\(\int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}} \, dx\) [265]

Optimal result

Integrand size = 25, antiderivative size = 75 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}} \, dx=\frac {2 (c \sin (a+b x))^{3/2}}{7 b c d (d \cos (a+b x))^{7/2}}+\frac {8 (c \sin (a+b x))^{3/2}}{21 b c d^3 (d \cos (a+b x))^{3/2}} \]

[Out]

2/7*(c*sin(b*x+a))^(3/2)/b/c/d/(d*cos(b*x+a))^(7/2)+8/21*(c*sin(b*x+a))^(3/2)/b/c/d^3/(d*cos(b*x+a))^(3/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2651, 2643} \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}} \, dx=\frac {8 (c \sin (a+b x))^{3/2}}{21 b c d^3 (d \cos (a+b x))^{3/2}}+\frac {2 (c \sin (a+b x))^{3/2}}{7 b c d (d \cos (a+b x))^{7/2}} \]

[In]

Int[Sqrt[c*Sin[a + b*x]]/(d*Cos[a + b*x])^(9/2),x]

[Out]

(2*(c*Sin[a + b*x])^(3/2))/(7*b*c*d*(d*Cos[a + b*x])^(7/2)) + (8*(c*Sin[a + b*x])^(3/2))/(21*b*c*d^3*(d*Cos[a
+ b*x])^(3/2))

Rule 2643

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

Rule 2651

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 (c \sin (a+b x))^{3/2}}{7 b c d (d \cos (a+b x))^{7/2}}+\frac {4 \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}} \, dx}{7 d^2} \\ & = \frac {2 (c \sin (a+b x))^{3/2}}{7 b c d (d \cos (a+b x))^{7/2}}+\frac {8 (c \sin (a+b x))^{3/2}}{21 b c d^3 (d \cos (a+b x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}} \, dx=\frac {2 \sqrt {d \cos (a+b x)} (5+2 \cos (2 (a+b x))) \sec ^4(a+b x) (c \sin (a+b x))^{3/2}}{21 b c d^5} \]

[In]

Integrate[Sqrt[c*Sin[a + b*x]]/(d*Cos[a + b*x])^(9/2),x]

[Out]

(2*Sqrt[d*Cos[a + b*x]]*(5 + 2*Cos[2*(a + b*x)])*Sec[a + b*x]^4*(c*Sin[a + b*x])^(3/2))/(21*b*c*d^5)

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.72

method	result	size
default	\(\frac {2 \sqrt {c \sin \left (b x +a \right )}\, \left (4 \tan \left (b x +a \right )+3 \tan \left (b x +a \right ) \left (\sec ^{2}\left (b x +a \right )\right )\right )}{21 b \,d^{4} \sqrt {d \cos \left (b x +a \right )}}\)	\(54\)

[In]

int((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(9/2),x,method=_RETURNVERBOSE)

[Out]

2/21/b*(c*sin(b*x+a))^(1/2)/d^4/(d*cos(b*x+a))^(1/2)*(4*tan(b*x+a)+3*tan(b*x+a)*sec(b*x+a)^2)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}} \, dx=\frac {2 \, \sqrt {d \cos \left (b x + a\right )} {\left (4 \, \cos \left (b x + a\right )^{2} + 3\right )} \sqrt {c \sin \left (b x + a\right )} \sin \left (b x + a\right )}{21 \, b d^{5} \cos \left (b x + a\right )^{4}} \]

[In]

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(9/2),x, algorithm="fricas")

[Out]

2/21*sqrt(d*cos(b*x + a))*(4*cos(b*x + a)^2 + 3)*sqrt(c*sin(b*x + a))*sin(b*x + a)/(b*d^5*cos(b*x + a)^4)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}} \, dx=\text {Timed out} \]

[In]

integrate((c*sin(b*x+a))**(1/2)/(d*cos(b*x+a))**(9/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}} \, dx=\int { \frac {\sqrt {c \sin \left (b x + a\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {9}{2}}} \,d x } \]

[In]

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(9/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*sin(b*x + a))/(d*cos(b*x + a))^(9/2), x)

Giac [F]

\[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}} \, dx=\int { \frac {\sqrt {c \sin \left (b x + a\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {9}{2}}} \,d x } \]

[In]

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(9/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*sin(b*x + a))/(d*cos(b*x + a))^(9/2), x)

Mupad [B] (verification not implemented)

Time = 1.92 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}} \, dx=\frac {8\,\sqrt {c\,\sin \left (a+b\,x\right )}\,\left (11\,\sin \left (2\,a+2\,b\,x\right )+7\,\sin \left (4\,a+4\,b\,x\right )+\sin \left (6\,a+6\,b\,x\right )\right )}{21\,b\,d^4\,\sqrt {d\,\cos \left (a+b\,x\right )}\,\left (15\,\cos \left (2\,a+2\,b\,x\right )+6\,\cos \left (4\,a+4\,b\,x\right )+\cos \left (6\,a+6\,b\,x\right )+10\right )} \]

[In]

int((c*sin(a + b*x))^(1/2)/(d*cos(a + b*x))^(9/2),x)

[Out]

(8*(c*sin(a + b*x))^(1/2)*(11*sin(2*a + 2*b*x) + 7*sin(4*a + 4*b*x) + sin(6*a + 6*b*x)))/(21*b*d^4*(d*cos(a +
b*x))^(1/2)*(15*cos(2*a + 2*b*x) + 6*cos(4*a + 4*b*x) + cos(6*a + 6*b*x) + 10))