\(\int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [566]

Optimal result

Integrand size = 21, antiderivative size = 236 \[ \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b \sqrt {a+b} d}+\frac {2 \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b \sqrt {a+b} d}-\frac {2 b \tan (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \]

[Out]

-2*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)
*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b/d/(a+b)^(1/2)+2*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a
+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b/d/(a+b)^(1/2)-2*b*tan(d*x+c
)/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3918, 21, 3914, 3917, 4089} \[ \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 b \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {2 \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d \sqrt {a+b}}-\frac {2 \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d \sqrt {a+b}} \]

[In]

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

[Out]

(-2*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c +
d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*Sqrt[a + b]*d) + (2*Cot[c + d*x]*EllipticF[ArcSin
[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + S
ec[c + d*x]))/(a - b))])/(b*Sqrt[a + b]*d) - (2*b*Tan[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3914

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[a - b, Int[Csc[e + f
*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[b, Int[Csc[e + f*x]*((1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]),
x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3917

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(Rt[a + b, 2]/(b*
f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin
[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3918

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-b)*Cot[e + f*x]*(
(a + b*Csc[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1)*(a*(m + 1) - b*(m + 2)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 -
b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 4089

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 b \tan (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 \int \frac {\sec (c+d x) \left (-\frac {a}{2}-\frac {1}{2} b \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{a^2-b^2} \\ & = -\frac {2 b \tan (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {\int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \, dx}{a^2-b^2} \\ & = -\frac {2 b \tan (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {\int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{a+b}+\frac {b \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{a^2-b^2} \\ & = -\frac {2 \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b \sqrt {a+b} d}+\frac {2 \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b \sqrt {a+b} d}-\frac {2 b \tan (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.38 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.03 \[ \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (4 (a+b) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}}-4 (a+b) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}}+(a-b) \left (\sin \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{\left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \]

[In]

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

[Out]

-((Sec[(c + d*x)/2]*Sec[c + d*x]*(4*(a + b)*Cos[(c + d*x)/2]^3*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c +
d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)] - 4*(a + b)*Cos[(c
+ d*x)/2]^3*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a -
b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)] + (a - b)*(Sin[(c + d*x)/2] - Sin[(3*(c + d*x))/2])))/((a^2 - b^2)*d
*Sqrt[a + b*Sec[c + d*x]]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1081\) vs. \(2(216)=432\).

Time = 3.96 (sec) , antiderivative size = 1082, normalized size of antiderivative = 4.58

method	result	size
default	\(\text {Expression too large to display}\)	\(1082\)

[In]

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

[Out]

2/d/(a-b)/(a+b)*(EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+
b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*cos(d*x+c)^2+EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*
(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b*cos(d*x+c)^2-EllipticF(cot
(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c
)+1))^(1/2)*a*cos(d*x+c)^2-EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1
/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b*cos(d*x+c)^2+2*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a
+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*cos(d*x+c)+2*E
llipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c)
)/(cos(d*x+c)+1))^(1/2)*b*cos(d*x+c)-2*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d
*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*cos(d*x+c)-2*EllipticF(cot(d*x+c)-csc(d*x+c)
,((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b*cos(
d*x+c)+EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*co
s(d*x+c))/(cos(d*x+c)+1))^(1/2)*a+EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)
+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b-EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/
2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a-EllipticF(cot(d*x+c)-c
sc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/
2)*b+a*sin(d*x+c)*cos(d*x+c)-cos(d*x+c)*sin(d*x+c)*b)*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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