Integrand size = 23, antiderivative size = 231 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=-\frac {(2 a-7 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{4 (a-b)^{7/2} d}+\frac {(2 a+7 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{4 (a+b)^{7/2} d}-\frac {b \left (3 a^2+7 b^2\right )}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac {a b \left (a^2+19 b^2\right )}{2 \left (a^2-b^2\right )^3 d \sqrt {a+b \sin (c+d x)}} \]
-1/4*(2*a-7*b)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/(a-b)^(7/2)/d+1 /4*(2*a+7*b)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))/(a+b)^(7/2)/d-1/6 *b*(3*a^2+7*b^2)/(a^2-b^2)^2/d/(a+b*sin(d*x+c))^(3/2)-1/2*sec(d*x+c)^2*(b- a*sin(d*x+c))/(a^2-b^2)/d/(a+b*sin(d*x+c))^(3/2)-1/2*a*b*(a^2+19*b^2)/(a^2 -b^2)^3/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 = 0.57 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\frac {-\left (\left (3 a^3+3 a^2 b+7 a b^2+7 b^3\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \sin (c+d x)}{a-b}\right )\right )+\left (3 a^3-3 a^2 b+7 a b^2-7 b^3\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \sin (c+d x)}{a+b}\right )+15 a (a+b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sin (c+d x)}{a-b}\right ) (a+b \sin (c+d x))-3 (a-b) \left (-2 (a+b) \sec ^2(c+d x) (-b+a \sin (c+d x))+5 a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sin (c+d x)}{a+b}\right ) (a+b \sin (c+d x))\right )}{12 (a-b)^2 (a+b)^2 d (a+b \sin (c+d x))^{3/2}} \]
(-((3*a^3 + 3*a^2*b + 7*a*b^2 + 7*b^3)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Sin[c + d*x])/(a - b)]) + (3*a^3 - 3*a^2*b + 7*a*b^2 - 7*b^3)*Hyperge ometric2F1[-3/2, 1, -1/2, (a + b*Sin[c + d*x])/(a + b)] + 15*a*(a + b)*Hyp ergeometric2F1[-1/2, 1, 1/2, (a + b*Sin[c + d*x])/(a - b)]*(a + b*Sin[c + d*x]) - 3*(a - b)*(-2*(a + b)*Sec[c + d*x]^2*(-b + a*Sin[c + d*x]) + 5*a*H ypergeometric2F1[-1/2, 1, 1/2, (a + b*Sin[c + d*x])/(a + b)]*(a + b*Sin[c + d*x])))/(12*(a - b)^2*(a + b)^2*d*(a + b*Sin[c + d*x])^(3/2))
Time = 0.57 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.30, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 3147, 496, 27, 655, 25, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^3 (a+b \sin (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {b^3 \int \frac {1}{(a+b \sin (c+d x))^{5/2} \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 496 |
| \(\displaystyle \frac {b^3 \left (\frac {\int \frac {2 a^2+5 b \sin (c+d x) a-7 b^2}{2 (a+b \sin (c+d x))^{5/2} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^3 \left (\frac {\int \frac {2 a^2+5 b \sin (c+d x) a-7 b^2}{(a+b \sin (c+d x))^{5/2} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{4 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}\right )}{d}\) |
| \(\Big \downarrow \) 655 |
| \(\displaystyle \frac {b^3 \left (\frac {-\frac {\int -\frac {2 a \left (a^2-6 b^2\right )+b \left (3 a^2+7 b^2\right ) \sin (c+d x)}{(a+b \sin (c+d x))^{3/2} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{a^2-b^2}-\frac {2 \left (3 a^2+7 b^2\right )}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}}{4 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^3 \left (\frac {\frac {\int \frac {2 a \left (a^2-6 b^2\right )+b \left (3 a^2+7 b^2\right ) \sin (c+d x)}{(a+b \sin (c+d x))^{3/2} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{a^2-b^2}-\frac {2 \left (3 a^2+7 b^2\right )}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}}{4 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}\right )}{d}\) |
| \(\Big \downarrow \) 655 |
| \(\displaystyle \frac {b^3 \left (\frac {\frac {-\frac {\int -\frac {2 a^4-15 b^2 a^2+b \left (a^2+19 b^2\right ) \sin (c+d x) a-7 b^4}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{a^2-b^2}-\frac {2 a \left (a^2+19 b^2\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}-\frac {2 \left (3 a^2+7 b^2\right )}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}}{4 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^3 \left (\frac {\frac {\frac {\int \frac {2 a^4-15 b^2 a^2+b \left (a^2+19 b^2\right ) \sin (c+d x) a-7 b^4}{\sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{a^2-b^2}-\frac {2 a \left (a^2+19 b^2\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}-\frac {2 \left (3 a^2+7 b^2\right )}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}}{4 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}\right )}{d}\) |
| \(\Big \downarrow \) 654 |
| \(\displaystyle \frac {b^3 \left (\frac {\frac {\frac {2 \int -\frac {a^4-34 b^2 a^2+b^2 \left (a^2+19 b^2\right ) \sin ^2(c+d x) a-7 b^4}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{a^2-b^2}-\frac {2 a \left (a^2+19 b^2\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}-\frac {2 \left (3 a^2+7 b^2\right )}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}}{4 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^3 \left (\frac {\frac {-\frac {2 \int \frac {a^4-34 b^2 a^2+b^2 \left (a^2+19 b^2\right ) \sin ^2(c+d x) a-7 b^4}{b^4 \sin ^4(c+d x)-2 a b^2 \sin ^2(c+d x)+a^2-b^2}d\sqrt {a+b \sin (c+d x)}}{a^2-b^2}-\frac {2 a \left (a^2+19 b^2\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}-\frac {2 \left (3 a^2+7 b^2\right )}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}}{4 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}\right )}{d}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {b^3 \left (\frac {\frac {\frac {2 \left (\frac {(2 a-7 b) (a+b)^3 \int \frac {1}{b^2 \sin ^2(c+d x)-a+b}d\sqrt {a+b \sin (c+d x)}}{2 b}-\frac {(a-b)^3 (2 a+7 b) \int \frac {1}{b^2 \sin ^2(c+d x)-a-b}d\sqrt {a+b \sin (c+d x)}}{2 b}\right )}{a^2-b^2}-\frac {2 a \left (a^2+19 b^2\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}-\frac {2 \left (3 a^2+7 b^2\right )}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}}{4 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}\right )}{d}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {b^3 \left (\frac {\frac {\frac {2 \left (\frac {(a-b)^3 (2 a+7 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{2 b \sqrt {a+b}}-\frac {(2 a-7 b) (a+b)^3 \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{2 b \sqrt {a-b}}\right )}{a^2-b^2}-\frac {2 a \left (a^2+19 b^2\right )}{\left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}}{a^2-b^2}-\frac {2 \left (3 a^2+7 b^2\right )}{3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}}{4 b^2 \left (a^2-b^2\right )}-\frac {b^2-a b \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right ) (a+b \sin (c+d x))^{3/2}}\right )}{d}\) |
(b^3*(-1/2*(b^2 - a*b*Sin[c + d*x])/(b^2*(a^2 - b^2)*(a + b*Sin[c + d*x])^ (3/2)*(b^2 - b^2*Sin[c + d*x]^2)) + ((-2*(3*a^2 + 7*b^2))/(3*(a^2 - b^2)*( a + b*Sin[c + d*x])^(3/2)) + ((2*(-1/2*((2*a - 7*b)*(a + b)^3*ArcTanh[Sqrt [a + b*Sin[c + d*x]]/Sqrt[a - b]])/(Sqrt[a - b]*b) + ((a - b)^3*(2*a + 7*b )*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(2*b*Sqrt[a + b])))/(a^2 - b^2) - (2*a*(a^2 + 19*b^2))/((a^2 - b^2)*Sqrt[a + b*Sin[c + d*x]]))/(a^2 - b^2))/(4*b^2*(a^2 - b^2))))/d
3.6.31.3.1 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 = 1.03 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.14
| method | result | size |
| default | \(\frac {-\frac {2 b^{3}}{3 \left (a -b \right )^{2} \left (a +b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {8 b^{3} a}{\left (a -b \right )^{3} \left (a +b \right )^{3} \sqrt {a +b \sin \left (d x +c \right )}}-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{4 \left (a -b \right )^{3} \left (b \sin \left (d x +c \right )+b \right )}+\frac {\arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a}{2 \left (a -b \right )^{3} \sqrt {-a +b}}-\frac {7 b \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{4 \left (a -b \right )^{3} \sqrt {-a +b}}-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{4 \left (a +b \right )^{3} \left (b \sin \left (d x +c \right )-b \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a}{2 \left (a +b \right )^{\frac {7}{2}}}+\frac {7 b \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{4 \left (a +b \right )^{\frac {7}{2}}}}{d}\) | \(263\) |
(-2/3*b^3/(a-b)^2/(a+b)^2/(a+b*sin(d*x+c))^(3/2)-8*b^3*a/(a-b)^3/(a+b)^3/( a+b*sin(d*x+c))^(1/2)-1/4*b/(a-b)^3*(a+b*sin(d*x+c))^(1/2)/(b*sin(d*x+c)+b )+1/2/(a-b)^3/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*a-7 /4*b/(a-b)^3/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))-1/4* b/(a+b)^3*(a+b*sin(d*x+c))^(1/2)/(b*sin(d*x+c)-b)+1/2/(a+b)^(7/2)*arctanh( (a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a+7/4*b/(a+b)^(7/2)*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. 945 vs. \(2 (204) = 408\).
Time = 0.87 (sec) , antiderivative size = 4329, normalized size of antiderivative = 18.74 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
[1/96*(3*((2*a^5*b^2 - a^4*b^3 - 16*a^3*b^4 + 34*a^2*b^5 - 26*a*b^6 + 7*b^ 7)*cos(d*x + c)^4 - 2*(2*a^6*b - a^5*b^2 - 16*a^4*b^3 + 34*a^3*b^4 - 26*a^ 2*b^5 + 7*a*b^6)*cos(d*x + c)^2*sin(d*x + c) - (2*a^7 - a^6*b - 14*a^5*b^2 + 33*a^4*b^3 - 42*a^3*b^4 + 41*a^2*b^5 - 26*a*b^6 + 7*b^7)*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 + 1 4*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*((2*a^5*b ^2 + a^4*b^3 - 16*a^3*b^4 - 34*a^2*b^5 - 26*a*b^6 - 7*b^7)*cos(d*x + c)^4 - 2*(2*a^6*b + a^5*b^2 - 16*a^4*b^3 - 34*a^3*b^4 - 26*a^2*b^5 - 7*a*b^6)*c os(d*x + c)^2*sin(d*x + c) - (2*a^7 + a^6*b - 14*a^5*b^2 - 33*a^4*b^3 - 42 *a^3*b^4 - 41*a^2*b^5 - 26*a*b^6 - 7*b^7)*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^...
\[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{3}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]