\(\int (d \cos (a+b x))^{5/2} (c \sin (a+b x))^{5/2} \, dx\) [277]

Optimal result

Integrand size = 25, antiderivative size = 131 \[ \int (d \cos (a+b x))^{5/2} (c \sin (a+b x))^{5/2} \, dx=\frac {c d (d \cos (a+b x))^{3/2} (c \sin (a+b x))^{3/2}}{10 b}-\frac {c (d \cos (a+b x))^{7/2} (c \sin (a+b x))^{3/2}}{5 b d}+\frac {3 c^2 d^2 \sqrt {d \cos (a+b x)} E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {c \sin (a+b x)}}{20 b \sqrt {\sin (2 a+2 b x)}} \]

[Out]

1/10*c*d*(d*cos(b*x+a))^(3/2)*(c*sin(b*x+a))^(3/2)/b-1/5*c*(d*cos(b*x+a))^(7/2)*(c*sin(b*x+a))^(3/2)/b/d-3/20*
c^2*d^2*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))*(d*cos(b*x+a))^(1/2
)*(c*sin(b*x+a))^(1/2)/b/sin(2*b*x+2*a)^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2648, 2649, 2652, 2719} \[ \int (d \cos (a+b x))^{5/2} (c \sin (a+b x))^{5/2} \, dx=\frac {3 c^2 d^2 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}{20 b \sqrt {\sin (2 a+2 b x)}}-\frac {c (c \sin (a+b x))^{3/2} (d \cos (a+b x))^{7/2}}{5 b d}+\frac {c d (c \sin (a+b x))^{3/2} (d \cos (a+b x))^{3/2}}{10 b} \]

[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 2648

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 2649

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 2652

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 2719

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

Rubi steps \begin{align*} \text {integral}& = -\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] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.53 \[ \int (d \cos (a+b x))^{5/2} (c \sin (a+b x))^{5/2} \, dx=\frac {2 d^2 \sqrt {d \cos (a+b x)} \sqrt [4]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {7}{4},\frac {11}{4},\sin ^2(a+b x)\right ) (c \sin (a+b x))^{5/2} \tan (a+b x)}{7 b} \]

[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)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(425\) vs. \(2(136)=272\).

Time = 1.42 (sec) , antiderivative size = 426, normalized size of antiderivative = 3.25

method	result	size
default	\(\frac {\sqrt {2}\, d^{2} c^{2} \left (4 \sqrt {2}\, \left (\cos ^{6}\left (b x +a \right )\right )-6 \sqrt {2}\, \left (\cos ^{4}\left (b x +a \right )\right )-6 \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, E\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \cos \left (b x +a \right )+3 \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, F\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \cos \left (b x +a \right )-6 \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, E\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )+3 \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, F\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {2}\, \left (\cos ^{2}\left (b x +a \right )\right )+3 \sqrt {2}\, \cos \left (b x +a \right )\right ) \sqrt {c \sin \left (b x +a \right )}\, \sqrt {d \cos \left (b x +a \right )}\, \sec \left (b x +a \right ) \csc \left (b x +a \right )}{40 b}\)	\(426\)

[In]

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

[Out]

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

Fricas [F]

\[ \int (d \cos (a+b x))^{5/2} (c \sin (a+b x))^{5/2} \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} \left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}} \,d x } \]

[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)

Sympy [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{5/2} (c \sin (a+b x))^{5/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int (d \cos (a+b x))^{5/2} (c \sin (a+b x))^{5/2} \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} \left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}} \,d x } \]

[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)

Giac [F]

\[ \int (d \cos (a+b x))^{5/2} (c \sin (a+b x))^{5/2} \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} \left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{5/2} (c \sin (a+b x))^{5/2} \, dx=\int {\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}\,{\left (c\,\sin \left (a+b\,x\right )\right )}^{5/2} \,d x \]

[In]

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

[Out]

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