3.50 \(\int (a+b \sin (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=256 \[ -\frac {2 b \left (71 a^2+25 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{105 d}+\frac {32 a \left (11 a^2+13 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{105 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 \left (71 a^4-46 a^2 b^2-25 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{105 d \sqrt {a+b \sin (c+d x)}}-\frac {2 b \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{7 d}-\frac {24 a b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 d} \]

[Out]

-24/35*a*b*cos(d*x+c)*(a+b*sin(d*x+c))^(3/2)/d-2/7*b*cos(d*x+c)*(a+b*sin(d*x+c))^(5/2)/d-2/105*b*(71*a^2+25*b^
2)*cos(d*x+c)*(a+b*sin(d*x+c))^(1/2)/d-32/105*a*(11*a^2+13*b^2)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+
1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*(a+b*sin(d*x+c))^(1/2)/d/((a+b*si
n(d*x+c))/(a+b))^(1/2)+2/105*(71*a^4-46*a^2*b^2-25*b^4)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1
/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*((a+b*sin(d*x+c))/(a+b))^(1/2)/d/(a+b*sin
(d*x+c))^(1/2)

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Rubi [A]  time = 0.38, antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2656, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac {2 b \left (71 a^2+25 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{105 d}-\frac {2 \left (-46 a^2 b^2+71 a^4-25 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{105 d \sqrt {a+b \sin (c+d x)}}+\frac {32 a \left (11 a^2+13 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{105 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 b \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{7 d}-\frac {24 a b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x])^(7/2),x]

[Out]

(-2*b*(71*a^2 + 25*b^2)*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(105*d) - (24*a*b*Cos[c + d*x]*(a + b*Sin[c + d
*x])^(3/2))/(35*d) - (2*b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(5/2))/(7*d) + (32*a*(11*a^2 + 13*b^2)*EllipticE[(
c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(105*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (2*(7
1*a^4 - 46*a^2*b^2 - 25*b^4)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/
(105*d*Sqrt[a + b*Sin[c + d*x]])

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2656

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n -
1))/(d*n), x] + Dist[1/n, Int[(a + b*Sin[c + d*x])^(n - 2)*Simp[a^2*n + b^2*(n - 1) + a*b*(2*n - 1)*Sin[c + d*
x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2752

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

Rule 2753

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int (a+b \sin (c+d x))^{7/2} \, dx &=-\frac {2 b \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{7 d}+\frac {2}{7} \int (a+b \sin (c+d x))^{3/2} \left (\frac {1}{2} \left (7 a^2+5 b^2\right )+6 a b \sin (c+d x)\right ) \, dx\\ &=-\frac {24 a b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 d}-\frac {2 b \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{7 d}+\frac {4}{35} \int \sqrt {a+b \sin (c+d x)} \left (\frac {1}{4} a \left (35 a^2+61 b^2\right )+\frac {1}{4} b \left (71 a^2+25 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac {2 b \left (71 a^2+25 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{105 d}-\frac {24 a b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 d}-\frac {2 b \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{7 d}+\frac {8}{105} \int \frac {\frac {1}{8} \left (105 a^4+254 a^2 b^2+25 b^4\right )+2 a b \left (11 a^2+13 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\\ &=-\frac {2 b \left (71 a^2+25 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{105 d}-\frac {24 a b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 d}-\frac {2 b \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{7 d}+\frac {1}{105} \left (16 a \left (11 a^2+13 b^2\right )\right ) \int \sqrt {a+b \sin (c+d x)} \, dx+\frac {1}{105} \left (-71 a^4+46 a^2 b^2+25 b^4\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx\\ &=-\frac {2 b \left (71 a^2+25 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{105 d}-\frac {24 a b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 d}-\frac {2 b \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{7 d}+\frac {\left (16 a \left (11 a^2+13 b^2\right ) \sqrt {a+b \sin (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}} \, dx}{105 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (\left (-71 a^4+46 a^2 b^2+25 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}} \, dx}{105 \sqrt {a+b \sin (c+d x)}}\\ &=-\frac {2 b \left (71 a^2+25 b^2\right ) \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{105 d}-\frac {24 a b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{35 d}-\frac {2 b \cos (c+d x) (a+b \sin (c+d x))^{5/2}}{7 d}+\frac {32 a \left (11 a^2+13 b^2\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{105 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 \left (71 a^4-46 a^2 b^2-25 b^4\right ) F\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{105 d \sqrt {a+b \sin (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 1.06, size = 220, normalized size = 0.86 \[ \frac {4 \left (71 a^4-46 a^2 b^2-25 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )-b \cos (c+d x) \left (488 a^3+b \left (752 a^2+145 b^2\right ) \sin (c+d x)-162 a b^2 \cos (2 (c+d x))+262 a b^2-15 b^3 \sin (3 (c+d x))\right )-64 a \left (11 a^3+11 a^2 b+13 a b^2+13 b^3\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} E\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )}{210 d \sqrt {a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])^(7/2),x]

[Out]

(-64*a*(11*a^3 + 11*a^2*b + 13*a*b^2 + 13*b^3)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin
[c + d*x])/(a + b)] + 4*(71*a^4 - 46*a^2*b^2 - 25*b^4)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a
+ b*Sin[c + d*x])/(a + b)] - b*Cos[c + d*x]*(488*a^3 + 262*a*b^2 - 162*a*b^2*Cos[2*(c + d*x)] + b*(752*a^2 +
145*b^2)*Sin[c + d*x] - 15*b^3*Sin[3*(c + d*x)]))/(210*d*Sqrt[a + b*Sin[c + d*x]])

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - 3 \, a b^{2} + {\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

integral(-(3*a*b^2*cos(d*x + c)^2 - a^3 - 3*a*b^2 + (b^3*cos(d*x + c)^2 - 3*a^2*b - b^3)*sin(d*x + c))*sqrt(b*
sin(d*x + c) + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^(7/2), x)

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maple [B]  time = 0.30, size = 1040, normalized size = 4.06 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x+c))^(7/2),x)

[Out]

2/105*(15*b^5*sin(d*x+c)^5+105*a^5*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x
+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))+71*((a+b*sin(d*x+c))/(a-b))^
(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2)
,((a-b)/(a+b))^(1/2))*a^4*b+78*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))
*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3*b^2-46*((a+b*sin(d*x+c))/(a-
b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(
1/2),((a-b)/(a+b))^(1/2))*a^2*b^3-183*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(
d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a*b^4-25*((a+b*sin(d*x+c)
)/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-
b))^(1/2),((a-b)/(a+b))^(1/2))*b^5-176*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin
(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^5-32*((a+b*sin(d*x+c))
/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b
))^(1/2),((a-b)/(a+b))^(1/2))*a^3*b^2+208*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+
sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a*b^4+81*a*b^4*sin(d*
x+c)^4+188*a^2*b^3*sin(d*x+c)^3+10*b^5*sin(d*x+c)^3+122*a^3*b^2*sin(d*x+c)^2-56*a*b^4*sin(d*x+c)^2-188*a^2*b^3
*sin(d*x+c)-25*b^5*sin(d*x+c)-122*a^3*b^2-25*a*b^4)/b/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+b\,\sin \left (c+d\,x\right )\right )}^{7/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x))^(7/2),x)

[Out]

int((a + b*sin(c + d*x))^(7/2), 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((a+b*sin(d*x+c))**(7/2),x)

[Out]

Timed out

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