\(\int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx\) [520]

Optimal result

Integrand size = 23, antiderivative size = 186 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {(2 a-5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{4 (a-b)^{5/2} d}+\frac {(2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{4 (a+b)^{5/2} d}-\frac {b \left (a^2+5 b^2\right )}{2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}} \]

[Out]

-1/4*(2*a-5*b)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/(a-b)^(5/2)/d+1/4*(2*a+5*b)*arctanh((a+b*sin(d*x+c)
)^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)/d-1/2*b*(a^2+5*b^2)/(a^2-b^2)^2/d/(a+b*sin(d*x+c))^(1/2)-1/2*sec(d*x+c)^2*(b-
a*sin(d*x+c))/(a^2-b^2)/d/(a+b*sin(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2747, 755, 843, 841, 1180, 212} \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {b \left (a^2+5 b^2\right )}{2 d \left (a^2-b^2\right )^2 \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}-\frac {(2 a-5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{4 d (a-b)^{5/2}}+\frac {(2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{4 d (a+b)^{5/2}} \]

[In]

Int[Sec[c + d*x]^3/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

-1/4*((2*a - 5*b)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/((a - b)^(5/2)*d) + ((2*a + 5*b)*ArcTanh[Sqrt
[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(4*(a + b)^(5/2)*d) - (b*(a^2 + 5*b^2))/(2*(a^2 - b^2)^2*d*Sqrt[a + b*Sin[c
+ d*x]]) - (Sec[c + d*x]^2*(b - a*Sin[c + d*x]))/(2*(a^2 - b^2)*d*Sqrt[a + b*Sin[c + d*x]])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 755

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(a*e + c*d*x)*
((a + c*x^2)^(p + 1)/(2*a*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^
m*Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[
{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 841

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[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]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 843

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] + Dist[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, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&
FractionQ[m] && LtQ[m, -1]

Rule 1180

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[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]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[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] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {b^3 \text {Subst}\left (\int \frac {1}{(a+x)^{3/2} \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = -\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}+\frac {b \text {Subst}\left (\int \frac {\frac {1}{2} \left (2 a^2-5 b^2\right )+\frac {3 a x}{2}}{(a+x)^{3/2} \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{2 \left (a^2-b^2\right ) d} \\ & = -\frac {b \left (a^2+5 b^2\right )}{2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {b \text {Subst}\left (\int \frac {-a \left (a^2-4 b^2\right )-\frac {1}{2} \left (a^2+5 b^2\right ) x}{\sqrt {a+x} \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 d} \\ & = -\frac {b \left (a^2+5 b^2\right )}{2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {b \text {Subst}\left (\int \frac {-\frac {1}{2} a \left (-a^2-5 b^2\right )-a \left (a^2-4 b^2\right )+\frac {1}{2} \left (-a^2-5 b^2\right ) x^2}{-a^2+b^2+2 a x^2-x^4} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{\left (a^2-b^2\right )^2 d} \\ & = -\frac {b \left (a^2+5 b^2\right )}{2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {(2 a-5 b) \text {Subst}\left (\int \frac {1}{a-b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{4 (a-b)^2 d}+\frac {(2 a+5 b) \text {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{4 (a+b)^2 d} \\ & = -\frac {(2 a-5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{4 (a-b)^{5/2} d}+\frac {(2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{4 (a+b)^{5/2} d}-\frac {b \left (a^2+5 b^2\right )}{2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.75 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.19 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\frac {\frac {3 a \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}-\frac {3 a \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+\frac {\left (a^2+5 b^2\right ) \left ((a+b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sin (c+d x)}{a-b}\right )+(-a+b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sin (c+d x)}{a+b}\right )\right )}{(a-b) (a+b) \sqrt {a+b \sin (c+d x)}}+\frac {2 \sec ^2(c+d x) (b-a \sin (c+d x))}{\sqrt {a+b \sin (c+d x)}}}{4 \left (-a^2+b^2\right ) d} \]

[In]

Integrate[Sec[c + d*x]^3/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

((3*a*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/Sqrt[a - b] - (3*a*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[
a + b]])/Sqrt[a + b] + ((a^2 + 5*b^2)*((a + b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Sin[c + d*x])/(a - b)] +
(-a + b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Sin[c + d*x])/(a + b)]))/((a - b)*(a + b)*Sqrt[a + b*Sin[c +
d*x]]) + (2*Sec[c + d*x]^2*(b - a*Sin[c + d*x]))/Sqrt[a + b*Sin[c + d*x]])/(4*(-a^2 + b^2)*d)

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.25

method	result	size
default	\(\frac {-\frac {2 b^{3}}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a +b \sin \left (d x +c \right )}}-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{4 \left (a -b \right )^{2} \left (b \sin \left (d x +c \right )+b \right )}+\frac {\arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a}{2 \left (a -b \right )^{2} \sqrt {-a +b}}-\frac {5 b \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{4 \left (a -b \right )^{2} \sqrt {-a +b}}-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{4 \left (a +b \right )^{2} \left (b \sin \left (d x +c \right )-b \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a}{2 \left (a +b \right )^{\frac {5}{2}}}+\frac {5 b \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{4 \left (a +b \right )^{\frac {5}{2}}}}{d}\)	\(233\)

[In]

int(sec(d*x+c)^3/(a+b*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

(-2*b^3/(a-b)^2/(a+b)^2/(a+b*sin(d*x+c))^(1/2)-1/4*b/(a-b)^2*(a+b*sin(d*x+c))^(1/2)/(b*sin(d*x+c)+b)+1/2/(a-b)
^2/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*a-5/4*b/(a-b)^2/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c
))^(1/2)/(-a+b)^(1/2))-1/4*b/(a+b)^2*(a+b*sin(d*x+c))^(1/2)/(b*sin(d*x+c)-b)+1/2/(a+b)^(5/2)*arctanh((a+b*sin(
d*x+c))^(1/2)/(a+b)^(1/2))*a+5/4*b/(a+b)^(5/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2)))/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 641 vs. \(2 (163) = 326\).

Time = 0.77 (sec) , antiderivative size = 3117, normalized size of antiderivative = 16.76 \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate(sec(d*x+c)^3/(a+b*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/32*(((2*a^4*b - a^3*b^2 - 9*a^2*b^3 + 13*a*b^4 - 5*b^5)*cos(d*x + c)^2*sin(d*x + c) + (2*a^5 - a^4*b - 9*a^
3*b^2 + 13*a^2*b^3 - 5*a*b^4)*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^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^4*b + a^3*b^2 - 9*a^2*b^3 - 13*a*b^4 - 5*b^5)*cos(d*x + c)^2*sin(d*x + c) + (2*a^5 + a^4*b - 9*a^3*b^2
- 13*a^2*b^3 - 5*a*b^4)*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))*s
qrt(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
^4*b - 2*a^2*b^3 + b^5 + (a^4*b + 4*a^2*b^3 - 5*b^5)*cos(d*x + c)^2 - (a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c))*
sqrt(b*sin(d*x + c) + a))/((a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*d*cos(d*x + c)^2*sin(d*x + c) + (a^7 - 3*a^5*
b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^2), -1/32*(2*((2*a^4*b - a^3*b^2 - 9*a^2*b^3 + 13*a*b^4 - 5*b^5)*cos(d
*x + c)^2*sin(d*x + c) + (2*a^5 - a^4*b - 9*a^3*b^2 + 13*a^2*b^3 - 5*a*b^4)*cos(d*x + c)^2)*sqrt(-a - b)*arcta
n(-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))) + ((2*a^4*b + a^3*b^2 - 9*a^2*b^3 - 13*a*b^4 - 5*b^5)*cos(d*x + c)^2*sin(d*x + c) + (2*a^5 + a^4*b -
9*a^3*b^2 - 13*a^2*b^3 - 5*a*b^4)*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^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^4*b - 2*a^2*b^3 + b^5 + (a^4*b + 4*a^2*b^3 - 5*b^5)*cos(d*x + c)^2 - (a^5 - 2*a^3*b^2 + a*b^4)*sin
(d*x + c))*sqrt(b*sin(d*x + c) + a))/((a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*d*cos(d*x + c)^2*sin(d*x + c) + (a
^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^2), -1/32*(2*((2*a^4*b + a^3*b^2 - 9*a^2*b^3 - 13*a*b^4 - 5
*b^5)*cos(d*x + c)^2*sin(d*x + c) + (2*a^5 + a^4*b - 9*a^3*b^2 - 13*a^2*b^3 - 5*a*b^4)*cos(d*x + c)^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))) - ((2*a^4*b - a^3*b^2 - 9*a^2*b^3 + 13*a*b^4 - 5*b^5)*cos(d*x + c)^2*sin(d*x + c) + (2*a^5
- a^4*b - 9*a^3*b^2 + 13*a^2*b^3 - 5*a*b^4)*cos(d*x + c)^2)*sqrt(a + b)*log((b^4*cos(d*x + c)^4 + 128*a^4 + 2
56*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 + 2
4*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^4*b - 2*a^2*b^3 + b^5 + (a^4*b + 4*a^2*b^3 - 5*b^5)*cos(d*x + c)^2 - (a^5 - 2*a^3*b^2 +
a*b^4)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a))/((a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*d*cos(d*x + c)^2*sin(d*x
+ c) + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^2), -1/16*(((2*a^4*b + a^3*b^2 - 9*a^2*b^3 - 13*a
*b^4 - 5*b^5)*cos(d*x + c)^2*sin(d*x + c) + (2*a^5 + a^4*b - 9*a^3*b^2 - 13*a^2*b^3 - 5*a*b^4)*cos(d*x + c)^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))) + ((2*a^4*b - a^3*b^2 - 9*a^2*b^3 + 13*a*b^4 - 5*b^5)*cos(d*x + c)^2*sin(d*x + c)
+ (2*a^5 - a^4*b - 9*a^3*b^2 + 13*a^2*b^3 - 5*a*b^4)*cos(d*x + c)^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))) + 8*(a^4*b -
2*a^2*b^3 + b^5 + (a^4*b + 4*a^2*b^3 - 5*b^5)*cos(d*x + c)^2 - (a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c))*sqrt(b*
sin(d*x + c) + a))/((a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*d*cos(d*x + c)^2*sin(d*x + c) + (a^7 - 3*a^5*b^2 + 3
*a^3*b^4 - a*b^6)*d*cos(d*x + c)^2)]

Sympy [F]

\[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(sec(d*x+c)**3/(a+b*sin(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)**3/(a + b*sin(c + d*x))**(3/2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(sec(d*x+c)^3/(a+b*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a-4*b>0)', see `assume?` for
more detail

Giac [F]

\[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{3}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^3/(a+b*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^3/(b*sin(d*x + c) + a)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[In]

int(1/(cos(c + d*x)^3*(a + b*sin(c + d*x))^(3/2)),x)

[Out]

int(1/(cos(c + d*x)^3*(a + b*sin(c + d*x))^(3/2)), x)