Integrand size = 23, antiderivative size = 144 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=-\frac {(2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{4 (a-b)^{3/2} d}+\frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{4 (a+b)^{3/2} d}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{2 \left (a^2-b^2\right ) d} \]
-1/4*(2*a-3*b)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/(a-b)^(3/2)/d+1 /4*(2*a+3*b)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))/(a+b)^(3/2)/d-1/2 *sec(d*x+c)^2*(b-a*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/(a^2-b^2)/d
Time = 0.35 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.22 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {\sqrt {a+b} \left (2 a^2-a b-3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )-\sqrt {a-b} \left (\left (2 a^2+a b-3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )+2 \sqrt {a+b} \sec ^2(c+d x) (-b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}\right )}{4 \sqrt {a-b} \sqrt {a+b} \left (-a^2+b^2\right ) d} \]
(Sqrt[a + b]*(2*a^2 - a*b - 3*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]] - Sqrt[a - b]*((2*a^2 + a*b - 3*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x ]]/Sqrt[a + b]] + 2*Sqrt[a + b]*Sec[c + d*x]^2*(-b + a*Sin[c + d*x])*Sqrt[ a + b*Sin[c + d*x]]))/(4*Sqrt[a - b]*Sqrt[a + b]*(-a^2 + b^2)*d)
Time = 0.40 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.33, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3147, 496, 27, 654, 25, 1480, 220}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sec ^3(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (c+d x)^3 \sqrt {a+b \sin (c+d x)}}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {b^3 \int \frac {1}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 496 |
\(\displaystyle \frac {b^3 \left (\frac {\int \frac {2 a^2+b \sin (c+d x) a-3 b^2}{2 \sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2-a b \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b^3 \left (\frac {\int \frac {2 a^2+b \sin (c+d x) a-3 b^2}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{4 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2-a b \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 654 |
\(\displaystyle \frac {b^3 \left (\frac {\int -\frac {a^2+b^2 \sin ^2(c+d x) a-3 b^2}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{2 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2-a b \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b^3 \left (-\frac {\int \frac {a^2+b^2 \sin ^2(c+d x) a-3 b^2}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{2 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2-a b \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle \frac {b^3 \left (\frac {-\frac {1}{2} \left (-\frac {2 a^2}{b}+a+3 b\right ) \int \frac {1}{b^2 \sin ^2(c+d x)-a+b}d\sqrt {a+b \sin (c+d x)}-\frac {(a-b) (2 a+3 b) \int \frac {1}{b^2 \sin ^2(c+d x)-a-b}d\sqrt {a+b \sin (c+d x)}}{2 b}}{2 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2-a b \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {b^3 \left (\frac {\frac {\left (-\frac {2 a^2}{b}+a+3 b\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{2 \sqrt {a-b}}+\frac {(a-b) (2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{2 b \sqrt {a+b}}}{2 b^2 \left (a^2-b^2\right )}-\frac {\sqrt {a+b \sin (c+d x)} \left (b^2-a b \sin (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
(b^3*((((a - (2*a^2)/b + 3*b)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b] ])/(2*Sqrt[a - b]) + ((a - b)*(2*a + 3*b)*ArcTanh[Sqrt[a + b*Sin[c + d*x]] /Sqrt[a + b]])/(2*b*Sqrt[a + b]))/(2*b^2*(a^2 - b^2)) - (Sqrt[a + b*Sin[c + d*x]]*(b^2 - a*b*Sin[c + d*x]))/(2*b^2*(a^2 - b^2)*(b^2 - b^2*Sin[c + d* x]^2))))/d
3.6.10.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 *p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad raticQ[a, 0, b, c, d, n, p, x]
Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Simp[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]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(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]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[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] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.82 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.16
method | result | size |
default | \(\frac {2 b^{3} \left (\frac {-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{2 \left (a -b \right ) \left (b \sin \left (d x +c \right )+b \right )}+\frac {\left (2 a -3 b \right ) \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{2 \left (a -b \right ) \sqrt {-a +b}}}{4 b^{3}}-\frac {\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{2 \left (a +b \right ) \left (b \sin \left (d x +c \right )-b \right )}-\frac {\left (2 a +3 b \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}}{4 b^{3}}\right )}{d}\) | \(167\) |
2*b^3*(1/4/b^3*(-1/2*b/(a-b)*(a+b*sin(d*x+c))^(1/2)/(b*sin(d*x+c)+b)+1/2*( 2*a-3*b)/(a-b)/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2)))-1 /4/b^3*(1/2*b/(a+b)*(a+b*sin(d*x+c))^(1/2)/(b*sin(d*x+c)-b)-1/2*(2*a+3*b)/ (a+b)^(3/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))))/d
Leaf count of result is larger than twice the leaf count of optimal. 423 vs. \(2 (125) = 250\).
Time = 0.63 (sec) , antiderivative size = 2245, normalized size of antiderivative = 15.59 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\text {Too large to display} \]
[1/32*((2*a^3 - a^2*b - 4*a*b^2 + 3*b^3)*sqrt(a + b)*cos(d*x + c)^2*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)) + (2*a^3 + a^2*b - 4*a*b^2 - 3*b^ 3)*sqrt(a - b)*cos(d*x + c)^2*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)*c os(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))/(c os(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 - (a^3 - a*b^2)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a ))/((a^4 - 2*a^2*b^2 + b^4)*d*cos(d*x + c)^2), -1/32*(2*(2*a^3 - a^2*b - 4 *a*b^2 + 3*b^3)*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)^...
\[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\sqrt {a + b \sin {\left (c + d x \right )}}}\, dx \]
Exception generated. \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a-4*b>0)', see `assume?` for m ore detail
Time = 0.42 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.47 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {b^{3} {\left (\frac {{\left (2 \, a - 3 \, b\right )} \arctan \left (\frac {\sqrt {b \sin \left (d x + c\right ) + a}}{\sqrt {-a + b}}\right )}{{\left (a b^{3} - b^{4}\right )} \sqrt {-a + b}} - \frac {{\left (2 \, a + 3 \, b\right )} \arctan \left (\frac {\sqrt {b \sin \left (d x + c\right ) + a}}{\sqrt {-a - b}}\right )}{{\left (a b^{3} + b^{4}\right )} \sqrt {-a - b}} - \frac {2 \, {\left ({\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a - \sqrt {b \sin \left (d x + c\right ) + a} a^{2} - \sqrt {b \sin \left (d x + c\right ) + a} b^{2}\right )}}{{\left (a^{2} b^{2} - b^{4}\right )} {\left ({\left (b \sin \left (d x + c\right ) + a\right )}^{2} - 2 \, {\left (b \sin \left (d x + c\right ) + a\right )} a + a^{2} - b^{2}\right )}}\right )}}{4 \, d} \]
1/4*b^3*((2*a - 3*b)*arctan(sqrt(b*sin(d*x + c) + a)/sqrt(-a + b))/((a*b^3 - b^4)*sqrt(-a + b)) - (2*a + 3*b)*arctan(sqrt(b*sin(d*x + c) + a)/sqrt(- a - b))/((a*b^3 + b^4)*sqrt(-a - b)) - 2*((b*sin(d*x + c) + a)^(3/2)*a - s qrt(b*sin(d*x + c) + a)*a^2 - sqrt(b*sin(d*x + c) + a)*b^2)/((a^2*b^2 - b^ 4)*((b*sin(d*x + c) + a)^2 - 2*(b*sin(d*x + c) + a)*a + a^2 - b^2)))/d
Timed out. \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^3\,\sqrt {a+b\,\sin \left (c+d\,x\right )}} \,d x \]