\(\int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx\) [208]

Optimal result

Integrand size = 21, antiderivative size = 43 \[ \int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {2}{5 b d (d \cos (a+b x))^{5/2}}-\frac {2}{b d^3 \sqrt {d \cos (a+b x)}} \]

[Out]

2/5/b/d/(d*cos(b*x+a))^(5/2)-2/b/d^3/(d*cos(b*x+a))^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2645, 14} \[ \int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {2}{5 b d (d \cos (a+b x))^{5/2}}-\frac {2}{b d^3 \sqrt {d \cos (a+b x)}} \]

[In]

Int[Sin[a + b*x]^3/(d*Cos[a + b*x])^(7/2),x]

[Out]

2/(5*b*d*(d*Cos[a + b*x])^(5/2)) - 2/(b*d^3*Sqrt[d*Cos[a + b*x]])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2645

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {1-\frac {x^2}{d^2}}{x^{7/2}} \, dx,x,d \cos (a+b x)\right )}{b d} \\ & = -\frac {\text {Subst}\left (\int \left (\frac {1}{x^{7/2}}-\frac {1}{d^2 x^{3/2}}\right ) \, dx,x,d \cos (a+b x)\right )}{b d} \\ & = \frac {2}{5 b d (d \cos (a+b x))^{5/2}}-\frac {2}{b d^3 \sqrt {d \cos (a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.63 \[ \int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {2 \left (5-4 \sqrt [4]{\cos ^2(a+b x)}+4 \left (-1+\sqrt [4]{\cos ^2(a+b x)}\right ) \csc ^2(a+b x)\right ) \tan ^2(a+b x)}{5 b d^3 \sqrt {d \cos (a+b x)}} \]

[In]

Integrate[Sin[a + b*x]^3/(d*Cos[a + b*x])^(7/2),x]

[Out]

(2*(5 - 4*(Cos[a + b*x]^2)^(1/4) + 4*(-1 + (Cos[a + b*x]^2)^(1/4))*Csc[a + b*x]^2)*Tan[a + b*x]^2)/(5*b*d^3*Sq
rt[d*Cos[a + b*x]])

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {\frac {2 d^{2}}{5 \left (d \cos \left (b x +a \right )\right )^{\frac {5}{2}}}-\frac {2}{\sqrt {d \cos \left (b x +a \right )}}}{b \,d^{3}}\)	\(37\)
default	\(\frac {\frac {2 d^{2}}{5 \left (d \cos \left (b x +a \right )\right )^{\frac {5}{2}}}-\frac {2}{\sqrt {d \cos \left (b x +a \right )}}}{b \,d^{3}}\)	\(37\)

[In]

int(sin(b*x+a)^3/(d*cos(b*x+a))^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/b/d^3*(1/5*d^2/(d*cos(b*x+a))^(5/2)-1/(d*cos(b*x+a))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.88 \[ \int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=-\frac {2 \, \sqrt {d \cos \left (b x + a\right )} {\left (5 \, \cos \left (b x + a\right )^{2} - 1\right )}}{5 \, b d^{4} \cos \left (b x + a\right )^{3}} \]

[In]

integrate(sin(b*x+a)^3/(d*cos(b*x+a))^(7/2),x, algorithm="fricas")

[Out]

-2/5*sqrt(d*cos(b*x + a))*(5*cos(b*x + a)^2 - 1)/(b*d^4*cos(b*x + a)^3)

Sympy [A] (verification not implemented)

Time = 30.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.65 \[ \int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\begin {cases} \frac {2 \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{5 b \left (d \cos {\left (a + b x \right )}\right )^{\frac {7}{2}}} - \frac {8 \cos ^{3}{\left (a + b x \right )}}{5 b \left (d \cos {\left (a + b x \right )}\right )^{\frac {7}{2}}} & \text {for}\: b \neq 0 \\\frac {x \sin ^{3}{\left (a \right )}}{\left (d \cos {\left (a \right )}\right )^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate(sin(b*x+a)**3/(d*cos(b*x+a))**(7/2),x)

[Out]

Piecewise((2*sin(a + b*x)**2*cos(a + b*x)/(5*b*(d*cos(a + b*x))**(7/2)) - 8*cos(a + b*x)**3/(5*b*(d*cos(a + b*
x))**(7/2)), Ne(b, 0)), (x*sin(a)**3/(d*cos(a))**(7/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=-\frac {2 \, {\left (5 \, d^{2} \cos \left (b x + a\right )^{2} - d^{2}\right )}}{5 \, \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} b d^{3}} \]

[In]

integrate(sin(b*x+a)^3/(d*cos(b*x+a))^(7/2),x, algorithm="maxima")

[Out]

-2/5*(5*d^2*cos(b*x + a)^2 - d^2)/((d*cos(b*x + a))^(5/2)*b*d^3)

Giac [F]

\[ \int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\int { \frac {\sin \left (b x + a\right )^{3}}{\left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate(sin(b*x+a)^3/(d*cos(b*x+a))^(7/2),x, algorithm="giac")

[Out]

integrate(sin(b*x + a)^3/(d*cos(b*x + a))^(7/2), x)

Mupad [B] (verification not implemented)

Time = 3.46 (sec) , antiderivative size = 93, normalized size of antiderivative = 2.16 \[ \int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=-\frac {4\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {d\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{2}\right )}\,\left (6\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+5\,{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+5\right )}{5\,b\,d^4\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^3} \]

[In]

int(sin(a + b*x)^3/(d*cos(a + b*x))^(7/2),x)

[Out]

-(4*exp(a*1i + b*x*1i)*(d*(exp(- a*1i - b*x*1i)/2 + exp(a*1i + b*x*1i)/2))^(1/2)*(6*exp(a*2i + b*x*2i) + 5*exp
(a*4i + b*x*4i) + 5))/(5*b*d^4*(exp(a*2i + b*x*2i) + 1)^3)