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

Optimal. Leaf size=126 \[ \frac{20 d^3 \sin (a+b x) \sqrt{d \cos (a+b x)}}{231 b}+\frac{20 d^4 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{231 b \sqrt{d \cos (a+b x)}}-\frac{2 \sin (a+b x) (d \cos (a+b x))^{9/2}}{11 b d}+\frac{4 d \sin (a+b x) (d \cos (a+b x))^{5/2}}{77 b} \]

[Out]

(20*d^4*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2])/(231*b*Sqrt[d*Cos[a + b*x]]) + (20*d^3*Sqrt[d*Cos[a + b*
x]]*Sin[a + b*x])/(231*b) + (4*d*(d*Cos[a + b*x])^(5/2)*Sin[a + b*x])/(77*b) - (2*(d*Cos[a + b*x])^(9/2)*Sin[a
+ b*x])/(11*b*d)

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Rubi [A]  time = 0.0977436, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2568, 2635, 2642, 2641} \[ \frac{20 d^3 \sin (a+b x) \sqrt{d \cos (a+b x)}}{231 b}+\frac{20 d^4 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{231 b \sqrt{d \cos (a+b x)}}-\frac{2 \sin (a+b x) (d \cos (a+b x))^{9/2}}{11 b d}+\frac{4 d \sin (a+b x) (d \cos (a+b x))^{5/2}}{77 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(20*d^4*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2])/(231*b*Sqrt[d*Cos[a + b*x]]) + (20*d^3*Sqrt[d*Cos[a + b*
x]]*Sin[a + b*x])/(231*b) + (4*d*(d*Cos[a + b*x])^(5/2)*Sin[a + b*x])/(77*b) - (2*(d*Cos[a + b*x])^(9/2)*Sin[a
+ b*x])/(11*b*d)

Rule 2568

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

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 (d \cos (a+b x))^{7/2} \sin ^2(a+b x) \, dx &=-\frac{2 (d \cos (a+b x))^{9/2} \sin (a+b x)}{11 b d}+\frac{2}{11} \int (d \cos (a+b x))^{7/2} \, dx\\ &=\frac{4 d (d \cos (a+b x))^{5/2} \sin (a+b x)}{77 b}-\frac{2 (d \cos (a+b x))^{9/2} \sin (a+b x)}{11 b d}+\frac{1}{77} \left (10 d^2\right ) \int (d \cos (a+b x))^{3/2} \, dx\\ &=\frac{20 d^3 \sqrt{d \cos (a+b x)} \sin (a+b x)}{231 b}+\frac{4 d (d \cos (a+b x))^{5/2} \sin (a+b x)}{77 b}-\frac{2 (d \cos (a+b x))^{9/2} \sin (a+b x)}{11 b d}+\frac{1}{231} \left (10 d^4\right ) \int \frac{1}{\sqrt{d \cos (a+b x)}} \, dx\\ &=\frac{20 d^3 \sqrt{d \cos (a+b x)} \sin (a+b x)}{231 b}+\frac{4 d (d \cos (a+b x))^{5/2} \sin (a+b x)}{77 b}-\frac{2 (d \cos (a+b x))^{9/2} \sin (a+b x)}{11 b d}+\frac{\left (10 d^4 \sqrt{\cos (a+b x)}\right ) \int \frac{1}{\sqrt{\cos (a+b x)}} \, dx}{231 \sqrt{d \cos (a+b x)}}\\ &=\frac{20 d^4 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{231 b \sqrt{d \cos (a+b x)}}+\frac{20 d^3 \sqrt{d \cos (a+b x)} \sin (a+b x)}{231 b}+\frac{4 d (d \cos (a+b x))^{5/2} \sin (a+b x)}{77 b}-\frac{2 (d \cos (a+b x))^{9/2} \sin (a+b x)}{11 b d}\\ \end{align*}

Mathematica [C]  time = 0.133755, size = 60, normalized size = 0.48 \[ \frac{d^2 \cos ^2(a+b x)^{3/4} \tan ^3(a+b x) (d \cos (a+b x))^{3/2} \, _2F_1\left (-\frac{5}{4},\frac{3}{2};\frac{5}{2};\sin ^2(a+b x)\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(d*Cos[a + b*x])^(3/2)*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[-5/4, 3/2, 5/2, Sin[a + b*x]^2]*Tan[a + b
*x]^3)/(3*b)

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Maple [A]  time = 0.059, size = 236, normalized size = 1.9 \begin{align*}{\frac{4\,{d}^{4}}{231\,b}\sqrt{d \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}} \left ( 672\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{13}-2352\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{11}+3312\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{9}-2400\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{7}+922\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{5}-159\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{3}-5\,\sqrt{ \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,bx+a/2 \right ) ,\sqrt{2} \right ) +5\,\cos \left ( 1/2\,bx+a/2 \right ) \right ){\frac{1}{\sqrt{-d \left ( 2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}- \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2} \right ) }}} \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{d \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/231*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*d^4*(672*cos(1/2*b*x+1/2*a)^13-2352*cos(1/2*b*
x+1/2*a)^11+3312*cos(1/2*b*x+1/2*a)^9-2400*cos(1/2*b*x+1/2*a)^7+922*cos(1/2*b*x+1/2*a)^5-159*cos(1/2*b*x+1/2*a
)^3-5*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(-2*cos(1/2*b*x+1/2*a)^2+1)^(1/2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2))+5*c
os(1/2*b*x+1/2*a))/(-d*(2*sin(1/2*b*x+1/2*a)^4-sin(1/2*b*x+1/2*a)^2))^(1/2)/sin(1/2*b*x+1/2*a)/(d*(2*cos(1/2*b
*x+1/2*a)^2-1))^(1/2)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (d^{3} \cos \left (b x + a\right )^{5} - d^{3} \cos \left (b x + a\right )^{3}\right )} \sqrt{d \cos \left (b x + a\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(d^3*cos(b*x + a)^5 - d^3*cos(b*x + a)^3)*sqrt(d*cos(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((d*cos(b*x+a))**(7/2)*sin(b*x+a)**2,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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