3.267 \(\int (d \cos (a+b x))^{3/2} (c \sin (a+b x))^{3/2} \, dx\)

Optimal. Leaf size=131 \[ \frac{c^2 d^2 \sqrt{\sin (2 a+2 b x)} F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{12 b \sqrt{c \sin (a+b x)} \sqrt{d \cos (a+b x)}}-\frac{c \sqrt{c \sin (a+b x)} (d \cos (a+b x))^{5/2}}{3 b d}+\frac{c d \sqrt{c \sin (a+b x)} \sqrt{d \cos (a+b x)}}{6 b} \]

[Out]

(c*d*Sqrt[d*Cos[a + b*x]]*Sqrt[c*Sin[a + b*x]])/(6*b) - (c*(d*Cos[a + b*x])^(5/2)*Sqrt[c*Sin[a + b*x]])/(3*b*d
) + (c^2*d^2*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]])/(12*b*Sqrt[d*Cos[a + b*x]]*Sqrt[c*Sin[a + b*
x]])

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Rubi [A]  time = 0.178903, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2568, 2569, 2573, 2641} \[ \frac{c^2 d^2 \sqrt{\sin (2 a+2 b x)} F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{12 b \sqrt{c \sin (a+b x)} \sqrt{d \cos (a+b x)}}-\frac{c \sqrt{c \sin (a+b x)} (d \cos (a+b x))^{5/2}}{3 b d}+\frac{c d \sqrt{c \sin (a+b x)} \sqrt{d \cos (a+b x)}}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cos[a + b*x])^(3/2)*(c*Sin[a + b*x])^(3/2),x]

[Out]

(c*d*Sqrt[d*Cos[a + b*x]]*Sqrt[c*Sin[a + b*x]])/(6*b) - (c*(d*Cos[a + b*x])^(5/2)*Sqrt[c*Sin[a + b*x]])/(3*b*d
) + (c^2*d^2*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]])/(12*b*Sqrt[d*Cos[a + b*x]]*Sqrt[c*Sin[a + b*
x]])

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 2569

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(b*Sin[e +
f*x])^(n + 1)*(a*Cos[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Sin[e + f*x])^
n*(a*Cos[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 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, 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))^{3/2} (c \sin (a+b x))^{3/2} \, dx &=-\frac{c (d \cos (a+b x))^{5/2} \sqrt{c \sin (a+b x)}}{3 b d}+\frac{1}{6} c^2 \int \frac{(d \cos (a+b x))^{3/2}}{\sqrt{c \sin (a+b x)}} \, dx\\ &=\frac{c d \sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}}{6 b}-\frac{c (d \cos (a+b x))^{5/2} \sqrt{c \sin (a+b x)}}{3 b d}+\frac{1}{12} \left (c^2 d^2\right ) \int \frac{1}{\sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}} \, dx\\ &=\frac{c d \sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}}{6 b}-\frac{c (d \cos (a+b x))^{5/2} \sqrt{c \sin (a+b x)}}{3 b d}+\frac{\left (c^2 d^2 \sqrt{\sin (2 a+2 b x)}\right ) \int \frac{1}{\sqrt{\sin (2 a+2 b x)}} \, dx}{12 \sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}}\\ &=\frac{c d \sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}}{6 b}-\frac{c (d \cos (a+b x))^{5/2} \sqrt{c \sin (a+b x)}}{3 b d}+\frac{c^2 d^2 F\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sqrt{\sin (2 a+2 b x)}}{12 b \sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}}\\ \end{align*}

Mathematica [C]  time = 0.120046, size = 71, normalized size = 0.54 \[ \frac{2 c d \cos ^2(a+b x)^{3/4} \tan ^2(a+b x) \sqrt{c \sin (a+b x)} \sqrt{d \cos (a+b x)} \, _2F_1\left (-\frac{1}{4},\frac{5}{4};\frac{9}{4};\sin ^2(a+b x)\right )}{5 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Cos[a + b*x])^(3/2)*(c*Sin[a + b*x])^(3/2),x]

[Out]

(2*c*d*Sqrt[d*Cos[a + b*x]]*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[-1/4, 5/4, 9/4, Sin[a + b*x]^2]*Sqrt[c*Si
n[a + b*x]]*Tan[a + b*x]^2)/(5*b)

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Maple [A]  time = 0.132, size = 216, normalized size = 1.7 \begin{align*} -{\frac{\sqrt{2}}{12\,b\sin \left ( bx+a \right ) \left ( -1+\cos \left ( bx+a \right ) \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{2}} \left ( \sin \left ( bx+a \right ) \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{-1+\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{-1+\cos \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},{\frac{\sqrt{2}}{2}} \right ) +2\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}\sqrt{2}-2\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}\sqrt{2}- \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sqrt{2}+\cos \left ( bx+a \right ) \sqrt{2} \right ) \left ( d\cos \left ( bx+a \right ) \right ) ^{{\frac{3}{2}}} \left ( c\sin \left ( bx+a \right ) \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(b*x+a))^(3/2)*(c*sin(b*x+a))^(3/2),x)

[Out]

-1/12/b*2^(1/2)*(sin(b*x+a)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a
))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2)
)+2*cos(b*x+a)^4*2^(1/2)-2*cos(b*x+a)^3*2^(1/2)-cos(b*x+a)^2*2^(1/2)+cos(b*x+a)*2^(1/2))*(d*cos(b*x+a))^(3/2)*
(c*sin(b*x+a))^(3/2)/sin(b*x+a)/(-1+cos(b*x+a))/cos(b*x+a)^2

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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{3}{2}} \left (c \sin \left (b x + a\right )\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(3/2)*(c*sin(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate((d*cos(b*x + a))^(3/2)*(c*sin(b*x + a))^(3/2), x)

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

[Out]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(3/2)*(c*sin(b*x+a))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError