3.109 \(\int \sec ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=72 \[ \frac {\sec (c+d x) \sqrt {a \sin (c+d x)+a}}{d}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{\sqrt {2} d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2675, 2649, 206} \[ \frac {\sec (c+d x) \sqrt {a \sin (c+d x)+a}}{d}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{\sqrt {2} d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

-((Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(Sqrt[2]*d)) + (Sec[c + d*x]*Sq
rt[a + a*Sin[c + d*x]])/d

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])

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

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]

Rubi steps

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

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Mathematica [C]  time = 0.23, size = 106, normalized size = 1.47 \[ \frac {\sec (c+d x) \sqrt {a (\sin (c+d x)+1)} \left (1-(1+i) (-1)^{3/4} \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \sec \left (\frac {d x}{4}\right ) \left (\cos \left (\frac {1}{4} (2 c+d x)\right )-\sin \left (\frac {1}{4} (2 c+d x)\right )\right )\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

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

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fricas [B]  time = 0.55, size = 159, normalized size = 2.21 \[ \frac {\sqrt {2} \sqrt {a} \cos \left (d x + c\right ) \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, \sqrt {a \sin \left (d x + c\right ) + a}}{4 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.52, size = 102, normalized size = 1.42 \[ \frac {\sqrt {2} {\left (\log \left ({\left | \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - \log \left ({\left | \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - \frac {2 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}\right )} \sqrt {a}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/4*sqrt(2)*(log(abs(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - log(abs(sin(-1
/4*pi + 1/2*d*x + 1/2*c) - 1))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/sin
(-1/4*pi + 1/2*d*x + 1/2*c))*sqrt(a)/d

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maple [A]  time = 0.23, size = 83, normalized size = 1.15 \[ -\frac {\left (1+\sin \left (d x +c \right )\right ) \left (\sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \sqrt {a -a \sin \left (d x +c \right )}-2 a^{\frac {3}{2}}\right )}{2 \sqrt {a}\, \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sin(d*x + c) + a)*sec(d*x + c)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^(1/2)/cos(c + d*x)^2,x)

[Out]

int((a + a*sin(c + d*x))^(1/2)/cos(c + d*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sin(c + d*x) + 1))*sec(c + d*x)**2, x)

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