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

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Rubi [A]  time = 0.16, antiderivative size = 105, normalized size of antiderivative = 1.00, 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 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 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 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]

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*}

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Mathematica [C]  time = 0.09, 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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fricas [F]  time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {b \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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]

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

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maple [A]  time = 0.56, size = 99, normalized size = 0.94 \[ \frac {2 b}{d \left (a -b \right ) \left (a +b \right ) \sqrt {a +b \sin \left (d x +c \right )}}+\frac {\arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{d \left (a -b \right ) \sqrt {-a +b}}+\frac {\arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{\left (a +b \right )^{\frac {3}{2}} d} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a-4*b>0)', see `assume?` for
more details)Is 4*a-4*b positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\cos \left (c+d\,x\right )\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec {\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

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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