3.111 \(\int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=137 \[ -\frac {5 a^2 \cos (c+d x)}{8 d (a \sin (c+d x)+a)^{3/2}}+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}+\frac {5 a \sec (c+d x)}{6 d \sqrt {a \sin (c+d x)+a}}-\frac {5 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{8 \sqrt {2} d} \]

[Out]

-5/8*a^2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)-5/16*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)+5/6*a*sec(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+1/3*sec(d*x+c)^3*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.16, antiderivative size = 137, normalized size of antiderivative = 1.00, 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, 2687, 2650, 2649, 206} \[ -\frac {5 a^2 \cos (c+d x)}{8 d (a \sin (c+d x)+a)^{3/2}}+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}+\frac {5 a \sec (c+d x)}{6 d \sqrt {a \sin (c+d x)+a}}-\frac {5 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{8 \sqrt {2} d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-5*Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(8*Sqrt[2]*d) - (5*a^2*Cos[c +
d*x])/(8*d*(a + a*Sin[c + d*x])^(3/2)) + (5*a*Sec[c + d*x])/(6*d*Sqrt[a + a*Sin[c + d*x]]) + (Sec[c + d*x]^3*
Sqrt[a + a*Sin[c + d*x]])/(3*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 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

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 2687

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Simp[(b*(g*
Cos[e + f*x])^(p + 1))/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(a*(2*p + 1))/(2*g^2*(p + 1)), Int[
(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[p, -1] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx &=\frac {\sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}+\frac {1}{6} (5 a) \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=\frac {5 a \sec (c+d x)}{6 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}+\frac {1}{4} \left (5 a^2\right ) \int \frac {1}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {5 a^2 \cos (c+d x)}{8 d (a+a \sin (c+d x))^{3/2}}+\frac {5 a \sec (c+d x)}{6 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}+\frac {1}{16} (5 a) \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {5 a^2 \cos (c+d x)}{8 d (a+a \sin (c+d x))^{3/2}}+\frac {5 a \sec (c+d x)}{6 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}-\frac {(5 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 )}{8 d}\\ &=-\frac {5 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{8 \sqrt {2} d}-\frac {5 a^2 \cos (c+d x)}{8 d (a+a \sin (c+d x))^{3/2}}+\frac {5 a \sec (c+d x)}{6 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}\\ \end {align*}

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Mathematica [C]  time = 0.42, size = 302, normalized size = 2.20 \[ \frac {\sqrt {a (\sin (c+d x)+1)} \left (\frac {12 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {4 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {3 \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )}+\frac {6 \sin \left (\frac {d x}{2}\right )}{\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )}+(-15-15 i) (-1)^{3/4} \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2 \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \sec \left (\frac {d x}{4}\right ) \left (\cos \left (\frac {1}{4} (2 c+d x)\right )-\sin \left (\frac {1}{4} (2 c+d x)\right )\right )\right )\right )}{24 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(((6*Sin[(d*x)/2])/(Cos[c/2] + Sin[c/2]) - (3*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))/(Co
s[c/2] + Sin[c/2]) - (15 + 15*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])^2 + (4*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(Co
s[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (12*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2]))*Sqrt[a*(1 + Sin[c + d*x])])/(24*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3)

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fricas [A]  time = 0.69, size = 188, normalized size = 1.37 \[ \frac {15 \, \sqrt {2} \sqrt {a} \cos \left (d x + c\right )^{3} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\sqrt {2} \cos \left (d x + c\right ) - \sqrt {2} \sin \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {a} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (15 \, \cos \left (d x + c\right )^{2} + 10 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{96 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/96*(15*sqrt(2)*sqrt(a)*cos(d*x + c)^3*log(-(a*cos(d*x + c)^2 - 2*sqrt(a*sin(d*x + c) + a)*(sqrt(2)*cos(d*x +
c) - sqrt(2)*sin(d*x + c) + sqrt(2))*sqrt(a) + 3*a*cos(d*x + c) - (a*cos(d*x + c) - 2*a)*sin(d*x + c) + 2*a)/
(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)) + 4*(15*cos(d*x + c)^2 + 10*sin(d*x + c
) - 2)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c)^3)

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giac [A]  time = 0.77, size = 178, normalized size = 1.30 \[ \frac {\sqrt {2} {\left (15 \, \log \left ({\left | \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 15 \, \log \left ({\left | \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - \frac {6 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - \frac {4 \, {\left (6 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )}}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}\right )} \sqrt {a}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/96*sqrt(2)*(15*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)) - 15*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)) - 6*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*
c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)/(sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1) - 4*(6*sgn(cos(-1/4*pi + 1/2*d*x + 1
/2*c))*sin(-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)^
3)*sqrt(a)/d

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maple [A]  time = 0.24, size = 153, normalized size = 1.12 \[ \frac {\sin \left (d x +c \right ) \left (15 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a -20 a^{\frac {5}{2}}\right )-30 a^{\frac {5}{2}} \left (\cos ^{2}\left (d x +c \right )\right )+15 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a +4 a^{\frac {5}{2}}}{48 a^{\frac {3}{2}} \left (\sin \left (d x +c \right )-1\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x)

[Out]

1/48/a^(3/2)*(sin(d*x+c)*(15*(a-a*sin(d*x+c))^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2)
)*a-20*a^(5/2))-30*a^(5/2)*cos(d*x+c)^2+15*(a-a*sin(d*x+c))^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2
^(1/2)/a^(1/2))*a+4*a^(5/2))/(sin(d*x+c)-1)/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 \[ \int \sqrt {a \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sin(d*x + c) + a)*sec(d*x + c)^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+a\,\sin \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^(1/2)/cos(c + d*x)^4,x)

[Out]

int((a + a*sin(c + d*x))^(1/2)/cos(c + d*x)^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \sec ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sin(c + d*x) + 1))*sec(c + d*x)**4, x)

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