\(\int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx\) [498]

Optimal result

Integrand size = 23, antiderivative size = 155 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=-\frac {(a-b)^{3/2} (2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{4 d}+\frac {(2 a-3 b) (a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{4 d}+\frac {a b \sqrt {a+b \sin (c+d x)}}{2 d}+\frac {\sec ^2(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{2 d} \]

[Out]

-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

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 155, 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, 753, 839, 841, 1180, 212} \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=-\frac {(a-b)^{3/2} (2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{4 d}+\frac {(2 a-3 b) (a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{4 d}+\frac {a b \sqrt {a+b \sin (c+d x)}}{2 d}+\frac {\sec ^2(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{2 d} \]

[In]

Int[Sec[c + d*x]^3*(a + b*Sin[c + d*x])^(5/2),x]

[Out]

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

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 753

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*c*(p + 1))), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 839

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[g*((d + e*x)^m/(
c*m)), x] + Dist[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] /
; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && FractionQ[m] && GtQ[m, 0]

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

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.95 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\frac {-\sqrt {a-b} \left (2 a^2+a b-3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )+\sqrt {a+b} \left (2 a^2-a b-3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )+2 \sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (2 a b+\left (a^2+b^2\right ) \sin (c+d x)\right )}{4 d} \]

[In]

Integrate[Sec[c + d*x]^3*(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(-(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)

Maple [A] (verified)

Time = 34.96 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.03

method	result	size
default	\(\frac {2 b^{3} \left (\frac {\left (a -b \right )^{2} \left (-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{2 \left (b \sin \left (d x +c \right )+b \right )}+\frac {\left (2 a +3 b \right ) \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{2 \sqrt {-a +b}}\right )}{4 b^{3}}-\frac {\left (a +b \right )^{2} \left (\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{2 b \sin \left (d x +c \right )-2 b}-\frac {\left (2 a -3 b \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{2 \sqrt {a +b}}\right )}{4 b^{3}}\right )}{d}\)	\(160\)

[In]

int(sec(d*x+c)^3*(a+b*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

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 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Too large to display} \]

[In]

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[-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)*c
os(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)*sqr
t(-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^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^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), -1/16*((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^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 - 8*(2*a*b + (a^2 + b^2)*sin(d*x
+ c))*sqrt(b*sin(d*x + c) + a))/(d*cos(d*x + c)^2)]

Sympy [F(-1)]

Timed out. \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**3*(a+b*sin(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^(5/2),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(-1)]

Timed out. \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \]

[In]

int((a + b*sin(c + d*x))^(5/2)/cos(c + d*x)^3,x)

[Out]

int((a + b*sin(c + d*x))^(5/2)/cos(c + d*x)^3, x)