Integrand size = 23, antiderivative size = 186 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {(2 a-5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{4 (a-b)^{5/2} d}+\frac {(2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{4 (a+b)^{5/2} d}-\frac {b \left (a^2+5 b^2\right )}{2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}} \]
-1/4*(2*a-5*b)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/(a-b)^(5/2)/d+1 /4*(2*a+5*b)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)/d-1/2 *b*(a^2+5*b^2)/(a^2-b^2)^2/d/(a+b*sin(d*x+c))^(1/2)-1/2*sec(d*x+c)^2*(b-a* sin(d*x+c))/(a^2-b^2)/d/(a+b*sin(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.75 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.19 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\frac {\frac {3 a \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}-\frac {3 a \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+\frac {\left (a^2+5 b^2\right ) \left ((a+b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sin (c+d x)}{a-b}\right )+(-a+b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sin (c+d x)}{a+b}\right )\right )}{(a-b) (a+b) \sqrt {a+b \sin (c+d x)}}+\frac {2 \sec ^2(c+d x) (b-a \sin (c+d x))}{\sqrt {a+b \sin (c+d x)}}}{4 \left (-a^2+b^2\right ) d} \]
((3*a*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/Sqrt[a - b] - (3*a*Ar cTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/Sqrt[a + b] + ((a^2 + 5*b^2)* ((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)*Sqrt[a + b*Sin[c + d*x]]) + (2*Sec[c + d*x]^2*(b - a*Sin[c + d*x]))/Sqrt[a + b*Sin[c + d*x]])/(4*(-a^2 + b^2)*d)
Time = 0.49 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.32, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 3147, 496, 27, 655, 25, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^3 (a+b \sin (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {b^3 \int \frac {1}{(a+b \sin (c+d x))^{3/2} \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 496 |
| \(\displaystyle \frac {b^3 \left (\frac {\int \frac {2 a^2+3 b \sin (c+d x) a-5 b^2}{2 (a+b \sin (c+d x))^{3/2} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) \sqrt {a+b \sin (c+d x)}}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^3 \left (\frac {\int \frac {2 a^2+3 b \sin (c+d x) a-5 b^2}{(a+b \sin (c+d x))^{3/2} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{4 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) \sqrt {a+b \sin (c+d x)}}\right )}{d}\) |
| \(\Big \downarrow \) 655 |
| \(\displaystyle \frac {b^3 \left (\frac {-\frac {\int -\frac {2 a \left (a^2-4 b^2\right )+b \left (a^2+5 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{a^2-b^2}-\frac {2 \left (a^2+5 b^2\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{4 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) \sqrt {a+b \sin (c+d x)}}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^3 \left (\frac {\frac {\int \frac {2 a \left (a^2-4 b^2\right )+b \left (a^2+5 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{a^2-b^2}-\frac {2 \left (a^2+5 b^2\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{4 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) \sqrt {a+b \sin (c+d x)}}\right )}{d}\) |
| \(\Big \downarrow \) 654 |
| \(\displaystyle \frac {b^3 \left (\frac {\frac {2 \int -\frac {b^2 \left (a^2+5 b^2\right ) \sin ^2(c+d x)+a \left (a^2-13 b^2\right )}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{a^2-b^2}-\frac {2 \left (a^2+5 b^2\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{4 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) \sqrt {a+b \sin (c+d x)}}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^3 \left (\frac {-\frac {2 \int \frac {b^2 \left (a^2+5 b^2\right ) \sin ^2(c+d x)+a \left (a^2-13 b^2\right )}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{a^2-b^2}-\frac {2 \left (a^2+5 b^2\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{4 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) \sqrt {a+b \sin (c+d x)}}\right )}{d}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {b^3 \left (\frac {\frac {2 \left (\frac {(2 a-5 b) (a+b)^2 \int \frac {1}{b^2 \sin ^2(c+d x)-a+b}d\sqrt {a+b \sin (c+d x)}}{2 b}-\frac {(a-b)^2 (2 a+5 b) \int \frac {1}{b^2 \sin ^2(c+d x)-a-b}d\sqrt {a+b \sin (c+d x)}}{2 b}\right )}{a^2-b^2}-\frac {2 \left (a^2+5 b^2\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{4 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) \sqrt {a+b \sin (c+d x)}}\right )}{d}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {b^3 \left (\frac {\frac {2 \left (\frac {(a-b)^2 (2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{2 b \sqrt {a+b}}-\frac {(2 a-5 b) (a+b)^2 \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{2 b \sqrt {a-b}}\right )}{a^2-b^2}-\frac {2 \left (a^2+5 b^2\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{4 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) \sqrt {a+b \sin (c+d x)}}\right )}{d}\) |
(b^3*(-1/2*(b^2 - a*b*Sin[c + d*x])/(b^2*(a^2 - b^2)*Sqrt[a + b*Sin[c + d* x]]*(b^2 - b^2*Sin[c + d*x]^2)) + ((2*(-1/2*((2*a - 5*b)*(a + b)^2*ArcTanh [Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(Sqrt[a - b]*b) + ((a - b)^2*(2*a + 5*b)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(2*b*Sqrt[a + b])))/ (a^2 - b^2) - (2*(a^2 + 5*b^2))/((a^2 - b^2)*Sqrt[a + b*Sin[c + d*x]]))/(4 *b^2*(a^2 - b^2))))/d
3.6.20.3.1 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_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*f - d*g)*((d + e*x)^(m + 1)/((m + 1)*(c*d^2 + a*e^2)) ), x] + Simp[1/(c*d^2 + a*e^2) Int[(d + e*x)^(m + 1)*(Simp[c*d*f + a*e*g - c*(e*f - d*g)*x, x]/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && FractionQ[m] && LtQ[m, -1]
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 = 0.85 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.25
| method | result | size |
| default | \(\frac {-\frac {2 b^{3}}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a +b \sin \left (d x +c \right )}}-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{4 \left (a -b \right )^{2} \left (b \sin \left (d x +c \right )+b \right )}+\frac {\arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a}{2 \left (a -b \right )^{2} \sqrt {-a +b}}-\frac {5 b \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{4 \left (a -b \right )^{2} \sqrt {-a +b}}-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{4 \left (a +b \right )^{2} \left (b \sin \left (d x +c \right )-b \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a}{2 \left (a +b \right )^{\frac {5}{2}}}+\frac {5 b \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{4 \left (a +b \right )^{\frac {5}{2}}}}{d}\) | \(233\) |
(-2*b^3/(a-b)^2/(a+b)^2/(a+b*sin(d*x+c))^(1/2)-1/4*b/(a-b)^2*(a+b*sin(d*x+ c))^(1/2)/(b*sin(d*x+c)+b)+1/2/(a-b)^2/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c) )^(1/2)/(-a+b)^(1/2))*a-5/4*b/(a-b)^2/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c)) ^(1/2)/(-a+b)^(1/2))-1/4*b/(a+b)^2*(a+b*sin(d*x+c))^(1/2)/(b*sin(d*x+c)-b) +1/2/(a+b)^(5/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a+5/4*b/(a+b) ^(5/2)*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. 641 vs. \(2 (163) = 326\).
Time = 0.77 (sec) , antiderivative size = 3117, normalized size of antiderivative = 16.76 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
[1/32*(((2*a^4*b - a^3*b^2 - 9*a^2*b^3 + 13*a*b^4 - 5*b^5)*cos(d*x + c)^2* sin(d*x + c) + (2*a^5 - a^4*b - 9*a^3*b^2 + 13*a^2*b^3 - 5*a*b^4)*cos(d*x + c)^2)*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)) - ((2*a ^4*b + a^3*b^2 - 9*a^2*b^3 - 13*a*b^4 - 5*b^5)*cos(d*x + c)^2*sin(d*x + c) + (2*a^5 + a^4*b - 9*a^3*b^2 - 13*a^2*b^3 - 5*a*b^4)*cos(d*x + c)^2)*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^4*b - 2*a^2 *b^3 + b^5 + (a^4*b + 4*a^2*b^3 - 5*b^5)*cos(d*x + c)^2 - (a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a))/((a^6*b - 3*a^4*b^3 +...
\[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{3}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]