Integrand size = 23, antiderivative size = 130 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=-\frac {\sqrt {a-b} (2 a+b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{4 d}+\frac {(2 a-b) \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{4 d}+\frac {\sec ^2(c+d x) (b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{2 d} \]
-1/4*(2*a+b)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))*(a-b)^(1/2)/d+1/4 *(2*a-b)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*(a+b)^(1/2)/d+1/2*sec (d*x+c)^2*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/d
Time = 0.44 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.93 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=\frac {-\sqrt {a-b} (2 a+b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )+(2 a-b) \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )+2 \sec ^2(c+d x) (b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 d} \]
(-(Sqrt[a - b]*(2*a + b)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]]) + (2*a - b)*Sqrt[a + b]*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]] + 2*Se c[c + d*x]^2*(b + a*Sin[c + d*x])*Sqrt[a + b*Sin[c + d*x]])/(4*d)
Time = 0.38 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.22, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3147, 495, 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.
\(\displaystyle \int \sec ^3(c+d x) (a+b \sin (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \sin (c+d x))^{3/2}}{\cos (c+d x)^3}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {b^3 \int \frac {(a+b \sin (c+d x))^{3/2}}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 495 |
\(\displaystyle \frac {b^3 \left (\frac {\sqrt {a+b \sin (c+d x)} \left (a b \sin (c+d x)+b^2\right )}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int -\frac {2 a^2+b \sin (c+d x) a-b^2}{2 \sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}\right )}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b^3 \left (\frac {\int \frac {2 a^2+b \sin (c+d x) a-b^2}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{4 b^2}+\frac {\sqrt {a+b \sin (c+d x)} \left (a b \sin (c+d x)+b^2\right )}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 654 |
\(\displaystyle \frac {b^3 \left (\frac {\int -\frac {a^2+b^2 \sin ^2(c+d x) a-b^2}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{2 b^2}+\frac {\sqrt {a+b \sin (c+d x)} \left (a b \sin (c+d x)+b^2\right )}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b^3 \left (\frac {\sqrt {a+b \sin (c+d x)} \left (a b \sin (c+d x)+b^2\right )}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int \frac {a^2+b^2 \sin ^2(c+d x) a-b^2}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{2 b^2}\right )}{d}\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle \frac {b^3 \left (\frac {\frac {(a-b) (2 a+b) \int \frac {1}{b^2 \sin ^2(c+d x)-a+b}d\sqrt {a+b \sin (c+d x)}}{2 b}-\frac {(2 a-b) (a+b) \int \frac {1}{b^2 \sin ^2(c+d x)-a-b}d\sqrt {a+b \sin (c+d x)}}{2 b}}{2 b^2}+\frac {\sqrt {a+b \sin (c+d x)} \left (a b \sin (c+d x)+b^2\right )}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {b^3 \left (\frac {\frac {(2 a-b) \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{2 b}-\frac {\sqrt {a-b} (2 a+b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{2 b}}{2 b^2}+\frac {\sqrt {a+b \sin (c+d x)} \left (a b \sin (c+d x)+b^2\right )}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
(b^3*((-1/2*(Sqrt[a - b]*(2*a + b)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/b + ((2*a - b)*Sqrt[a + b]*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(2*b))/(2*b^2) + (Sqrt[a + b*Sin[c + d*x]]*(b^2 + a*b*Sin[c + d*x] ))/(2*b^2*(b^2 - b^2*Sin[c + d*x]^2))))/d
3.5.87.3.1 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*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(c + d*x)^(n - 2)*(a + b*x^2)^(p + 1)*Simp[a* d^2*(n - 1) - b*c^2*(2*p + 3) - b*c*d*(n + 2*p + 2)*x, x], x], x] /; FreeQ[ {a, b, c, d}, x] && LtQ[p, -1] && GtQ[n, 1] && IntQuadraticQ[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 = 1.34 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.18
method | result | size |
default | \(\frac {2 b^{3} \left (\frac {\left (a -b \right ) \left (-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{2 \left (b \sin \left (d x +c \right )+b \right )}+\frac {\left (2 a +b \right ) \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{2 \sqrt {-a +b}}\right )}{4 b^{3}}-\frac {\left (a +b \right ) \left (\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{2 b \sin \left (d x +c \right )-2 b}-\frac {\left (2 a -b \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{2 \sqrt {a +b}}\right )}{4 b^{3}}\right )}{d}\) | \(154\) |
2*b^3*(1/4*(a-b)/b^3*(-1/2*b*(a+b*sin(d*x+c))^(1/2)/(b*sin(d*x+c)+b)+1/2*( 2*a+b)/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2)))-1/4*(a+b) /b^3*(1/2*b*(a+b*sin(d*x+c))^(1/2)/(b*sin(d*x+c)-b)-1/2*(2*a-b)/(a+b)^(1/2 )*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. 354 vs. \(2 (110) = 220\).
Time = 0.58 (sec) , antiderivative size = 1969, normalized size of antiderivative = 15.15 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=\text {Too large to display} \]
[-1/32*((2*a - b)*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)*s in(d*x + c))/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2)*s in(d*x + c) + 8)) - (2*a + b)*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)) - 16*(a*sin(d*x + c) + b)*sqrt(b*sin(d*x + c ) + a))/(d*cos(d*x + c)^2), -1/32*(2*(2*a - b)*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)))*c os(d*x + c)^2 - (2*a + b)*sqrt(a - b)*cos(d*x + c)^2*log((b^4*cos(d*x +...
Timed out. \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a-4*b>0)', see `assume?` for m ore detail
Timed out. \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \]