*x + c) + a)*(3*sin(d*x + c) - 1))/(d*cos(d*x + c)^2)

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

[Out]

Integral(sqrt(a*(sin(c + d*x) + 1))*sec(c + d*x)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \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))^(1/2),x, algorithm="giac")

[Out]

sage2