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*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]])

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Rubi [A]  time = 0.133365, antiderivative size = 116, normalized size of antiderivative = 1., 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 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]

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 \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*}

Mathematica [C]  time = 0.0654945, 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]

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

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Maple [A]  time = 0.185, size = 107, normalized size = 0.9 \begin{align*} 2\,{\frac{{a}^{3}}{d} \left ( -1/4\,{\frac{1}{{a}^{3}} \left ( 1/4\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }}{a\sin \left ( dx+c \right ) -a}}-5/8\,{\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}^{3}\sqrt{a+a\sin \left ( dx+c \right ) }}}-1/12\,{\frac{1}{{a}^{2} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{3/2}}} \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^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)))-1/4/a^3/(a+a*sin(d*x+c))^(1/2)-1/12/a^2/(a+a*sin(d*x+c))^(3/2))/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 = 2.27795, size = 396, normalized size = 3.41 \begin{align*} \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 )}} \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]

1/96*(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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{3}{\left (c + d x \right )}}{\sqrt{a \left (\sin{\left (c + d x \right )} + 1\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(sec(c + d*x)**3/sqrt(a*(sin(c + d*x) + 1)), x)

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Giac [A]  time = 1.10862, size = 143, normalized size = 1.23 \begin{align*} -\frac{a^{3}{\left (\frac{15 \, \sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{a \sin \left (d x + c\right ) + a}}{2 \, \sqrt{-a}}\right )}{\sqrt{-a} a^{3}} + \frac{8 \,{\left (3 \, a \sin \left (d x + c\right ) + 4 \, a\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a^{3}} + \frac{6 \, \sqrt{a \sin \left (d x + c\right ) + a}}{{\left (a \sin \left (d x + c\right ) - a\right )} a^{3}}\right )}}{48 \, d} \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]

-1/48*a^3*(15*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(a*sin(d*x + c) + a)/sqrt(-a))/(sqrt(-a)*a^3) + 8*(3*a*sin(d*x +
c) + 4*a)/((a*sin(d*x + c) + a)^(3/2)*a^3) + 6*sqrt(a*sin(d*x + c) + a)/((a*sin(d*x + c) - a)*a^3))/d