3.297 \(\int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{13/2}} \, dx\)

Optimal. Leaf size=150 \[ \frac {32 (a \sin (c+d x)+a)^{11/2}}{77 a^3 d e (e \cos (c+d x))^{11/2}}-\frac {16 (a \sin (c+d x)+a)^{9/2}}{7 a^2 d e (e \cos (c+d x))^{11/2}}+\frac {4 (a \sin (c+d x)+a)^{7/2}}{a d e (e \cos (c+d x))^{11/2}}-\frac {2 (a \sin (c+d x)+a)^{5/2}}{d e (e \cos (c+d x))^{11/2}} \]

[Out]

-2*(a+a*sin(d*x+c))^(5/2)/d/e/(e*cos(d*x+c))^(11/2)+4*(a+a*sin(d*x+c))^(7/2)/a/d/e/(e*cos(d*x+c))^(11/2)-16/7*
(a+a*sin(d*x+c))^(9/2)/a^2/d/e/(e*cos(d*x+c))^(11/2)+32/77*(a+a*sin(d*x+c))^(11/2)/a^3/d/e/(e*cos(d*x+c))^(11/
2)

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Rubi [A]  time = 0.31, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2672, 2671} \[ \frac {32 (a \sin (c+d x)+a)^{11/2}}{77 a^3 d e (e \cos (c+d x))^{11/2}}-\frac {16 (a \sin (c+d x)+a)^{9/2}}{7 a^2 d e (e \cos (c+d x))^{11/2}}+\frac {4 (a \sin (c+d x)+a)^{7/2}}{a d e (e \cos (c+d x))^{11/2}}-\frac {2 (a \sin (c+d x)+a)^{5/2}}{d e (e \cos (c+d x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + a*Sin[c + d*x])^(5/2))/(d*e*(e*Cos[c + d*x])^(11/2)) + (4*(a + a*Sin[c + d*x])^(7/2))/(a*d*e*(e*Cos[c
+ d*x])^(11/2)) - (16*(a + a*Sin[c + d*x])^(9/2))/(7*a^2*d*e*(e*Cos[c + d*x])^(11/2)) + (32*(a + a*Sin[c + d*
x])^(11/2))/(77*a^3*d*e*(e*Cos[c + d*x])^(11/2))

Rule 2671

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

Rule 2672

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m
, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{13/2}} \, dx &=-\frac {2 (a+a \sin (c+d x))^{5/2}}{d e (e \cos (c+d x))^{11/2}}+\frac {6 \int \frac {(a+a \sin (c+d x))^{7/2}}{(e \cos (c+d x))^{13/2}} \, dx}{a}\\ &=-\frac {2 (a+a \sin (c+d x))^{5/2}}{d e (e \cos (c+d x))^{11/2}}+\frac {4 (a+a \sin (c+d x))^{7/2}}{a d e (e \cos (c+d x))^{11/2}}-\frac {8 \int \frac {(a+a \sin (c+d x))^{9/2}}{(e \cos (c+d x))^{13/2}} \, dx}{a^2}\\ &=-\frac {2 (a+a \sin (c+d x))^{5/2}}{d e (e \cos (c+d x))^{11/2}}+\frac {4 (a+a \sin (c+d x))^{7/2}}{a d e (e \cos (c+d x))^{11/2}}-\frac {16 (a+a \sin (c+d x))^{9/2}}{7 a^2 d e (e \cos (c+d x))^{11/2}}+\frac {16 \int \frac {(a+a \sin (c+d x))^{11/2}}{(e \cos (c+d x))^{13/2}} \, dx}{7 a^3}\\ &=-\frac {2 (a+a \sin (c+d x))^{5/2}}{d e (e \cos (c+d x))^{11/2}}+\frac {4 (a+a \sin (c+d x))^{7/2}}{a d e (e \cos (c+d x))^{11/2}}-\frac {16 (a+a \sin (c+d x))^{9/2}}{7 a^2 d e (e \cos (c+d x))^{11/2}}+\frac {32 (a+a \sin (c+d x))^{11/2}}{77 a^3 d e (e \cos (c+d x))^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 74, normalized size = 0.49 \[ \frac {2 \left (16 \sin ^3(c+d x)-40 \sin ^2(c+d x)+26 \sin (c+d x)+5\right ) \sec ^6(c+d x) (a (\sin (c+d x)+1))^{5/2} \sqrt {e \cos (c+d x)}}{77 d e^7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[e*Cos[c + d*x]]*Sec[c + d*x]^6*(a*(1 + Sin[c + d*x]))^(5/2)*(5 + 26*Sin[c + d*x] - 40*Sin[c + d*x]^2 +
16*Sin[c + d*x]^3))/(77*d*e^7)

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fricas [A]  time = 1.28, size = 120, normalized size = 0.80 \[ -\frac {2 \, {\left (40 \, a^{2} \cos \left (d x + c\right )^{2} - 35 \, a^{2} - 2 \, {\left (8 \, a^{2} \cos \left (d x + c\right )^{2} - 21 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{77 \, {\left (d e^{7} \cos \left (d x + c\right )^{4} + 2 \, d e^{7} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, d e^{7} \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/77*(40*a^2*cos(d*x + c)^2 - 35*a^2 - 2*(8*a^2*cos(d*x + c)^2 - 21*a^2)*sin(d*x + c))*sqrt(e*cos(d*x + c))*s
qrt(a*sin(d*x + c) + a)/(d*e^7*cos(d*x + c)^4 + 2*d*e^7*cos(d*x + c)^2*sin(d*x + c) - 2*d*e^7*cos(d*x + c)^2)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.21, size = 70, normalized size = 0.47 \[ -\frac {2 \left (16 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-40 \left (\cos ^{2}\left (d x +c \right )\right )-42 \sin \left (d x +c \right )+35\right ) \left (a \left (1+\sin \left (d x +c \right )\right )\right )^{\frac {5}{2}} \cos \left (d x +c \right )}{77 d \left (e \cos \left (d x +c \right )\right )^{\frac {13}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/77/d*(16*cos(d*x+c)^2*sin(d*x+c)-40*cos(d*x+c)^2-42*sin(d*x+c)+35)*(a*(1+sin(d*x+c)))^(5/2)*cos(d*x+c)/(e*c
os(d*x+c))^(13/2)

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maxima [B]  time = 1.13, size = 357, normalized size = 2.38 \[ \frac {2 \, {\left (5 \, a^{\frac {5}{2}} \sqrt {e} + \frac {52 \, a^{\frac {5}{2}} \sqrt {e} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {150 \, a^{\frac {5}{2}} \sqrt {e} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {180 \, a^{\frac {5}{2}} \sqrt {e} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {180 \, a^{\frac {5}{2}} \sqrt {e} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {150 \, a^{\frac {5}{2}} \sqrt {e} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {52 \, a^{\frac {5}{2}} \sqrt {e} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {5 \, a^{\frac {5}{2}} \sqrt {e} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{4}}{77 \, {\left (e^{7} + \frac {4 \, e^{7} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, e^{7} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, e^{7} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {e^{7} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {3}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {13}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/77*(5*a^(5/2)*sqrt(e) + 52*a^(5/2)*sqrt(e)*sin(d*x + c)/(cos(d*x + c) + 1) - 150*a^(5/2)*sqrt(e)*sin(d*x + c
)^2/(cos(d*x + c) + 1)^2 + 180*a^(5/2)*sqrt(e)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 180*a^(5/2)*sqrt(e)*sin(d
*x + c)^5/(cos(d*x + c) + 1)^5 + 150*a^(5/2)*sqrt(e)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 52*a^(5/2)*sqrt(e)*
sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 5*a^(5/2)*sqrt(e)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8)*(sin(d*x + c)^2/(
cos(d*x + c) + 1)^2 + 1)^4/((e^7 + 4*e^7*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*e^7*sin(d*x + c)^4/(cos(d*x +
c) + 1)^4 + 4*e^7*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + e^7*sin(d*x + c)^8/(cos(d*x + c) + 1)^8)*d*(sin(d*x +
c)/(cos(d*x + c) + 1) + 1)^(3/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(13/2))

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mupad [B]  time = 11.17, size = 232, normalized size = 1.55 \[ \frac {30\,a^2\,\sqrt {a+a\,\sin \left (c+d\,x\right )}-40\,a^2\,\cos \left (2\,c+2\,d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )}+8\,a^2\,\sin \left (3\,c+3\,d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )}-76\,a^2\,\sin \left (c+d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )}}{\frac {77\,d\,e^6\,\cos \left (3\,c+3\,d\,x\right )\,\sqrt {\frac {e\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {e\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}{4}+77\,d\,e^6\,\sin \left (2\,c+2\,d\,x\right )\,\sqrt {\frac {e\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {e\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}-\frac {385\,d\,e^6\,\cos \left (c+d\,x\right )\,\sqrt {\frac {e\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {e\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(30*a^2*(a + a*sin(c + d*x))^(1/2) - 40*a^2*cos(2*c + 2*d*x)*(a + a*sin(c + d*x))^(1/2) + 8*a^2*sin(3*c + 3*d*
x)*(a + a*sin(c + d*x))^(1/2) - 76*a^2*sin(c + d*x)*(a + a*sin(c + d*x))^(1/2))/((77*d*e^6*cos(3*c + 3*d*x)*((
e*exp(- c*1i - d*x*1i))/2 + (e*exp(c*1i + d*x*1i))/2)^(1/2))/4 + 77*d*e^6*sin(2*c + 2*d*x)*((e*exp(- c*1i - d*
x*1i))/2 + (e*exp(c*1i + d*x*1i))/2)^(1/2) - (385*d*e^6*cos(c + d*x)*((e*exp(- c*1i - d*x*1i))/2 + (e*exp(c*1i
+ d*x*1i))/2)^(1/2))/4)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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