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

Optimal. Leaf size=131 \[ \frac{3 c^2 d^2 E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{c \sin (a+b x)} \sqrt{d \cos (a+b x)}}{20 b \sqrt{\sin (2 a+2 b x)}}-\frac{c (c \sin (a+b x))^{3/2} (d \cos (a+b x))^{7/2}}{5 b d}+\frac{c d (c \sin (a+b x))^{3/2} (d \cos (a+b x))^{3/2}}{10 b} \]

[Out]

(c*d*(d*Cos[a + b*x])^(3/2)*(c*Sin[a + b*x])^(3/2))/(10*b) - (c*(d*Cos[a + b*x])^(7/2)*(c*Sin[a + b*x])^(3/2))
/(5*b*d) + (3*c^2*d^2*Sqrt[d*Cos[a + b*x]]*EllipticE[a - Pi/4 + b*x, 2]*Sqrt[c*Sin[a + b*x]])/(20*b*Sqrt[Sin[2
*a + 2*b*x]])

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Rubi [A]  time = 0.175756, 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, 2572, 2639} \[ \frac{3 c^2 d^2 E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{c \sin (a+b x)} \sqrt{d \cos (a+b x)}}{20 b \sqrt{\sin (2 a+2 b x)}}-\frac{c (c \sin (a+b x))^{3/2} (d \cos (a+b x))^{7/2}}{5 b d}+\frac{c d (c \sin (a+b x))^{3/2} (d \cos (a+b x))^{3/2}}{10 b} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cos[a + b*x])^(5/2)*(c*Sin[a + b*x])^(5/2),x]

[Out]

(c*d*(d*Cos[a + b*x])^(3/2)*(c*Sin[a + b*x])^(3/2))/(10*b) - (c*(d*Cos[a + b*x])^(7/2)*(c*Sin[a + b*x])^(3/2))
/(5*b*d) + (3*c^2*d^2*Sqrt[d*Cos[a + b*x]]*EllipticE[a - Pi/4 + b*x, 2]*Sqrt[c*Sin[a + b*x]])/(20*b*Sqrt[Sin[2
*a + 2*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 2572

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(Sqrt[a*Sin[e +
f*x]]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[2*e + 2*f*x]], Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
x]

Rule 2639

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

Rubi steps

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

Mathematica [C]  time = 0.177644, size = 70, normalized size = 0.53 \[ \frac{2 d^2 \sqrt [4]{\cos ^2(a+b x)} \tan (a+b x) (c \sin (a+b x))^{5/2} \sqrt{d \cos (a+b x)} \, _2F_1\left (-\frac{3}{4},\frac{7}{4};\frac{11}{4};\sin ^2(a+b x)\right )}{7 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Cos[a + b*x])^(5/2)*(c*Sin[a + b*x])^(5/2),x]

[Out]

(2*d^2*Sqrt[d*Cos[a + b*x]]*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[-3/4, 7/4, 11/4, Sin[a + b*x]^2]*(c*Sin[a
+ b*x])^(5/2)*Tan[a + b*x])/(7*b)

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Maple [B]  time = 0.08, size = 532, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40/b*2^(1/2)*(4*cos(b*x+a)^6*2^(1/2)-6*cos(b*x+a)^4*2^(1/2)-6*cos(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)*EllipticE(((1-cos(b
*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+3*cos(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))-6*((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)*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1
/2*2^(1/2))+3*((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))-cos(b*x+a)^2
*2^(1/2)+3*cos(b*x+a)*2^(1/2))*(d*cos(b*x+a))^(5/2)*(c*sin(b*x+a))^(5/2)/sin(b*x+a)^3/cos(b*x+a)^3

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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{5}{2}} \left (c \sin \left (b x + a\right )\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(5/2)*(c*sin(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (c^{2} d^{2} \cos \left (b x + a\right )^{4} - c^{2} d^{2} \cos \left (b x + a\right )^{2}\right )} \sqrt{d \cos \left (b x + a\right )} \sqrt{c \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))^(5/2)*(c*sin(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

integral(-(c^2*d^2*cos(b*x + a)^4 - c^2*d^2*cos(b*x + a)^2)*sqrt(d*cos(b*x + a))*sqrt(c*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))**(5/2)*(c*sin(b*x+a))**(5/2),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{5}{2}} \left (c \sin \left (b x + a\right )\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(5/2)*(c*sin(b*x+a))^(5/2),x, algorithm="giac")

[Out]

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