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

Optimal. Leaf size=116 \[ -\frac {5}{8 d \sqrt {a \sin (c+d x)+a}}-\frac {5 a}{12 d (a \sin (c+d x)+a)^{3/2}}+\frac {5 \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} \sqrt {a} d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}} \]

[Out]

-5/12*a/d/(a+a*sin(d*x+c))^(3/2)+5/16*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d*2^(1/2)/a^(1/2)-5/
8/d/(a+a*sin(d*x+c))^(1/2)+1/2*sec(d*x+c)^2/d/(a+a*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2687, 2667, 51, 63, 206} \[ -\frac {5}{8 d \sqrt {a \sin (c+d x)+a}}-\frac {5 a}{12 d (a \sin (c+d x)+a)^{3/2}}+\frac {5 \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} \sqrt {a} d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(8*Sqrt[2]*Sqrt[a]*d) - (5*a)/(12*d*(a + a*Sin[c + d*x
])^(3/2)) - 5/(8*d*Sqrt[a + a*Sin[c + d*x]]) + Sec[c + d*x]^2/(2*d*Sqrt[a + a*Sin[c + d*x]])

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])

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 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 \frac {\sec ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx &=\frac {\sec ^2(c+d x)}{2 d \sqrt {a+a \sin (c+d x)}}+\frac {1}{4} (5 a) \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=\frac {\sec ^2(c+d x)}{2 d \sqrt {a+a \sin (c+d x)}}+\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,a \sin (c+d x)\right )}{4 d}\\ &=-\frac {5 a}{12 d (a+a \sin (c+d x))^{3/2}}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a+a \sin (c+d x)}}+\frac {(5 a) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,a \sin (c+d x)\right )}{8 d}\\ &=-\frac {5 a}{12 d (a+a \sin (c+d x))^{3/2}}-\frac {5}{8 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a+a \sin (c+d x)}}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{16 d}\\ &=-\frac {5 a}{12 d (a+a \sin (c+d x))^{3/2}}-\frac {5}{8 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a+a \sin (c+d x)}}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{8 d}\\ &=\frac {5 \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} \sqrt {a} d}-\frac {5 a}{12 d (a+a \sin (c+d x))^{3/2}}-\frac {5}{8 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 42, normalized size = 0.36 \[ -\frac {a \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};\frac {1}{2} (\sin (c+d x)+1)\right )}{6 d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(a*Hypergeometric2F1[-3/2, 2, -1/2, (1 + Sin[c + d*x])/2])/(d*(a + a*Sin[c + d*x])^(3/2))

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fricas [A]  time = 0.59, size = 145, normalized size = 1.25 \[ \frac {15 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 4 \, {\left (15 \, \cos \left (d x + c\right )^{2} - 10 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{96 \, {\left (a d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{2}\right )}} \]

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]

1/96*(15*sqrt(2)*(cos(d*x + c)^2*sin(d*x + c) + cos(d*x + c)^2)*sqrt(a)*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*(15*cos(d*x + c)^2 - 10*sin(d*x + c) - 2)*sqrt(a*si
n(d*x + c) + a))/(a*d*cos(d*x + c)^2*sin(d*x + c) + a*d*cos(d*x + c)^2)

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giac [B]  time = 2.81, size = 587, normalized size = 5.06 \[ -\frac {\frac {15 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} - \sqrt {a}\right )}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {6 \, {\left (3 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{3} - {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} \sqrt {a} - {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )} a - a^{\frac {3}{2}}\right )}}{{\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 2 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )} \sqrt {a} - a\right )}^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {16 \, {\left (3 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{5} + 6 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} \sqrt {a} - 4 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{3} a - 12 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} a^{\frac {3}{2}} + 9 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )} a^{2} - 2 \, a^{\frac {5}{2}}\right )}}{{\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 2 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )} \sqrt {a} - a\right )}^{3} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{24 \, d} \]

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]

-1/24*(15*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) - sqr
t(a))/sqrt(-a))/(sqrt(-a)*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 6*(3*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2
*d*x + 1/2*c)^2 + a))^3 - (sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*sqrt(a) - (sqr
t(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*a - a^(3/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) -
sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))
*sqrt(a) - a)^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 16*(3*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/
2*c)^2 + a))^5 + 6*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*sqrt(a) - 4*(sqrt(a)*
tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^3*a - 12*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan
(1/2*d*x + 1/2*c)^2 + a))^2*a^(3/2) + 9*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*a^
2 - 2*a^(5/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 2*(sqrt(a)*tan(1/2*d*
x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*sqrt(a) - a)^3*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d

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maple [A]  time = 0.27, size = 107, normalized size = 0.92 \[ \frac {2 a^{3} \left (-\frac {1}{4 a^{3} \sqrt {a +a \sin \left (d x +c \right )}}-\frac {1}{12 a^{2} \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {\frac {\sqrt {a +a \sin \left (d x +c \right )}}{4 a \sin \left (d x +c \right )-4 a}-\frac {5 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 \sqrt {a}}}{4 a^{3}}\right )}{d} \]

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^3/(a+a*sin(d*x+c))^(1/2)-1/12/a^2/(a+a*sin(d*x+c))^(3/2)-1/4/a^3*(1/4*(a+a*sin(d*x+c))^(1/2)/(a*
sin(d*x+c)-a)-5/8*2^(1/2)/a^(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 [A]  time = 0.74, size = 132, normalized size = 1.14 \[ -\frac {15 \, \sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (15 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a - 20 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{2} - 8 \, a^{3}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 2 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a}}{96 \, a d} \]

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]

-1/96*(15*sqrt(2)*sqrt(a)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x
+ c) + a))) + 4*(15*(a*sin(d*x + c) + a)^2*a - 20*(a*sin(d*x + c) + a)*a^2 - 8*a^3)/((a*sin(d*x + c) + a)^(5/2
) - 2*(a*sin(d*x + c) + a)^(3/2)*a))/(a*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^3\,\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^3*(a + a*sin(c + d*x))^(1/2)),x)

[Out]

int(1/(cos(c + d*x)^3*(a + a*sin(c + d*x))^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{3}{\left (c + d x \right )}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \]

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

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