Integrand size = 23, antiderivative size = 199 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=-\frac {3 \sqrt {a-b} \left (4 a^2+2 a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 d}+\frac {3 \sqrt {a+b} \left (4 a^2-2 a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 d}+\frac {\sec ^4(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{4 d}+\frac {3 \sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (a b+\left (2 a^2-b^2\right ) \sin (c+d x)\right )}{16 d} \]
1/4*sec(d*x+c)^4*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))^(3/2)/d-3/32*(4*a^2+2*a *b-b^2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))*(a-b)^(1/2)/d+3/32*(4* a^2-2*a*b-b^2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*(a+b)^(1/2)/d+3 /16*sec(d*x+c)^2*(a*b+(2*a^2-b^2)*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/d
Time = 2.14 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.54 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=-\frac {3 \sqrt {a-b} \left (a^2-b^2\right )^2 \left (4 a^2+2 a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )-3 \sqrt {a+b} \left (a^2-b^2\right )^2 \left (4 a^2-2 a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )+8 \left (-a^2+b^2\right ) \sec ^4(c+d x) (-b+a \sin (c+d x)) (a+b \sin (c+d x))^{7/2}-2 \sec ^2(c+d x) \left (-7 a^2 b+b^3+6 a^3 \sin (c+d x)\right ) (a+b \sin (c+d x))^{7/2}-2 b \sqrt {a+b \sin (c+d x)} \left (18 a^5-16 a^3 b^2+7 a b^4-3 a^3 b^2 \cos (2 (c+d x))+b \left (18 a^4-7 a^2 b^2+b^4\right ) \sin (c+d x)\right )}{32 \left (a^2-b^2\right )^2 d} \]
-1/32*(3*Sqrt[a - b]*(a^2 - b^2)^2*(4*a^2 + 2*a*b - b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]] - 3*Sqrt[a + b]*(a^2 - b^2)^2*(4*a^2 - 2*a*b - b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]] + 8*(-a^2 + b^2)*Sec[ c + d*x]^4*(-b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^(7/2) - 2*Sec[c + d* x]^2*(-7*a^2*b + b^3 + 6*a^3*Sin[c + d*x])*(a + b*Sin[c + d*x])^(7/2) - 2* b*Sqrt[a + b*Sin[c + d*x]]*(18*a^5 - 16*a^3*b^2 + 7*a*b^4 - 3*a^3*b^2*Cos[ 2*(c + d*x)] + b*(18*a^4 - 7*a^2*b^2 + b^4)*Sin[c + d*x]))/((a^2 - b^2)^2* d)
Time = 0.47 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.26, 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, 685, 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.
| \(\displaystyle \int \sec ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^{5/2}}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {b^5 \int \frac {(a+b \sin (c+d x))^{5/2}}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 495 |
| \(\displaystyle \frac {b^5 \left (\frac {(a+b \sin (c+d x))^{3/2} \left (a b \sin (c+d x)+b^2\right )}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int -\frac {3 \sqrt {a+b \sin (c+d x)} \left (2 a^2+b \sin (c+d x) a-b^2\right )}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^5 \left (\frac {3 \int \frac {\sqrt {a+b \sin (c+d x)} \left (2 a^2+b \sin (c+d x) a-b^2\right )}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{8 b^2}+\frac {\left (a b \sin (c+d x)+b^2\right ) (a+b \sin (c+d x))^{3/2}}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 685 |
| \(\displaystyle \frac {b^5 \left (\frac {3 \left (\frac {\sqrt {a+b \sin (c+d x)} \left (b \left (2 a^2-b^2\right ) \sin (c+d x)+a b^2\right )}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int -\frac {a \left (4 a^2-3 b^2\right )+b \left (2 a^2-b^2\right ) \sin (c+d x)}{2 \sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}\right )}{8 b^2}+\frac {\left (a b \sin (c+d x)+b^2\right ) (a+b \sin (c+d x))^{3/2}}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^5 \left (\frac {3 \left (\frac {\int \frac {a \left (4 a^2-3 b^2\right )+b \left (2 a^2-b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{4 b^2}+\frac {\sqrt {a+b \sin (c+d x)} \left (b \left (2 a^2-b^2\right ) \sin (c+d x)+a b^2\right )}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{8 b^2}+\frac {\left (a b \sin (c+d x)+b^2\right ) (a+b \sin (c+d x))^{3/2}}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 654 |
| \(\displaystyle \frac {b^5 \left (\frac {3 \left (\frac {\int -\frac {b^2 \left (2 a^2-b^2\right ) \sin ^2(c+d x)+2 a \left (a^2-b^2\right )}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{2 b^2}+\frac {\sqrt {a+b \sin (c+d x)} \left (b \left (2 a^2-b^2\right ) \sin (c+d x)+a b^2\right )}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{8 b^2}+\frac {\left (a b \sin (c+d x)+b^2\right ) (a+b \sin (c+d x))^{3/2}}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^5 \left (\frac {3 \left (\frac {\sqrt {a+b \sin (c+d x)} \left (b \left (2 a^2-b^2\right ) \sin (c+d x)+a b^2\right )}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int \frac {b^2 \left (2 a^2-b^2\right ) \sin ^2(c+d x)+2 a \left (a^2-b^2\right )}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{2 b^2}\right )}{8 b^2}+\frac {\left (a b \sin (c+d x)+b^2\right ) (a+b \sin (c+d x))^{3/2}}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {b^5 \left (\frac {3 \left (\frac {\frac {(a-b) \left (4 a^2+2 a b-b^2\right ) \int \frac {1}{b^2 \sin ^2(c+d x)-a+b}d\sqrt {a+b \sin (c+d x)}}{2 b}-\frac {(a+b) \left (4 a^2-2 a b-b^2\right ) \int \frac {1}{b^2 \sin ^2(c+d x)-a-b}d\sqrt {a+b \sin (c+d x)}}{2 b}}{2 b^2}+\frac {\sqrt {a+b \sin (c+d x)} \left (b \left (2 a^2-b^2\right ) \sin (c+d x)+a b^2\right )}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{8 b^2}+\frac {\left (a b \sin (c+d x)+b^2\right ) (a+b \sin (c+d x))^{3/2}}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {b^5 \left (\frac {3 \left (\frac {\frac {\sqrt {a+b} \left (4 a^2-2 a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{2 b}-\frac {\sqrt {a-b} \left (4 a^2+2 a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{2 b}}{2 b^2}+\frac {\sqrt {a+b \sin (c+d x)} \left (b \left (2 a^2-b^2\right ) \sin (c+d x)+a b^2\right )}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{8 b^2}+\frac {\left (a b \sin (c+d x)+b^2\right ) (a+b \sin (c+d x))^{3/2}}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
(b^5*(((a + b*Sin[c + d*x])^(3/2)*(b^2 + a*b*Sin[c + d*x]))/(4*b^2*(b^2 - b^2*Sin[c + d*x]^2)^2) + (3*((-1/2*(Sqrt[a - b]*(4*a^2 + 2*a*b - b^2)*ArcT anh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/b + (Sqrt[a + b]*(4*a^2 - 2*a*b - b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(2*b))/(2*b^2) + (S qrt[a + b*Sin[c + d*x]]*(a*b^2 + b*(2*a^2 - b^2)*Sin[c + d*x]))/(2*b^2*(b^ 2 - b^2*Sin[c + d*x]^2))))/(8*b^2)))/d
3.5.99.3.1 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_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> Simp[(d + e*x)^m*(a + c*x^2)^(p + 1)*((a*g - c*f*x)/(2*a*c *(p + 1))), x] - Simp[1/(2*a*c*(p + 1)) Int[(d + e*x)^(m - 1)*(a + c*x^2) ^(p + 1)*Simp[a*e*g*m - c*d*f*(2*p + 3) - c*e*f*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
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 = 1.17 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.12
\[-\frac {2 b^{5} \left (-\frac {\left (a -b \right ) \left (-\frac {\sqrt {a +b \sin \left (d x +c \right )}\, b^{2} \left (6 a \sin \left (d x +c \right )+3 b \sin \left (d x +c \right )+8 a +b \right )}{4 \left (b \sin \left (d x +c \right )+b \right )^{2}}+\frac {3 \left (4 a^{2}+2 a b -b^{2}\right ) \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{4 \sqrt {-a +b}}\right )}{16 b^{5}}+\frac {\left (a +b \right ) \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}\, b^{2} \left (6 a \sin \left (d x +c \right )-3 b \sin \left (d x +c \right )-8 a +b \right )}{4 \left (b \sin \left (d x +c \right )-b \right )^{2}}-\frac {3 \left (4 a^{2}-2 a b -b^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{4 \sqrt {a +b}}\right )}{16 b^{5}}\right )}{d}\]
-2*b^5*(-1/16*(a-b)/b^5*(-1/4*(a+b*sin(d*x+c))^(1/2)*b^2*(6*a*sin(d*x+c)+3 *b*sin(d*x+c)+8*a+b)/(b*sin(d*x+c)+b)^2+3/4*(4*a^2+2*a*b-b^2)/(-a+b)^(1/2) *arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2)))+1/16*(a+b)/b^5*(1/4*(a+b*sin (d*x+c))^(1/2)*b^2*(6*a*sin(d*x+c)-3*b*sin(d*x+c)-8*a+b)/(b*sin(d*x+c)-b)^ 2-3/4*(4*a^2-2*a*b-b^2)/(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. 420 vs. \(2 (175) = 350\).
Time = 0.69 (sec) , antiderivative size = 2229, normalized size of antiderivative = 11.20 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Too large to display} \]
[-1/256*(3*(4*a^2 - 2*a*b - b^2)*sqrt(a + b)*cos(d*x + c)^4*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)*co s(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*(4*a^2 + 2*a*b - b^2)*sqrt(a - b)*cos (d*x + c)^4*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*b*cos(d *x + c)^2 - 8*a*b - (3*(2*a^2 - b^2)*cos(d*x + c)^2 + 4*a^2 + 4*b^2)*sin(d *x + c))*sqrt(b*sin(d*x + c) + a))/(d*cos(d*x + c)^4), -1/256*(6*(4*a^2 - 2*a*b - 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 +...
Timed out. \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Exception raised: ValueError} \]
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. \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^5} \,d x \]