3.6.0 ∫ sec(c+dx) ______ (a+b sin(c+dx))3∕2 dx [519]

Integrand size = 21, antiderivative size = 105 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{(a-b)^{3/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2} d}+\frac {2 b}{\left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}} \]

-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)^(

Rubi [A] (veriﬁed)

Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2747, 724, 841, 1180, 212} \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\frac {2 b}{d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{d (a-b)^{3/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{d (a+b)^{3/2}} \]

-(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 +

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]

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,

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]

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^

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

Rubi steps \begin{align*} \text {integral}& = \frac {b \text {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 \text {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) \text {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 {\text {Subst}\left (\int \frac {1}{a-b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{(a-b) d}+\frac {\text {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{(a+b) d} \\ & = -\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{(a-b)^{3/2} d}+\frac {\text {arctanh}\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] (veriﬁed)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.87 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\frac {(a+b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sin (c+d x)}{a-b}\right )+(-a+b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sin (c+d x)}{a+b}\right )}{(a-b) (a+b) d \sqrt {a+b \sin (c+d x)}} \]

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

Maple [A] (veriﬁed)

Time = 0.60 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90


method	result	size

default	\(\frac {\frac {2 b}{\left (a -b \right ) \left (a +b \right ) \sqrt {a +b \sin \left (d x +c \right )}}+\frac {\arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{\left (a -b \right ) \sqrt {-a +b}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{\left (a +b \right )^{\frac {3}{2}}}}{d}\)	\(94\)

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

Fricas [B] (veriﬁcation not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 437 vs. \(2 (91) = 182\).

Time = 0.59 (sec) , antiderivative size = 2301, normalized size of antiderivative = 21.91 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

[1/8*((a^3 - 2*a^2*b + a*b^2 + (a^2*b - 2*a*b^2 + b^3)*sin(d*x + c))*sqrt(a + b)*log((b^4*cos(d*x + c)^4 + 128

*a^4 + 256*a^3*b + 320*a^2*b^2 + 256*a*b^3 + 72*b^4 - 8*(20*a^2*b^2 + 28*a*b^3 + 9*b^4)*cos(d*x + c)^2 + 8*(16

*a^3 + 24*a^2*b + 20*a*b^2 + 8*b^3 - (10*a*b^2 + 7*b^3)*cos(d*x + c)^2 - (b^3*cos(d*x + c)^2 - 24*a^2*b - 28*a

*b^2 - 8*b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(a + b) + 4*(64*a^3*b + 112*a^2*b^2 + 64*a*b^3 + 14*b

^4 - (8*a*b^3 + 7*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 -

2)*sin(d*x + c) + 8)) - (a^3 + 2*a^2*b + a*b^2 + (a^2*b + 2*a*b^2 + b^3)*sin(d*x + c))*sqrt(a - b)*log((b^4*co

s(d*x + c)^4 + 128*a^4 - 256*a^3*b + 320*a^2*b^2 - 256*a*b^3 + 72*b^4 - 8*(20*a^2*b^2 - 28*a*b^3 + 9*b^4)*cos(

d*x + c)^2 + 8*(16*a^3 - 24*a^2*b + 20*a*b^2 - 8*b^3 - (10*a*b^2 - 7*b^3)*cos(d*x + c)^2 - (b^3*cos(d*x + c)^2

- 24*a^2*b + 28*a*b^2 - 8*b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(a - b) + 4*(64*a^3*b - 112*a^2*b^2

+ 64*a*b^3 - 14*b^4 - (8*a*b^3 - 7*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*

(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) + 16*(a^2*b - b^3)*sqrt(b*sin(d*x + c) + a))/((a^4*b - 2*a^2*b^3 + b^5

)*d*sin(d*x + c) + (a^5 - 2*a^3*b^2 + a*b^4)*d), -1/8*(2*(a^3 - 2*a^2*b + a*b^2 + (a^2*b - 2*a*b^2 + b^3)*sin(

d*x + c))*sqrt(-a - b)*arctan(-1/4*(b^2*cos(d*x + c)^2 - 8*a^2 - 8*a*b - 2*b^2 - 2*(4*a*b + 3*b^2)*sin(d*x + c

))*sqrt(b*sin(d*x + c) + a)*sqrt(-a - b)/(2*a^3 + 3*a^2*b + 2*a*b^2 + b^3 - (a*b^2 + b^3)*cos(d*x + c)^2 + (3*

a^2*b + 4*a*b^2 + b^3)*sin(d*x + c))) + (a^3 + 2*a^2*b + a*b^2 + (a^2*b + 2*a*b^2 + b^3)*sin(d*x + c))*sqrt(a

- b)*log((b^4*cos(d*x + c)^4 + 128*a^4 - 256*a^3*b + 320*a^2*b^2 - 256*a*b^3 + 72*b^4 - 8*(20*a^2*b^2 - 28*a*b

^3 + 9*b^4)*cos(d*x + c)^2 + 8*(16*a^3 - 24*a^2*b + 20*a*b^2 - 8*b^3 - (10*a*b^2 - 7*b^3)*cos(d*x + c)^2 - (b^

3*cos(d*x + c)^2 - 24*a^2*b + 28*a*b^2 - 8*b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(a - b) + 4*(64*a^3

*b - 112*a^2*b^2 + 64*a*b^3 - 14*b^4 - (8*a*b^3 - 7*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(cos(d*x + c)^4 - 8*cos

(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) - 16*(a^2*b - b^3)*sqrt(b*sin(d*x + c) + a))/((a^4*b -

2*a^2*b^3 + b^5)*d*sin(d*x + c) + (a^5 - 2*a^3*b^2 + a*b^4)*d), -1/8*(2*(a^3 + 2*a^2*b + a*b^2 + (a^2*b + 2*a

*b^2 + b^3)*sin(d*x + c))*sqrt(-a + b)*arctan(1/4*(b^2*cos(d*x + c)^2 - 8*a^2 + 8*a*b - 2*b^2 - 2*(4*a*b - 3*b

^2)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(-a + b)/(2*a^3 - 3*a^2*b + 2*a*b^2 - b^3 - (a*b^2 - b^3)*cos(d

*x + c)^2 + (3*a^2*b - 4*a*b^2 + b^3)*sin(d*x + c))) - (a^3 - 2*a^2*b + a*b^2 + (a^2*b - 2*a*b^2 + b^3)*sin(d*

x + c))*sqrt(a + b)*log((b^4*cos(d*x + c)^4 + 128*a^4 + 256*a^3*b + 320*a^2*b^2 + 256*a*b^3 + 72*b^4 - 8*(20*a

^2*b^2 + 28*a*b^3 + 9*b^4)*cos(d*x + c)^2 + 8*(16*a^3 + 24*a^2*b + 20*a*b^2 + 8*b^3 - (10*a*b^2 + 7*b^3)*cos(d

*x + c)^2 - (b^3*cos(d*x + c)^2 - 24*a^2*b - 28*a*b^2 - 8*b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(a +

b) + 4*(64*a^3*b + 112*a^2*b^2 + 64*a*b^3 + 14*b^4 - (8*a*b^3 + 7*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(cos(d*x

+ c)^4 - 8*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) - 16*(a^2*b - b^3)*sqrt(b*sin(d*x + c)

+ a))/((a^4*b - 2*a^2*b^3 + b^5)*d*sin(d*x + c) + (a^5 - 2*a^3*b^2 + a*b^4)*d), -1/4*((a^3 + 2*a^2*b + a*b^2 +

(a^2*b + 2*a*b^2 + b^3)*sin(d*x + c))*sqrt(-a + b)*arctan(1/4*(b^2*cos(d*x + c)^2 - 8*a^2 + 8*a*b - 2*b^2 - 2

*(4*a*b - 3*b^2)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(-a + b)/(2*a^3 - 3*a^2*b + 2*a*b^2 - b^3 - (a*b^2

- b^3)*cos(d*x + c)^2 + (3*a^2*b - 4*a*b^2 + b^3)*sin(d*x + c))) + (a^3 - 2*a^2*b + a*b^2 + (a^2*b - 2*a*b^2

+ b^3)*sin(d*x + c))*sqrt(-a - b)*arctan(-1/4*(b^2*cos(d*x + c)^2 - 8*a^2 - 8*a*b - 2*b^2 - 2*(4*a*b + 3*b^2)*

sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(-a - b)/(2*a^3 + 3*a^2*b + 2*a*b^2 + b^3 - (a*b^2 + b^3)*cos(d*x +

c)^2 + (3*a^2*b + 4*a*b^2 + b^3)*sin(d*x + c))) - 8*(a^2*b - b^3)*sqrt(b*sin(d*x + c) + a))/((a^4*b - 2*a^2*b

Sympy [F]

\[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\text {Exception raised: ValueError} \]

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

Giac [F]

\[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\int \frac {1}{\cos \left (c+d\,x\right )\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

\(\int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx\) [519]

Optimal result