3.499 \(\int \sec ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx\)
Optimal. Leaf size=199 \[ -\frac {3 \sqrt {a-b} \left (4 a^2+2 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 d}+\frac {3 \sqrt {a+b} \left (4 a^2-2 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 d}+\frac {3 \sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (\left (2 a^2-b^2\right ) \sin (c+d x)+a b\right )}{16 d}+\frac {\sec ^4(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{4 d} \]
[Out]
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
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Rubi [A] time = 0.28, antiderivative size = 199, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used =
{2668, 739, 821, 827, 1166, 206} \[ -\frac {3 \sqrt {a-b} \left (4 a^2+2 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 d}+\frac {3 \sqrt {a+b} \left (4 a^2-2 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 d}+\frac {3 \sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (\left (2 a^2-b^2\right ) \sin (c+d x)+a b\right )}{16 d}+\frac {\sec ^4(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{4 d} \]
Antiderivative was successfully verified.
[In]
Int[Sec[c + d*x]^5*(a + b*Sin[c + d*x])^(5/2),x]
[Out]
(-3*Sqrt[a - b]*(4*a^2 + 2*a*b - b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(32*d) + (3*Sqrt[a + b]*(
4*a^2 - 2*a*b - b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(32*d) + (Sec[c + d*x]^4*(b + a*Sin[c + d*
x])*(a + b*Sin[c + d*x])^(3/2))/(4*d) + (3*Sec[c + d*x]^2*Sqrt[a + b*Sin[c + d*x]]*(a*b + (2*a^2 - b^2)*Sin[c
+ d*x]))/(16*d)
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 739
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 821
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] - Dist[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
] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Rule 827
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 1166
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 2668
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*} \int \sec ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {(a+x)^{5/2}}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\sec ^4(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{4 d}-\frac {b^3 \operatorname {Subst}\left (\int \frac {\sqrt {a+x} \left (-\frac {3}{2} \left (2 a^2-b^2\right )-\frac {3 a x}{2}\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 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}+\frac {b \operatorname {Subst}\left (\int \frac {\frac {3}{4} a \left (4 a^2-3 b^2\right )+\frac {3}{4} \left (2 a^2-b^2\right ) x}{\sqrt {a+x} \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 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}+\frac {b \operatorname {Subst}\left (\int \frac {\frac {3}{4} a \left (4 a^2-3 b^2\right )-\frac {3}{4} a \left (2 a^2-b^2\right )+\frac {3}{4} \left (2 a^2-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 )}{4 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}+\frac {\left (3 (a+b) \left (4 a^2-2 a b-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{32 d}-\frac {\left (3 (a-b) \left (4 a^2+2 a b-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a-b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{32 d}\\ &=-\frac {3 \sqrt {a-b} \left (4 a^2+2 a b-b^2\right ) \tanh ^{-1}\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 ) \tanh ^{-1}\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}\\ \end {align*}
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Mathematica [A] time = 3.38, size = 307, normalized size = 1.54 \[ -\frac {3 \sqrt {a-b} \left (a^2-b^2\right )^2 \left (4 a^2+2 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )-3 \sqrt {a+b} \left (a^2-b^2\right )^2 \left (4 a^2-2 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )+8 \left (b^2-a^2\right ) \sec ^4(c+d x) (a \sin (c+d x)-b) (a+b \sin (c+d x))^{7/2}-2 \sec ^2(c+d x) \left (6 a^3 \sin (c+d x)-7 a^2 b+b^3\right ) (a+b \sin (c+d x))^{7/2}-2 b \sqrt {a+b \sin (c+d x)} \left (18 a^5-3 a^3 b^2 \cos (2 (c+d x))-16 a^3 b^2+b \left (18 a^4-7 a^2 b^2+b^4\right ) \sin (c+d x)+7 a b^4\right )}{32 d \left (a^2-b^2\right )^2} \]
Antiderivative was successfully verified.
[In]
Integrate[Sec[c + d*x]^5*(a + b*Sin[c + d*x])^(5/2),x]
[Out]
-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*Sqr
t[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)*Se
c[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)
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fricas [B] time = 1.53, size = 2229, normalized size = 11.20 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
[-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 + 32
0*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*(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 + (3*a^2*b + 4*a*b^2 + b^3)*sin(d*x + c)))*cos(d*x + c)^4 + 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 - 2
56*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*co
s(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 + (3*a^2*b - 4*a*b^2 + b^3)*sin(d*x + c)))*cos(d*x + c)^4 + 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), -1/128*(3*(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 + (3*a^2*b - 4*a*b^2 + b^3)*sin(d*x + c)))*cos(d*x + c)^4 + 3*(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))*sqr
t(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)^4 + 8*(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)]
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")
[Out]
Timed out
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maple [B] time = 1.79, size = 538, normalized size = 2.70 \[ \frac {4 \sqrt {-a +b}\, \sqrt {a +b}\, \sqrt {a +b \sin \left (d x +c \right )}\, b \left (3 a b \left (\cos ^{2}\left (d x +c \right )\right )+8 a^{2} \sin \left (d x +c \right )-b^{2} \sin \left (d x +c \right )-3 a b \right )+3 b \left (4 \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a^{3} \sqrt {-a +b}+2 b \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a^{2} \sqrt {-a +b}-3 b^{2} \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a \sqrt {-a +b}-b^{3} \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) \sqrt {-a +b}+4 \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a^{3} \sqrt {a +b}-2 b \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a^{2} \sqrt {a +b}-3 b^{2} \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a \sqrt {a +b}+b^{3} \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) \sqrt {a +b}\right ) \left (\cos ^{4}\left (d x +c \right )\right )+2 \sqrt {-a +b}\, \sqrt {a +b}\, \sqrt {a +b \sin \left (d x +c \right )}\, b \left (6 a^{2} \sin \left (d x +c \right )-3 b^{2} \sin \left (d x +c \right )-7 a b \right ) \left (\cos ^{2}\left (d x +c \right )\right )-24 \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} a^{2} \sqrt {-a +b}\, \sqrt {a +b}+12 \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} b^{2} \sqrt {-a +b}\, \sqrt {a +b}+24 \sqrt {a +b \sin \left (d x +c \right )}\, a^{3} \sqrt {-a +b}\, \sqrt {a +b}+16 a \sqrt {a +b \sin \left (d x +c \right )}\, b^{2} \sqrt {-a +b}\, \sqrt {a +b}}{32 \sqrt {-a +b}\, \sqrt {a +b}\, b \cos \left (d x +c \right )^{4} d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)^5*(a+b*sin(d*x+c))^(5/2),x)
[Out]
1/32*(4*(-a+b)^(1/2)*(a+b)^(1/2)*(a+b*sin(d*x+c))^(1/2)*b*(3*a*b*cos(d*x+c)^2+8*a^2*sin(d*x+c)-b^2*sin(d*x+c)-
3*a*b)+3*b*(4*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a^3*(-a+b)^(1/2)+2*b*arctanh((a+b*sin(d*x+c))^(1/2)/
(a+b)^(1/2))*a^2*(-a+b)^(1/2)-3*b^2*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a*(-a+b)^(1/2)-b^3*arctanh((a+
b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*(-a+b)^(1/2)+4*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*a^3*(a+b)^(1/2)-2*
b*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*a^2*(a+b)^(1/2)-3*b^2*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2)
)*a*(a+b)^(1/2)+b^3*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*(a+b)^(1/2))*cos(d*x+c)^4+2*(-a+b)^(1/2)*(a+b)
^(1/2)*(a+b*sin(d*x+c))^(1/2)*b*(6*a^2*sin(d*x+c)-3*b^2*sin(d*x+c)-7*a*b)*cos(d*x+c)^2-24*(a+b*sin(d*x+c))^(3/
2)*a^2*(-a+b)^(1/2)*(a+b)^(1/2)+12*(a+b*sin(d*x+c))^(3/2)*b^2*(-a+b)^(1/2)*(a+b)^(1/2)+24*(a+b*sin(d*x+c))^(1/
2)*a^3*(-a+b)^(1/2)*(a+b)^(1/2)+16*a*(a+b*sin(d*x+c))^(1/2)*b^2*(-a+b)^(1/2)*(a+b)^(1/2))/(-a+b)^(1/2)/(a+b)^(
1/2)/b/cos(d*x+c)^4/d
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^5*(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 details)Is 4*a-4*b positive or negative?
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^5} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*sin(c + d*x))^(5/2)/cos(c + d*x)^5,x)
[Out]
int((a + b*sin(c + d*x))^(5/2)/cos(c + d*x)^5, x)
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)**5*(a+b*sin(d*x+c))**(5/2),x)
[Out]
Timed out
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