Integrand size = 21, antiderivative size = 117 \[ \int \sec (c+d x) (a+b \sin (c+d x))^{5/2} \, dx=-\frac {(a-b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{d}+\frac {(a+b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{d}-\frac {4 a b \sqrt {a+b \sin (c+d x)}}{d}-\frac {2 b (a+b \sin (c+d x))^{3/2}}{3 d} \]
-(a-b)^(5/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/d+(a+b)^(5/2)*arc tanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))/d-2/3*b*(a+b*sin(d*x+c))^(3/2)/d- 4*a*b*(a+b*sin(d*x+c))^(1/2)/d
Time = 0.12 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.90 \[ \int \sec (c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\frac {-3 (a-b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )+3 (a+b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )-2 b \sqrt {a+b \sin (c+d x)} (7 a+b \sin (c+d x))}{3 d} \]
(-3*(a - b)^(5/2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]] + 3*(a + b )^(5/2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]] - 2*b*Sqrt[a + b*Sin [c + d*x]]*(7*a + b*Sin[c + d*x]))/(3*d)
Time = 0.38 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3147, 481, 25, 653, 25, 654, 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.
| \(\displaystyle \int \sec (c+d x) (a+b \sin (c+d x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^{5/2}}{\cos (c+d x)}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {b \int \frac {(a+b \sin (c+d x))^{5/2}}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 481 |
| \(\displaystyle \frac {b \left (-\int -\frac {\sqrt {a+b \sin (c+d x)} \left (a^2+2 b \sin (c+d x) a+b^2\right )}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))-\frac {2}{3} (a+b \sin (c+d x))^{3/2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b \left (\int \frac {\sqrt {a+b \sin (c+d x)} \left (a^2+2 b \sin (c+d x) a+b^2\right )}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))-\frac {2}{3} (a+b \sin (c+d x))^{3/2}\right )}{d}\) |
| \(\Big \downarrow \) 653 |
| \(\displaystyle \frac {b \left (-\int -\frac {a \left (a^2+3 b^2\right )+b \left (3 a^2+b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))-\frac {2}{3} (a+b \sin (c+d x))^{3/2}-4 a \sqrt {a+b \sin (c+d x)}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b \left (\int \frac {a \left (a^2+3 b^2\right )+b \left (3 a^2+b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))-\frac {2}{3} (a+b \sin (c+d x))^{3/2}-4 a \sqrt {a+b \sin (c+d x)}\right )}{d}\) |
| \(\Big \downarrow \) 654 |
| \(\displaystyle \frac {b \left (2 \int \frac {2 a \left (a^2-b^2\right )-b^2 \left (3 a^2+b^2\right ) \sin ^2(c+d x)}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}-\frac {2}{3} (a+b \sin (c+d x))^{3/2}-4 a \sqrt {a+b \sin (c+d x)}\right )}{d}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {b \left (2 \left (\frac {(a-b)^3 \int \frac {1}{b^2 \sin ^2(c+d x)-a+b}d\sqrt {a+b \sin (c+d x)}}{2 b}-\frac {(a+b)^3 \int \frac {1}{b^2 \sin ^2(c+d x)-a-b}d\sqrt {a+b \sin (c+d x)}}{2 b}\right )-\frac {2}{3} (a+b \sin (c+d x))^{3/2}-4 a \sqrt {a+b \sin (c+d x)}\right )}{d}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {b \left (2 \left (\frac {(a+b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{2 b}-\frac {(a-b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{2 b}\right )-\frac {2}{3} (a+b \sin (c+d x))^{3/2}-4 a \sqrt {a+b \sin (c+d x)}\right )}{d}\) |
(b*(2*(-1/2*((a - b)^(5/2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/ b + ((a + b)^(5/2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(2*b)) - 4*a*Sqrt[a + b*Sin[c + d*x]] - (2*(a + b*Sin[c + d*x])^(3/2))/3))/d
3.5.97.3.1 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)/(b*(n - 1))), x] + Simp[1/b Int[(c + d*x)^(n - 2)*(Simp[b *c^2 - a*d^2 + 2*b*c*d*x, x]/(a + b*x^2)), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 1]
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 = 1.89 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.19
| method | result | size |
| default | \(-\frac {2 b \left (\frac {\left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 a \sqrt {a +b \sin \left (d x +c \right )}+\frac {\left (-a^{3}+3 a^{2} b -3 a \,b^{2}+b^{3}\right ) \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{2 b \sqrt {-a +b}}-\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{2 b \sqrt {a +b}}\right )}{d}\) | \(139\) |
-2*b*(1/3*(a+b*sin(d*x+c))^(3/2)+2*a*(a+b*sin(d*x+c))^(1/2)+1/2*(-a^3+3*a^ 2*b-3*a*b^2+b^3)/b/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2) )-1/2*(a^3+3*a^2*b+3*a*b^2+b^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. 347 vs. \(2 (99) = 198\).
Time = 3.29 (sec) , antiderivative size = 1937, normalized size of antiderivative = 16.56 \[ \int \sec (c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Too large to display} \]
[1/24*(3*(a^2 + 2*a*b + b^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)) + 3*(a^2 - 2*a*b + b^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 + 2 8*a*b^2 - 8*b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)*sqrt(a - b) + 4*(6 4*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*(b^2*sin(d*x + c) + 7*a*b)*sqrt(b*sin(d*x + c) + a))/d, -1/24*(6*(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 + (3*a^2*b + 4*a*b^2 + b^3)*sin(d*x + c))) - 3*(a^2 - 2*a*b + b^2)*sqrt(a - b)*log((b^4*cos(d*x + c)^4 + 128*a^4 - 256*a^3*...
Timed out. \[ \int \sec (c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \sec (c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Exception raised: ValueError} \]
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
Timed out. \[ \int \sec (c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \sec (c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2}}{\cos \left (c+d\,x\right )} \,d x \]