3.143 \(\int (a+b \sin ^2(c+d x))^{5/2} \, dx\)

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

[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])

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Rubi [A]  time = 0.281686, antiderivative size = 210, normalized size of antiderivative = 1., 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 = {3180, 3170, 3172, 3178, 3177, 3183, 3182} \[ \frac{\left (23 a^2+23 a b+8 b^2\right ) \sqrt{a+b \sin ^2(c+d x)} E\left (c+d x\left |-\frac{b}{a}\right .\right )}{15 d \sqrt{\frac{b \sin ^2(c+d x)}{a}+1}}-\frac{b \sin (c+d x) \cos (c+d x) \left (a+b \sin ^2(c+d x)\right )^{3/2}}{5 d}-\frac{4 b (2 a+b) \sin (c+d x) \cos (c+d x) \sqrt{a+b \sin ^2(c+d x)}}{15 d}-\frac{4 a (a+b) (2 a+b) \sqrt{\frac{b \sin ^2(c+d x)}{a}+1} F\left (c+d x\left |-\frac{b}{a}\right .\right )}{15 d \sqrt{a+b \sin ^2(c+d x)}} \]

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 3180

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> -Simp[(b*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[
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 3170

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Sim
p[(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*Si
n[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 3172

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 3178

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 3177

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

Rule 3183

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]

Rule 3182

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

Mathematica [A]  time = 1.44033, size = 194, normalized size = 0.92 \[ \frac{-\sqrt{2} b \sin (2 (c+d x)) \left (88 a^2-28 b (2 a+b) \cos (2 (c+d x))+88 a b+3 b^2 \cos (4 (c+d x))+25 b^2\right )-64 a \left (2 a^2+3 a b+b^2\right ) \sqrt{\frac{2 a-b \cos (2 (c+d x))+b}{a}} F\left (c+d x\left |-\frac{b}{a}\right .\right )+16 a \left (23 a^2+23 a b+8 b^2\right ) \sqrt{\frac{2 a-b \cos (2 (c+d x))+b}{a}} E\left (c+d x\left |-\frac{b}{a}\right .\right )}{240 d \sqrt{2 a-b \cos (2 (c+d x))+b}} \]

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 = 1.256, size = 437, normalized size = 2.1 \begin{align*}{\frac{1}{d\cos \left ( dx+c \right ) } \left ( -{\frac{{b}^{3}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{5}}+{\frac{ \left ( 14\,a{b}^{2}+10\,{b}^{3} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}\sin \left ( dx+c \right ) }{15}}+{\frac{ \left ( -11\,{a}^{2}b-18\,a{b}^{2}-7\,{b}^{3} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) }{15}}-{\frac{8\,{a}^{3}}{15}\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}}\sqrt{-{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{a}}+{\frac{a+b}{a}}}{\it EllipticF} \left ( \sin \left ( dx+c \right ) ,\sqrt{-{\frac{b}{a}}} \right ) }-{\frac{4\,{a}^{2}b}{5}\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}}\sqrt{-{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{a}}+{\frac{a+b}{a}}}{\it EllipticF} \left ( \sin \left ( dx+c \right ) ,\sqrt{-{\frac{b}{a}}} \right ) }-{\frac{4\,a{b}^{2}}{15}\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}}\sqrt{-{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{a}}+{\frac{a+b}{a}}}{\it EllipticF} \left ( \sin \left ( dx+c \right ) ,\sqrt{-{\frac{b}{a}}} \right ) }+{\frac{23\,{a}^{3}}{15}\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}}\sqrt{-{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{a}}+{\frac{a+b}{a}}}{\it EllipticE} \left ( \sin \left ( dx+c \right ) ,\sqrt{-{\frac{b}{a}}} \right ) }+{\frac{23\,{a}^{2}b}{15}\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}}\sqrt{-{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{a}}+{\frac{a+b}{a}}}{\it EllipticE} \left ( \sin \left ( dx+c \right ) ,\sqrt{-{\frac{b}{a}}} \right ) }+{\frac{8\,a{b}^{2}}{15}\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}}\sqrt{-{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{a}}+{\frac{a+b}{a}}}{\it EllipticE} \left ( \sin \left ( dx+c \right ) ,\sqrt{-{\frac{b}{a}}} \right ) } \right ){\frac{1}{\sqrt{a+ \left ( \sin \left ( dx+c \right ) \right ) ^{2}b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right )^{2} + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\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{align*}

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right )^{2} + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}

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)