3.3.85 \(\int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx\) [285]

Optimal. Leaf size=106 \[ \frac {2 c (c \sin (a+b x))^{3/2}}{11 b d (d \cos (a+b x))^{11/2}}-\frac {6 c (c \sin (a+b x))^{3/2}}{77 b d^3 (d \cos (a+b x))^{7/2}}-\frac {8 c (c \sin (a+b x))^{3/2}}{77 b d^5 (d \cos (a+b x))^{3/2}} \]

[Out]

2/11*c*(c*sin(b*x+a))^(3/2)/b/d/(d*cos(b*x+a))^(11/2)-6/77*c*(c*sin(b*x+a))^(3/2)/b/d^3/(d*cos(b*x+a))^(7/2)-8
/77*c*(c*sin(b*x+a))^(3/2)/b/d^5/(d*cos(b*x+a))^(3/2)

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Rubi [A]
time = 0.11, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2646, 2651, 2643} \begin {gather*} -\frac {8 c (c \sin (a+b x))^{3/2}}{77 b d^5 (d \cos (a+b x))^{3/2}}-\frac {6 c (c \sin (a+b x))^{3/2}}{77 b d^3 (d \cos (a+b x))^{7/2}}+\frac {2 c (c \sin (a+b x))^{3/2}}{11 b d (d \cos (a+b x))^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x])^(5/2)/(d*Cos[a + b*x])^(13/2),x]

[Out]

(2*c*(c*Sin[a + b*x])^(3/2))/(11*b*d*(d*Cos[a + b*x])^(11/2)) - (6*c*(c*Sin[a + b*x])^(3/2))/(77*b*d^3*(d*Cos[
a + b*x])^(7/2)) - (8*c*(c*Sin[a + b*x])^(3/2))/(77*b*d^5*(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 2646

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

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*} \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{13/2}} \, dx &=\frac {2 c (c \sin (a+b x))^{3/2}}{11 b d (d \cos (a+b x))^{11/2}}-\frac {\left (3 c^2\right ) \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{9/2}} \, dx}{11 d^2}\\ &=\frac {2 c (c \sin (a+b x))^{3/2}}{11 b d (d \cos (a+b x))^{11/2}}-\frac {6 c (c \sin (a+b x))^{3/2}}{77 b d^3 (d \cos (a+b x))^{7/2}}-\frac {\left (12 c^2\right ) \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}} \, dx}{77 d^4}\\ &=\frac {2 c (c \sin (a+b x))^{3/2}}{11 b d (d \cos (a+b x))^{11/2}}-\frac {6 c (c \sin (a+b x))^{3/2}}{77 b d^3 (d \cos (a+b x))^{7/2}}-\frac {8 c (c \sin (a+b x))^{3/2}}{77 b d^5 (d \cos (a+b x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 57, normalized size = 0.54 \begin {gather*} \frac {2 c^4 (9+2 \cos (2 (a+b x))) \tan ^5(a+b x)}{77 b d^6 \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*Sin[a + b*x])^(5/2)/(d*Cos[a + b*x])^(13/2),x]

[Out]

(2*c^4*(9 + 2*Cos[2*(a + b*x)])*Tan[a + b*x]^5)/(77*b*d^6*Sqrt[d*Cos[a + b*x]]*(c*Sin[a + b*x])^(3/2))

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Maple [A]
time = 0.10, size = 50, normalized size = 0.47

method	result	size
default	\(\frac {2 \left (4 \left (\cos ^{2}\left (b x +a \right )\right )+7\right ) \cos \left (b x +a \right ) \left (c \sin \left (b x +a \right )\right )^{\frac {5}{2}} \sin \left (b x +a \right )}{77 b \left (d \cos \left (b x +a \right )\right )^{\frac {13}{2}}}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(13/2),x,method=_RETURNVERBOSE)

[Out]

2/77/b*(4*cos(b*x+a)^2+7)*cos(b*x+a)*(c*sin(b*x+a))^(5/2)*sin(b*x+a)/(d*cos(b*x+a))^(13/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(13/2),x, algorithm="maxima")

[Out]

integrate((c*sin(b*x + a))^(5/2)/(d*cos(b*x + a))^(13/2), x)

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Fricas [A]
time = 0.48, size = 74, normalized size = 0.70 \begin {gather*} -\frac {2 \, {\left (4 \, c^{2} \cos \left (b x + a\right )^{4} + 3 \, c^{2} \cos \left (b x + a\right )^{2} - 7 \, c^{2}\right )} \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} \sin \left (b x + a\right )}{77 \, b d^{7} \cos \left (b x + a\right )^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(13/2),x, algorithm="fricas")

[Out]

-2/77*(4*c^2*cos(b*x + a)^4 + 3*c^2*cos(b*x + a)^2 - 7*c^2)*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))*sin(b*x
+ a)/(b*d^7*cos(b*x + a)^6)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))**(5/2)/(d*cos(b*x+a))**(13/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(13/2),x, algorithm="giac")

[Out]

integrate((c*sin(b*x + a))^(5/2)/(d*cos(b*x + a))^(13/2), x)

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Mupad [B]
time = 6.34, size = 176, normalized size = 1.66 \begin {gather*} -\frac {{\mathrm {e}}^{-a\,5{}\mathrm {i}-b\,x\,5{}\mathrm {i}}\,\sqrt {c\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}\,\left (\frac {96\,c^2\,{\mathrm {e}}^{a\,5{}\mathrm {i}+b\,x\,5{}\mathrm {i}}\,\sin \left (3\,a+3\,b\,x\right )}{77\,b\,d^6}+\frac {16\,c^2\,{\mathrm {e}}^{a\,5{}\mathrm {i}+b\,x\,5{}\mathrm {i}}\,\sin \left (5\,a+5\,b\,x\right )}{77\,b\,d^6}-\frac {368\,c^2\,{\mathrm {e}}^{a\,5{}\mathrm {i}+b\,x\,5{}\mathrm {i}}\,\sin \left (a+b\,x\right )}{77\,b\,d^6}\right )}{32\,{\cos \left (a+b\,x\right )}^5\,\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(a + b*x))^(5/2)/(d*cos(a + b*x))^(13/2),x)

[Out]

-(exp(- a*5i - b*x*5i)*(c*((exp(- a*1i - b*x*1i)*1i)/2 - (exp(a*1i + b*x*1i)*1i)/2))^(1/2)*((96*c^2*exp(a*5i +
b*x*5i)*sin(3*a + 3*b*x))/(77*b*d^6) + (16*c^2*exp(a*5i + b*x*5i)*sin(5*a + 5*b*x))/(77*b*d^6) - (368*c^2*exp
(a*5i + b*x*5i)*sin(a + b*x))/(77*b*d^6)))/(32*cos(a + b*x)^5*(d*(exp(- a*1i - b*x*1i)/2 + exp(a*1i + b*x*1i)/
2))^(1/2))

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