3.2.43 \(\int (a+b \sin ^2(c+d x))^{5/2} \, dx\) [143]

Optimal. Leaf size=210 \[ -\frac {4 b (2 a+b) \cos (c+d x) \sin (c+d x) \sqrt {a+b \sin ^2(c+d x)}}{15 d}-\frac {b \cos (c+d x) \sin (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}+\frac {\left (23 a^2+23 a b+8 b^2\right ) E\left (c+d x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(c+d x)}}{15 d \sqrt {1+\frac {b \sin ^2(c+d x)}{a}}}-\frac {4 a (a+b) (2 a+b) F\left (c+d x\left |-\frac {b}{a}\right .\right ) \sqrt {1+\frac {b \sin ^2(c+d x)}{a}}}{15 d \sqrt {a+b \sin ^2(c+d x)}} \]

[Out]

-1/5*b*cos(d*x+c)*sin(d*x+c)*(a+b*sin(d*x+c)^2)^(3/2)/d-4/15*b*(2*a+b)*cos(d*x+c)*sin(d*x+c)*(a+b*sin(d*x+c)^2
)^(1/2)/d+1/15*(23*a^2+23*a*b+8*b^2)*(cos(d*x+c)^2)^(1/2)/cos(d*x+c)*EllipticE(sin(d*x+c),(-b/a)^(1/2))*(a+b*s
in(d*x+c)^2)^(1/2)/d/(1+b*sin(d*x+c)^2/a)^(1/2)-4/15*a*(a+b)*(2*a+b)*(cos(d*x+c)^2)^(1/2)/cos(d*x+c)*EllipticF
(sin(d*x+c),(-b/a)^(1/2))*(1+b*sin(d*x+c)^2/a)^(1/2)/d/(a+b*sin(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.19, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {3259, 3249, 3251, 3257, 3256, 3262, 3261} \begin {gather*} \frac {\left (23 a^2+23 a b+8 b^2\right ) \sqrt {a+b \sin ^2(c+d x)} E\left (c+d x\left |-\frac {b}{a}\right .\right )}{15 d \sqrt {\frac {b \sin ^2(c+d x)}{a}+1}}-\frac {b \sin (c+d x) \cos (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}-\frac {4 b (2 a+b) \sin (c+d x) \cos (c+d x) \sqrt {a+b \sin ^2(c+d x)}}{15 d}-\frac {4 a (a+b) (2 a+b) \sqrt {\frac {b \sin ^2(c+d x)}{a}+1} F\left (c+d x\left |-\frac {b}{a}\right .\right )}{15 d \sqrt {a+b \sin ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x]^2)^(5/2),x]

[Out]

(-4*b*(2*a + b)*Cos[c + d*x]*Sin[c + d*x]*Sqrt[a + b*Sin[c + d*x]^2])/(15*d) - (b*Cos[c + d*x]*Sin[c + d*x]*(a
 + b*Sin[c + d*x]^2)^(3/2))/(5*d) + ((23*a^2 + 23*a*b + 8*b^2)*EllipticE[c + d*x, -(b/a)]*Sqrt[a + b*Sin[c + d
*x]^2])/(15*d*Sqrt[1 + (b*Sin[c + d*x]^2)/a]) - (4*a*(a + b)*(2*a + b)*EllipticF[c + d*x, -(b/a)]*Sqrt[1 + (b*
Sin[c + d*x]^2)/a])/(15*d*Sqrt[a + b*Sin[c + d*x]^2])

Rule 3249

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp
[(-B)*Cos[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x]^2)^p/(2*f*(p + 1))), x] + Dist[1/(2*(p + 1)), Int[(a + b*
Sin[e + f*x]^2)^(p - 1)*Simp[a*B + 2*a*A*(p + 1) + (2*A*b*(p + 1) + B*(b + 2*a*p + 2*b*p))*Sin[e + f*x]^2, x],
 x], x] /; FreeQ[{a, b, e, f, A, B}, x] && GtQ[p, 0]

Rule 3251

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[
B/b, Int[Sqrt[a + b*Sin[e + f*x]^2], x], x] + Dist[(A*b - a*B)/b, Int[1/Sqrt[a + b*Sin[e + f*x]^2], x], x] /;
FreeQ[{a, b, e, f, A, B}, x]

Rule 3256

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(Sqrt[a]/f)*EllipticE[e + f*x, -b/a], x] /
; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rule 3257

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[1 + b*(Sin
[e + f*x]^2/a)], Int[Sqrt[1 + (b*Sin[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] &&  !GtQ[a, 0]

Rule 3259

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si
n[e + f*x]^2)^(p - 1)/(2*f*p)), x] + Dist[1/(2*p), Int[(a + b*Sin[e + f*x]^2)^(p - 2)*Simp[a*(b + 2*a*p) + b*(
2*a + b)*(2*p - 1)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a + b, 0] && GtQ[p, 1]

Rule 3261

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1/(Sqrt[a]*f))*EllipticF[e + f*x, -b/a]
, x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rule 3262

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[Sqrt[1 + b*(Sin[e + f*x]^2/a)]/Sqrt[a +
b*Sin[e + f*x]^2], Int[1/Sqrt[1 + (b*Sin[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int \left (a+b \sin ^2(c+d x)\right )^{5/2} \, dx &=-\frac {b \cos (c+d x) \sin (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}+\frac {1}{5} \int \sqrt {a+b \sin ^2(c+d x)} \left (a (5 a+b)+4 b (2 a+b) \sin ^2(c+d x)\right ) \, dx\\ &=-\frac {4 b (2 a+b) \cos (c+d x) \sin (c+d x) \sqrt {a+b \sin ^2(c+d x)}}{15 d}-\frac {b \cos (c+d x) \sin (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}+\frac {1}{15} \int \frac {a \left (15 a^2+11 a b+4 b^2\right )+b \left (23 a^2+23 a b+8 b^2\right ) \sin ^2(c+d x)}{\sqrt {a+b \sin ^2(c+d x)}} \, dx\\ &=-\frac {4 b (2 a+b) \cos (c+d x) \sin (c+d x) \sqrt {a+b \sin ^2(c+d x)}}{15 d}-\frac {b \cos (c+d x) \sin (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}-\frac {1}{15} (4 a (a+b) (2 a+b)) \int \frac {1}{\sqrt {a+b \sin ^2(c+d x)}} \, dx+\frac {1}{15} \left (23 a^2+23 a b+8 b^2\right ) \int \sqrt {a+b \sin ^2(c+d x)} \, dx\\ &=-\frac {4 b (2 a+b) \cos (c+d x) \sin (c+d x) \sqrt {a+b \sin ^2(c+d x)}}{15 d}-\frac {b \cos (c+d x) \sin (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}+\frac {\left (\left (23 a^2+23 a b+8 b^2\right ) \sqrt {a+b \sin ^2(c+d x)}\right ) \int \sqrt {1+\frac {b \sin ^2(c+d x)}{a}} \, dx}{15 \sqrt {1+\frac {b \sin ^2(c+d x)}{a}}}-\frac {\left (4 a (a+b) (2 a+b) \sqrt {1+\frac {b \sin ^2(c+d x)}{a}}\right ) \int \frac {1}{\sqrt {1+\frac {b \sin ^2(c+d x)}{a}}} \, dx}{15 \sqrt {a+b \sin ^2(c+d x)}}\\ &=-\frac {4 b (2 a+b) \cos (c+d x) \sin (c+d x) \sqrt {a+b \sin ^2(c+d x)}}{15 d}-\frac {b \cos (c+d x) \sin (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}+\frac {\left (23 a^2+23 a b+8 b^2\right ) E\left (c+d x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(c+d x)}}{15 d \sqrt {1+\frac {b \sin ^2(c+d x)}{a}}}-\frac {4 a (a+b) (2 a+b) F\left (c+d x\left |-\frac {b}{a}\right .\right ) \sqrt {1+\frac {b \sin ^2(c+d x)}{a}}}{15 d \sqrt {a+b \sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.97, size = 194, normalized size = 0.92 \begin {gather*} \frac {16 a \left (23 a^2+23 a b+8 b^2\right ) \sqrt {\frac {2 a+b-b \cos (2 (c+d x))}{a}} E\left (c+d x\left |-\frac {b}{a}\right .\right )-64 a \left (2 a^2+3 a b+b^2\right ) \sqrt {\frac {2 a+b-b \cos (2 (c+d x))}{a}} F\left (c+d x\left |-\frac {b}{a}\right .\right )-\sqrt {2} b \left (88 a^2+88 a b+25 b^2-28 b (2 a+b) \cos (2 (c+d x))+3 b^2 \cos (4 (c+d x))\right ) \sin (2 (c+d x))}{240 d \sqrt {2 a+b-b \cos (2 (c+d x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(16*a*(23*a^2 + 23*a*b + 8*b^2)*Sqrt[(2*a + b - b*Cos[2*(c + d*x)])/a]*EllipticE[c + d*x, -(b/a)] - 64*a*(2*a^
2 + 3*a*b + b^2)*Sqrt[(2*a + b - b*Cos[2*(c + d*x)])/a]*EllipticF[c + d*x, -(b/a)] - Sqrt[2]*b*(88*a^2 + 88*a*
b + 25*b^2 - 28*b*(2*a + b)*Cos[2*(c + d*x)] + 3*b^2*Cos[4*(c + d*x)])*Sin[2*(c + d*x)])/(240*d*Sqrt[2*a + b -
 b*Cos[2*(c + d*x)]])

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Maple [A]
time = 7.26, size = 437, normalized size = 2.08

method result size
default \(\frac {-\frac {b^{3} \sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{5}+\frac {\left (14 a \,b^{2}+10 b^{3}\right ) \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}+\frac {\left (-11 a^{2} b -18 a \,b^{2}-7 b^{3}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}-\frac {8 \sqrt {\frac {\cos \left (2 d x +2 c \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (d x +c \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticF \left (\sin \left (d x +c \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}}{15}-\frac {4 a^{2} \sqrt {\frac {\cos \left (2 d x +2 c \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (d x +c \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticF \left (\sin \left (d x +c \right ), \sqrt {-\frac {b}{a}}\right ) b}{5}-\frac {4 a \sqrt {\frac {\cos \left (2 d x +2 c \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (d x +c \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticF \left (\sin \left (d x +c \right ), \sqrt {-\frac {b}{a}}\right ) b^{2}}{15}+\frac {23 \sqrt {\frac {\cos \left (2 d x +2 c \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (d x +c \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticE \left (\sin \left (d x +c \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}}{15}+\frac {23 \sqrt {\frac {\cos \left (2 d x +2 c \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (d x +c \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticE \left (\sin \left (d x +c \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b}{15}+\frac {8 \sqrt {\frac {\cos \left (2 d x +2 c \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (d x +c \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticE \left (\sin \left (d x +c \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{2}}{15}}{\cos \left (d x +c \right ) \sqrt {a +\left (\sin ^{2}\left (d x +c \right )\right ) b}\, d}\) \(437\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/5*b^3*sin(d*x+c)*cos(d*x+c)^6+1/15*(14*a*b^2+10*b^3)*cos(d*x+c)^4*sin(d*x+c)+1/15*(-11*a^2*b-18*a*b^2-7*b^
3)*cos(d*x+c)^2*sin(d*x+c)-8/15*(cos(d*x+c)^2)^(1/2)*(-b/a*cos(d*x+c)^2+(a+b)/a)^(1/2)*EllipticF(sin(d*x+c),(-
1/a*b)^(1/2))*a^3-4/5*a^2*(cos(d*x+c)^2)^(1/2)*(-b/a*cos(d*x+c)^2+(a+b)/a)^(1/2)*EllipticF(sin(d*x+c),(-1/a*b)
^(1/2))*b-4/15*a*(cos(d*x+c)^2)^(1/2)*(-b/a*cos(d*x+c)^2+(a+b)/a)^(1/2)*EllipticF(sin(d*x+c),(-1/a*b)^(1/2))*b
^2+23/15*(cos(d*x+c)^2)^(1/2)*(-b/a*cos(d*x+c)^2+(a+b)/a)^(1/2)*EllipticE(sin(d*x+c),(-1/a*b)^(1/2))*a^3+23/15
*(cos(d*x+c)^2)^(1/2)*(-b/a*cos(d*x+c)^2+(a+b)/a)^(1/2)*EllipticE(sin(d*x+c),(-1/a*b)^(1/2))*a^2*b+8/15*(cos(d
*x+c)^2)^(1/2)*(-b/a*cos(d*x+c)^2+(a+b)/a)^(1/2)*EllipticE(sin(d*x+c),(-1/a*b)^(1/2))*a*b^2)/cos(d*x+c)/(a+sin
(d*x+c)^2*b)^(1/2)/d

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.12, size = 59, normalized size = 0.28 \begin {gather*} {\rm integral}\left ({\left (b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-b \cos \left (d x + c\right )^{2} + a + b}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)*sqrt(-b*cos(d*x + c)^2 + a +
b), x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3061 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (b\,{\sin \left (c+d\,x\right )}^2+a\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x)^2)^(5/2),x)

[Out]

int((a + b*sin(c + d*x)^2)^(5/2), x)

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