Integrand size = 21, antiderivative size = 139 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{(a-b)^{5/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{(a+b)^{5/2} d}+\frac {2 b}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac {4 a b}{\left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}} \]
-arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/(a-b)^(5/2)/d+arctanh((a+b*si n(d*x+c))^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)/d+2/3*b/(a^2-b^2)/d/(a+b*sin(d*x+ c))^(3/2)+4*a*b/(a^2-b^2)^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 = 0.05 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.68 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\frac {(a+b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \sin (c+d x)}{a-b}\right )+(-a+b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \sin (c+d x)}{a+b}\right )}{3 (a-b) (a+b) d (a+b \sin (c+d x))^{3/2}} \]
((a + b)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Sin[c + d*x])/(a - b)] + (-a + b)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Sin[c + d*x])/(a + b)])/( 3*(a - b)*(a + b)*d*(a + b*Sin[c + d*x])^(3/2))
Time = 0.41 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.29, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3147, 482, 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.
\(\displaystyle \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (c+d x) (a+b \sin (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {b \int \frac {1}{(a+b \sin (c+d x))^{5/2} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 482 |
\(\displaystyle \frac {b \left (\frac {\int \frac {a-b \sin (c+d x)}{(a+b \sin (c+d x))^{3/2} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{a^2-b^2}+\frac {2}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}\right )}{d}\) |
\(\Big \downarrow \) 655 |
\(\displaystyle \frac {b \left (\frac {\frac {4 a}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}-\frac {\int -\frac {a^2-2 b \sin (c+d x) a+b^2}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{a^2-b^2}}{a^2-b^2}+\frac {2}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}\right )}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b \left (\frac {\frac {\int \frac {a^2-2 b \sin (c+d x) a+b^2}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{a^2-b^2}+\frac {4 a}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}+\frac {2}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}\right )}{d}\) |
\(\Big \downarrow \) 654 |
\(\displaystyle \frac {b \left (\frac {\frac {2 \int -\frac {3 a^2-2 b^2 \sin ^2(c+d x) a+b^2}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{a^2-b^2}+\frac {4 a}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}+\frac {2}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}\right )}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b \left (\frac {\frac {4 a}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}-\frac {2 \int \frac {3 a^2-2 b^2 \sin ^2(c+d x) a+b^2}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{a^2-b^2}}{a^2-b^2}+\frac {2}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}\right )}{d}\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle \frac {b \left (\frac {\frac {2 \left (\frac {(a+b)^2 \int \frac {1}{b^2 \sin ^2(c+d x)-a+b}d\sqrt {a+b \sin (c+d x)}}{2 b}-\frac {(a-b)^2 \int \frac {1}{b^2 \sin ^2(c+d x)-a-b}d\sqrt {a+b \sin (c+d x)}}{2 b}\right )}{a^2-b^2}+\frac {4 a}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}+\frac {2}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}\right )}{d}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {b \left (\frac {\frac {2 \left (\frac {(a-b)^2 \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{2 b \sqrt {a+b}}-\frac {(a+b)^2 \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{2 b \sqrt {a-b}}\right )}{a^2-b^2}+\frac {4 a}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}+\frac {2}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}\right )}{d}\) |
(b*(2/(3*(a^2 - b^2)*(a + b*Sin[c + d*x])^(3/2)) + ((2*(-1/2*((a + b)^2*Ar cTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(Sqrt[a - b]*b) + ((a - b)^2* ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(2*b*Sqrt[a + b])))/(a^2 - b^2) + (4*a)/((a^2 - b^2)*Sqrt[a + b*Sin[c + d*x]]))/(a^2 - b^2)))/d
3.6.30.3.1 Defintions of rubi rules used
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), x_Symbol] :> Simp[d*((c + d*x)^(n + 1)/((n + 1)*(b*c^2 + a*d^2))), x] + Simp[b/(b*c^2 + a*d^2) I nt[(c + d*x)^(n + 1)*((c - d*x)/(a + b*x^2)), x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[n, -1]
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 = 0.69 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.88
method | result | size |
default | \(\frac {\frac {2 b}{3 \left (a -b \right ) \left (a +b \right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {4 b a}{\left (a +b \right )^{2} \left (a -b \right )^{2} \sqrt {a +b \sin \left (d x +c \right )}}+\frac {\arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{\left (a -b \right )^{2} \sqrt {-a +b}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{\left (a +b \right )^{\frac {5}{2}}}}{d}\) | \(122\) |
(2/3*b/(a-b)/(a+b)/(a+b*sin(d*x+c))^(3/2)+4*b*a/(a+b)^2/(a-b)^2/(a+b*sin(d *x+c))^(1/2)+1/(a-b)^2/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^( 1/2))+1/(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. 669 vs. \(2 (121) = 242\).
Time = 0.68 (sec) , antiderivative size = 3225, normalized size of antiderivative = 23.20 \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
[-1/24*(3*(a^5 - 3*a^4*b + 4*a^3*b^2 - 4*a^2*b^3 + 3*a*b^4 - b^5 - (a^3*b^ 2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*cos(d*x + c)^2 + 2*(a^4*b - 3*a^3*b^2 + 3*a ^2*b^3 - a*b^4)*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)) + 3*(a^5 + 3*a^4*b + 4*a^3*b^2 + 4*a^2*b^3 + 3*a*b^4 + b^5 - (a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*cos(d*x + c)^2 + 2*(a^4*b + 3*a^3* b^2 + 3*a^2*b^3 + a*b^4)*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*(7*a^4*b - 8*a^2*b^3 + b^5 + 6*(a^3*b^2 - a*b ^4)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a))/((a^6*b^2 - 3*a^4*b^4 + 3*a...
\[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\int \frac {1}{\cos \left (c+d\,x\right )\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]