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

Optimal. Leaf size=103 \[ -\frac{10 c^3 \cos (a+b x) \sqrt{c \sin (a+b x)}}{21 b}+\frac{10 c^4 \sqrt{\sin (a+b x)} F\left (\left .\frac{1}{2} \left (a+b x-\frac{\pi }{2}\right )\right |2\right )}{21 b \sqrt{c \sin (a+b x)}}-\frac{2 c \cos (a+b x) (c \sin (a+b x))^{5/2}}{7 b} \]

[Out]

(10*c^4*EllipticF[(a - Pi/2 + b*x)/2, 2]*Sqrt[Sin[a + b*x]])/(21*b*Sqrt[c*Sin[a + b*x]]) - (10*c^3*Cos[a + b*x
]*Sqrt[c*Sin[a + b*x]])/(21*b) - (2*c*Cos[a + b*x]*(c*Sin[a + b*x])^(5/2))/(7*b)

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Rubi [A]  time = 0.0523121, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2635, 2642, 2641} \[ -\frac{10 c^3 \cos (a+b x) \sqrt{c \sin (a+b x)}}{21 b}+\frac{10 c^4 \sqrt{\sin (a+b x)} F\left (\left .\frac{1}{2} \left (a+b x-\frac{\pi }{2}\right )\right |2\right )}{21 b \sqrt{c \sin (a+b x)}}-\frac{2 c \cos (a+b x) (c \sin (a+b x))^{5/2}}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x])^(7/2),x]

[Out]

(10*c^4*EllipticF[(a - Pi/2 + b*x)/2, 2]*Sqrt[Sin[a + b*x]])/(21*b*Sqrt[c*Sin[a + b*x]]) - (10*c^3*Cos[a + b*x
]*Sqrt[c*Sin[a + b*x]])/(21*b) - (2*c*Cos[a + b*x]*(c*Sin[a + b*x])^(5/2))/(7*b)

Rule 2635

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int (c \sin (a+b x))^{7/2} \, dx &=-\frac{2 c \cos (a+b x) (c \sin (a+b x))^{5/2}}{7 b}+\frac{1}{7} \left (5 c^2\right ) \int (c \sin (a+b x))^{3/2} \, dx\\ &=-\frac{10 c^3 \cos (a+b x) \sqrt{c \sin (a+b x)}}{21 b}-\frac{2 c \cos (a+b x) (c \sin (a+b x))^{5/2}}{7 b}+\frac{1}{21} \left (5 c^4\right ) \int \frac{1}{\sqrt{c \sin (a+b x)}} \, dx\\ &=-\frac{10 c^3 \cos (a+b x) \sqrt{c \sin (a+b x)}}{21 b}-\frac{2 c \cos (a+b x) (c \sin (a+b x))^{5/2}}{7 b}+\frac{\left (5 c^4 \sqrt{\sin (a+b x)}\right ) \int \frac{1}{\sqrt{\sin (a+b x)}} \, dx}{21 \sqrt{c \sin (a+b x)}}\\ &=\frac{10 c^4 F\left (\left .\frac{1}{2} \left (a-\frac{\pi }{2}+b x\right )\right |2\right ) \sqrt{\sin (a+b x)}}{21 b \sqrt{c \sin (a+b x)}}-\frac{10 c^3 \cos (a+b x) \sqrt{c \sin (a+b x)}}{21 b}-\frac{2 c \cos (a+b x) (c \sin (a+b x))^{5/2}}{7 b}\\ \end{align*}

Mathematica [A]  time = 0.153076, size = 80, normalized size = 0.78 \[ \frac{c^3 \sqrt{c \sin (a+b x)} \left (\sqrt{\sin (a+b x)} (3 \cos (3 (a+b x))-23 \cos (a+b x))-20 F\left (\left .\frac{1}{4} (-2 a-2 b x+\pi )\right |2\right )\right )}{42 b \sqrt{\sin (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Sin[a + b*x])^(7/2),x]

[Out]

(c^3*(-20*EllipticF[(-2*a + Pi - 2*b*x)/4, 2] + (-23*Cos[a + b*x] + 3*Cos[3*(a + b*x)])*Sqrt[Sin[a + b*x]])*Sq
rt[c*Sin[a + b*x]])/(42*b*Sqrt[Sin[a + b*x]])

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Maple [A]  time = 0.042, size = 108, normalized size = 1.1 \begin{align*} -{\frac{{c}^{4}}{21\,b\cos \left ( bx+a \right ) } \left ( -6\, \left ( \sin \left ( bx+a \right ) \right ) ^{5}+5\,\sqrt{-\sin \left ( bx+a \right ) +1}\sqrt{2\,\sin \left ( bx+a \right ) +2}\sqrt{\sin \left ( bx+a \right ) }{\it EllipticF} \left ( \sqrt{-\sin \left ( bx+a \right ) +1},1/2\,\sqrt{2} \right ) -4\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}+10\,\sin \left ( bx+a \right ) \right ){\frac{1}{\sqrt{c\sin \left ( bx+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/21*c^4*(-6*sin(b*x+a)^5+5*(-sin(b*x+a)+1)^(1/2)*(2*sin(b*x+a)+2)^(1/2)*sin(b*x+a)^(1/2)*EllipticF((-sin(b*x
+a)+1)^(1/2),1/2*2^(1/2))-4*sin(b*x+a)^3+10*sin(b*x+a))/cos(b*x+a)/(c*sin(b*x+a))^(1/2)/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \sin \left (b x + a\right )\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*sin(b*x + a))^(7/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (c^{3} \cos \left (b x + a\right )^{2} - c^{3}\right )} \sqrt{c \sin \left (b x + a\right )} \sin \left (b x + a\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(c^3*cos(b*x + a)^2 - c^3)*sqrt(c*sin(b*x + a))*sin(b*x + a), 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((c*sin(b*x+a))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \sin \left (b x + a\right )\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*sin(b*x + a))^(7/2), x)