3.519 \(\int \frac{\sec (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=105 \[ \frac{2 b}{d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{d (a-b)^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{d (a+b)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.157626, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2668, 710, 827, 1166, 206} \[ \frac{2 b}{d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{d (a-b)^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{d (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

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

Rule 2668

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*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 710

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 +
a*e^2)), x] + Dist[c/(c*d^2 + a*e^2), Int[((d + e*x)^(m + 1)*(d - e*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d,
e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

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

Mathematica [C]  time = 0.0792309, size = 91, normalized size = 0.87 \[ \frac{(a+b) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{a+b \sin (c+d x)}{a-b}\right )+(b-a) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{a+b \sin (c+d x)}{a+b}\right )}{d (a-b) (a+b) \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

((a + b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Sin[c + d*x])/(a - b)] + (-a + b)*Hypergeometric2F1[-1/2, 1, 1
/2, (a + b*Sin[c + d*x])/(a + b)])/((a - b)*(a + b)*d*Sqrt[a + b*Sin[c + d*x]])

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Maple [A]  time = 0.442, size = 99, normalized size = 0.9 \begin{align*} 2\,{\frac{b}{d \left ( a+b \right ) \left ( a-b \right ) \sqrt{a+b\sin \left ( dx+c \right ) }}}+{\frac{1}{d \left ( a-b \right ) }\arctan \left ({\sqrt{a+b\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}}+{\frac{1}{d}{\it Artanh} \left ({\sqrt{a+b\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{a+b}}}} \right ) \left ( a+b \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)/(a+b*sin(d*x+c))^(3/2),x)

[Out]

2/d*b/(a+b)/(a-b)/(a+b*sin(d*x+c))^(1/2)+1/d/(a-b)/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))+ar
ctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))/(a+b)^(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)/(a+b*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{b \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+b*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(b*sin(d*x + c) + a)*sec(d*x + c)/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+b*sin(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)/(a + b*sin(c + d*x))**(3/2), x)

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Giac [A]  time = 1.08793, size = 161, normalized size = 1.53 \begin{align*} b{\left (\frac{\arctan \left (\frac{\sqrt{b \sin \left (d x + c\right ) + a}}{\sqrt{-a + b}}\right )}{{\left (a b d - b^{2} d\right )} \sqrt{-a + b}} - \frac{\arctan \left (\frac{\sqrt{b \sin \left (d x + c\right ) + a}}{\sqrt{-a - b}}\right )}{{\left (a b d + b^{2} d\right )} \sqrt{-a - b}} + \frac{2}{{\left (a^{2} d - b^{2} d\right )} \sqrt{b \sin \left (d x + c\right ) + a}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+b*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

b*(arctan(sqrt(b*sin(d*x + c) + a)/sqrt(-a + b))/((a*b*d - b^2*d)*sqrt(-a + b)) - arctan(sqrt(b*sin(d*x + c) +
a)/sqrt(-a - b))/((a*b*d + b^2*d)*sqrt(-a - b)) + 2/((a^2*d - b^2*d)*sqrt(b*sin(d*x + c) + a)))