Integrand size = 23, antiderivative size = 155 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=-\frac {(a-b)^{3/2} (2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{4 d}+\frac {(2 a-3 b) (a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{4 d}+\frac {a b \sqrt {a+b \sin (c+d x)}}{2 d}+\frac {\sec ^2(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{2 d} \]
-1/4*(a-b)^(3/2)*(2*a+3*b)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/d+1 /4*(2*a-3*b)*(a+b)^(3/2)*arctanh((a+b*sin(d*x+c))^(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))^(3/2)/d+1/2*a*b*(a+b*sin(d *x+c))^(1/2)/d
Time = 0.61 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.95 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\frac {-\sqrt {a-b} \left (2 a^2+a b-3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )+\sqrt {a+b} \left (2 a^2-a b-3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )+2 \sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (2 a b+\left (a^2+b^2\right ) \sin (c+d x)\right )}{4 d} \]
(-(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*Sec[c + d*x]^2*Sqrt[a + b*Sin[c + d*x]]*(2*a*b + (a^ 2 + b^2)*Sin[c + d*x]))/(4*d)
Time = 0.45 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.17, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 3147, 495, 27, 653, 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.
\(\displaystyle \int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \sin (c+d x))^{5/2}}{\cos (c+d x)^3}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {b^3 \int \frac {(a+b \sin (c+d x))^{5/2}}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 495 |
\(\displaystyle \frac {b^3 \left (\frac {(a+b \sin (c+d x))^{3/2} \left (a b \sin (c+d x)+b^2\right )}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int -\frac {\sqrt {a+b \sin (c+d x)} \left (2 a^2-b \sin (c+d x) a-3 b^2\right )}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}\right )}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b^3 \left (\frac {\int \frac {\sqrt {a+b \sin (c+d x)} \left (2 a^2-b \sin (c+d x) a-3 b^2\right )}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{4 b^2}+\frac {\left (a b \sin (c+d x)+b^2\right ) (a+b \sin (c+d x))^{3/2}}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 653 |
\(\displaystyle \frac {b^3 \left (\frac {2 a \sqrt {a+b \sin (c+d x)}-\int -\frac {2 a \left (a^2-2 b^2\right )+b \left (a^2-3 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{4 b^2}+\frac {\left (a b \sin (c+d x)+b^2\right ) (a+b \sin (c+d x))^{3/2}}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b^3 \left (\frac {\int \frac {2 a \left (a^2-2 b^2\right )+b \left (a^2-3 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))+2 a \sqrt {a+b \sin (c+d x)}}{4 b^2}+\frac {\left (a b \sin (c+d x)+b^2\right ) (a+b \sin (c+d x))^{3/2}}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 654 |
\(\displaystyle \frac {b^3 \left (\frac {2 \int -\frac {b^2 \left (a^2-3 b^2\right ) \sin ^2(c+d x)+a \left (a^2-b^2\right )}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}+2 a \sqrt {a+b \sin (c+d x)}}{4 b^2}+\frac {\left (a b \sin (c+d x)+b^2\right ) (a+b \sin (c+d x))^{3/2}}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b^3 \left (\frac {2 a \sqrt {a+b \sin (c+d x)}-2 \int \frac {b^2 \left (a^2-3 b^2\right ) \sin ^2(c+d x)+a \left (a^2-b^2\right )}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{4 b^2}+\frac {\left (a b \sin (c+d x)+b^2\right ) (a+b \sin (c+d x))^{3/2}}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle \frac {b^3 \left (\frac {2 \left (\frac {(a-b)^2 (2 a+3 b) \int \frac {1}{b^2 \sin ^2(c+d x)-a+b}d\sqrt {a+b \sin (c+d x)}}{2 b}-\frac {(2 a-3 b) (a+b)^2 \int \frac {1}{b^2 \sin ^2(c+d x)-a-b}d\sqrt {a+b \sin (c+d x)}}{2 b}\right )+2 a \sqrt {a+b \sin (c+d x)}}{4 b^2}+\frac {\left (a b \sin (c+d x)+b^2\right ) (a+b \sin (c+d x))^{3/2}}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {b^3 \left (\frac {2 \left (\frac {(2 a-3 b) (a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{2 b}-\frac {(a-b)^{3/2} (2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{2 b}\right )+2 a \sqrt {a+b \sin (c+d x)}}{4 b^2}+\frac {\left (a b \sin (c+d x)+b^2\right ) (a+b \sin (c+d x))^{3/2}}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{d}\) |
(b^3*(((a + b*Sin[c + d*x])^(3/2)*(b^2 + a*b*Sin[c + d*x]))/(2*b^2*(b^2 - b^2*Sin[c + d*x]^2)) + (2*(-1/2*((a - b)^(3/2)*(2*a + 3*b)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/b + ((2*a - 3*b)*(a + b)^(3/2)*ArcTanh[Sqr t[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(2*b)) + 2*a*Sqrt[a + b*Sin[c + d*x]]) /(4*b^2)))/d
3.5.98.3.1 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[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[g*((d + e*x)^m/(c*m)), x] + Simp[1/c Int[(d + e*x)^(m - 1)*(Simp[c*d*f - a*e*g + (g*c*d + c*e*f)*x, x]/(a + c*x^2)), x], x] /; Fr eeQ[{a, c, d, e, f, g}, x] && FractionQ[m] && GtQ[m, 0]
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 = 34.96 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.03
method | result | size |
default | \(\frac {2 b^{3} \left (\frac {\left (a -b \right )^{2} \left (-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{2 \left (b \sin \left (d x +c \right )+b \right )}+\frac {\left (2 a +3 b \right ) \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{2 \sqrt {-a +b}}\right )}{4 b^{3}}-\frac {\left (a +b \right )^{2} \left (\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{2 b \sin \left (d x +c \right )-2 b}-\frac {\left (2 a -3 b \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{2 \sqrt {a +b}}\right )}{4 b^{3}}\right )}{d}\) | \(160\) |
2*b^3*(1/4*(a-b)^2/b^3*(-1/2*b*(a+b*sin(d*x+c))^(1/2)/(b*sin(d*x+c)+b)+1/2 *(2*a+3*b)/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2)))-1/4*( a+b)^2/b^3*(1/2*b*(a+b*sin(d*x+c))^(1/2)/(b*sin(d*x+c)-b)-1/2*(2*a-3*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. 380 vs. \(2 (131) = 262\).
Time = 0.60 (sec) , antiderivative size = 2071, normalized size of antiderivative = 13.36 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Too large to display} \]
[-1/32*((2*a^2 - a*b - 3*b^2)*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^2 + a*b - 3*b^2)*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*(2*a*b + (a^2 + b^2)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a))/(d*cos(d*x + c)^2), -1/32*(2 *(2*a^2 - a*b - 3*b^2)*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)))*cos(d*x + c)^2 + (2*a^...
Timed out. \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Exception raised: ValueError} \]
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
Timed out. \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \]