3.3.88 \(\int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{11/2}} \, dx\) [288]

Optimal. Leaf size=152 \[ -\frac {2 (a+a \sin (c+d x))^{3/2}}{3 d e (e \cos (c+d x))^{9/2}}+\frac {4 (a+a \sin (c+d x))^{5/2}}{a d e (e \cos (c+d x))^{9/2}}-\frac {16 (a+a \sin (c+d x))^{7/2}}{5 a^2 d e (e \cos (c+d x))^{9/2}}+\frac {32 (a+a \sin (c+d x))^{9/2}}{45 a^3 d e (e \cos (c+d x))^{9/2}} \]

[Out]

-2/3*(a+a*sin(d*x+c))^(3/2)/d/e/(e*cos(d*x+c))^(9/2)+4*(a+a*sin(d*x+c))^(5/2)/a/d/e/(e*cos(d*x+c))^(9/2)-16/5*
(a+a*sin(d*x+c))^(7/2)/a^2/d/e/(e*cos(d*x+c))^(9/2)+32/45*(a+a*sin(d*x+c))^(9/2)/a^3/d/e/(e*cos(d*x+c))^(9/2)

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Rubi [A]
time = 0.20, antiderivative size = 152, 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 = {2751, 2750} \begin {gather*} \frac {32 (a \sin (c+d x)+a)^{9/2}}{45 a^3 d e (e \cos (c+d x))^{9/2}}-\frac {16 (a \sin (c+d x)+a)^{7/2}}{5 a^2 d e (e \cos (c+d x))^{9/2}}+\frac {4 (a \sin (c+d x)+a)^{5/2}}{a d e (e \cos (c+d x))^{9/2}}-\frac {2 (a \sin (c+d x)+a)^{3/2}}{3 d e (e \cos (c+d x))^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^(3/2)/(e*Cos[c + d*x])^(11/2),x]

[Out]

(-2*(a + a*Sin[c + d*x])^(3/2))/(3*d*e*(e*Cos[c + d*x])^(9/2)) + (4*(a + a*Sin[c + d*x])^(5/2))/(a*d*e*(e*Cos[
c + d*x])^(9/2)) - (16*(a + a*Sin[c + d*x])^(7/2))/(5*a^2*d*e*(e*Cos[c + d*x])^(9/2)) + (32*(a + a*Sin[c + d*x
])^(9/2))/(45*a^3*d*e*(e*Cos[c + d*x])^(9/2))

Rule 2750

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 2751

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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))^{3/2}}{(e \cos (c+d x))^{11/2}} \, dx &=-\frac {2 (a+a \sin (c+d x))^{3/2}}{3 d e (e \cos (c+d x))^{9/2}}+\frac {2 \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{11/2}} \, dx}{a}\\ &=-\frac {2 (a+a \sin (c+d x))^{3/2}}{3 d e (e \cos (c+d x))^{9/2}}+\frac {4 (a+a \sin (c+d x))^{5/2}}{a d e (e \cos (c+d x))^{9/2}}-\frac {8 \int \frac {(a+a \sin (c+d x))^{7/2}}{(e \cos (c+d x))^{11/2}} \, dx}{a^2}\\ &=-\frac {2 (a+a \sin (c+d x))^{3/2}}{3 d e (e \cos (c+d x))^{9/2}}+\frac {4 (a+a \sin (c+d x))^{5/2}}{a d e (e \cos (c+d x))^{9/2}}-\frac {16 (a+a \sin (c+d x))^{7/2}}{5 a^2 d e (e \cos (c+d x))^{9/2}}+\frac {16 \int \frac {(a+a \sin (c+d x))^{9/2}}{(e \cos (c+d x))^{11/2}} \, dx}{5 a^3}\\ &=-\frac {2 (a+a \sin (c+d x))^{3/2}}{3 d e (e \cos (c+d x))^{9/2}}+\frac {4 (a+a \sin (c+d x))^{5/2}}{a d e (e \cos (c+d x))^{9/2}}-\frac {16 (a+a \sin (c+d x))^{7/2}}{5 a^2 d e (e \cos (c+d x))^{9/2}}+\frac {32 (a+a \sin (c+d x))^{9/2}}{45 a^3 d e (e \cos (c+d x))^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 74, normalized size = 0.49 \begin {gather*} \frac {2 \sqrt {e \cos (c+d x)} \sec ^5(c+d x) (a (1+\sin (c+d x)))^{3/2} (7+12 \cos (2 (c+d x))+6 \sin (c+d x)-4 \sin (3 (c+d x)))}{45 d e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^(3/2)/(e*Cos[c + d*x])^(11/2),x]

[Out]

(2*Sqrt[e*Cos[c + d*x]]*Sec[c + d*x]^5*(a*(1 + Sin[c + d*x]))^(3/2)*(7 + 12*Cos[2*(c + d*x)] + 6*Sin[c + d*x]
- 4*Sin[3*(c + d*x)]))/(45*d*e^6)

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Maple [A]
time = 0.17, size = 70, normalized size = 0.46

method	result	size
default	\(-\frac {2 \left (16 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-24 \left (\cos ^{2}\left (d x +c \right )\right )-10 \sin \left (d x +c \right )+5\right ) \cos \left (d x +c \right ) \left (a \left (1+\sin \left (d x +c \right )\right )\right )^{\frac {3}{2}}}{45 d \left (e \cos \left (d x +c \right )\right )^{\frac {11}{2}}}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^(3/2)/(e*cos(d*x+c))^(11/2),x,method=_RETURNVERBOSE)

[Out]

-2/45/d*(16*cos(d*x+c)^2*sin(d*x+c)-24*cos(d*x+c)^2-10*sin(d*x+c)+5)*cos(d*x+c)*(a*(1+sin(d*x+c)))^(3/2)/(e*co
s(d*x+c))^(11/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (118) = 236\).
time = 0.56, size = 321, normalized size = 2.11 \begin {gather*} \frac {2 \, {\left (19 \, a^{\frac {3}{2}} - \frac {12 \, a^{\frac {3}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {58 \, a^{\frac {3}{2}} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {116 \, a^{\frac {3}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {116 \, a^{\frac {3}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {58 \, a^{\frac {3}{2}} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {12 \, a^{\frac {3}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {19 \, a^{\frac {3}{2}} \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} e^{\left (-\frac {11}{2}\right )}}{45 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {11}{2}} {\left (\frac {4 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {\sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^(3/2)/(e*cos(d*x+c))^(11/2),x, algorithm="maxima")

[Out]

2/45*(19*a^(3/2) - 12*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 58*a^(3/2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2
+ 116*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 116*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 58*a^(3
/2)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 12*a^(3/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 19*a^(3/2)*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^(-11/2)/(d*(sin(d*x + c)/(cos(d*x +
c) + 1) + 1)^(5/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*s
in(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + sin(d*x + c)^8/(cos(d*x + c) + 1)
^8 + 1))

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Fricas [A]
time = 0.36, size = 94, normalized size = 0.62 \begin {gather*} -\frac {2 \, {\left (24 \, a \cos \left (d x + c\right )^{2} - 2 \, {\left (8 \, a \cos \left (d x + c\right )^{2} - 5 \, a\right )} \sin \left (d x + c\right ) - 5 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{45 \, {\left (d \cos \left (d x + c\right )^{3} e^{\frac {11}{2}} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{3} e^{\frac {11}{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^(3/2)/(e*cos(d*x+c))^(11/2),x, algorithm="fricas")

[Out]

-2/45*(24*a*cos(d*x + c)^2 - 2*(8*a*cos(d*x + c)^2 - 5*a)*sin(d*x + c) - 5*a)*sqrt(a*sin(d*x + c) + a)*sqrt(co
s(d*x + c))/(d*cos(d*x + c)^3*e^(11/2)*sin(d*x + c) - d*cos(d*x + c)^3*e^(11/2))

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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((a+a*sin(d*x+c))**(3/2)/(e*cos(d*x+c))**(11/2),x)

[Out]

Timed out

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Giac [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((a+a*sin(d*x+c))^(3/2)/(e*cos(d*x+c))^(11/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 10.96, size = 261, normalized size = 1.72 \begin {gather*} \frac {14\,a\,\sqrt {a+a\,\sin \left (c+d\,x\right )}+12\,a\,\sin \left (c+d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )}+24\,a\,\cos \left (2\,c+2\,d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )}-8\,a\,\sin \left (3\,c+3\,d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )}}{\frac {45\,d\,e^5\,\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}}}{2}+\frac {45\,d\,e^5\,\cos \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}}}{2}-\frac {45\,d\,e^5\,\sin \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}-\frac {45\,d\,e^5\,\sin \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^(3/2)/(e*cos(c + d*x))^(11/2),x)

[Out]

(14*a*(a + a*sin(c + d*x))^(1/2) + 12*a*sin(c + d*x)*(a + a*sin(c + d*x))^(1/2) + 24*a*cos(2*c + 2*d*x)*(a + a
*sin(c + d*x))^(1/2) - 8*a*sin(3*c + 3*d*x)*(a + a*sin(c + d*x))^(1/2))/((45*d*e^5*((e*exp(- c*1i - d*x*1i))/2
+ (e*exp(c*1i + d*x*1i))/2)^(1/2))/2 + (45*d*e^5*cos(2*c + 2*d*x)*((e*exp(- c*1i - d*x*1i))/2 + (e*exp(c*1i +
d*x*1i))/2)^(1/2))/2 - (45*d*e^5*sin(3*c + 3*d*x)*((e*exp(- c*1i - d*x*1i))/2 + (e*exp(c*1i + d*x*1i))/2)^(1/
2))/4 - (45*d*e^5*sin(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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