3.232 \(\int (a \cos (c+d x)+b \sin (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=186 \[ \frac {10 \left (a^2+b^2\right )^2 \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}} F\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right )}{21 d \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {10 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{21 d}-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d} \]

[Out]

-2/7*(b*cos(d*x+c)-a*sin(d*x+c))*(a*cos(d*x+c)+b*sin(d*x+c))^(5/2)/d-10/21*(a^2+b^2)*(b*cos(d*x+c)-a*sin(d*x+c
))*(a*cos(d*x+c)+b*sin(d*x+c))^(1/2)/d+10/21*(a^2+b^2)^2*(cos(1/2*c+1/2*d*x-1/2*arctan(a,b))^2)^(1/2)/cos(1/2*
c+1/2*d*x-1/2*arctan(a,b))*EllipticF(sin(1/2*c+1/2*d*x-1/2*arctan(a,b)),2^(1/2))*((a*cos(d*x+c)+b*sin(d*x+c))/
(a^2+b^2)^(1/2))^(1/2)/d/(a*cos(d*x+c)+b*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3073, 3078, 2641} \[ \frac {10 \left (a^2+b^2\right )^2 \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}} F\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right )}{21 d \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {10 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{21 d}-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*(a^2 + b^2)*(b*Cos[c + d*x] - a*Sin[c + d*x])*Sqrt[a*Cos[c + d*x] + b*Sin[c + d*x]])/(21*d) - (2*(b*Cos[c
+ d*x] - a*Sin[c + d*x])*(a*Cos[c + d*x] + b*Sin[c + d*x])^(5/2))/(7*d) + (10*(a^2 + b^2)^2*EllipticF[(c + d*
x - ArcTan[a, b])/2, 2]*Sqrt[(a*Cos[c + d*x] + b*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(21*d*Sqrt[a*Cos[c + d*x] + b
*Sin[c + d*x]])

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 3073

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[((b*Cos[c + d*x]
- a*Sin[c + d*x])*(a*Cos[c + d*x] + b*Sin[c + d*x])^(n - 1))/(d*n), x] + Dist[((n - 1)*(a^2 + b^2))/n, Int[(a*
Cos[c + d*x] + b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] &&  !IntegerQ[(n
- 1)/2] && GtQ[n, 1]

Rule 3078

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[c + d*x] +
b*Sin[c + d*x])^n/((a*Cos[c + d*x] + b*Sin[c + d*x])/Sqrt[a^2 + b^2])^n, Int[Cos[c + d*x - ArcTan[a, b]]^n, x]
, x] /; FreeQ[{a, b, c, d, n}, x] &&  !(GeQ[n, 1] || LeQ[n, -1]) &&  !(GtQ[a^2 + b^2, 0] || EqQ[a^2 + b^2, 0])

Rubi steps

\begin {align*} \int (a \cos (c+d x)+b \sin (c+d x))^{7/2} \, dx &=-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d}+\frac {1}{7} \left (5 \left (a^2+b^2\right )\right ) \int (a \cos (c+d x)+b \sin (c+d x))^{3/2} \, dx\\ &=-\frac {10 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{21 d}-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d}+\frac {1}{21} \left (5 \left (a^2+b^2\right )^2\right ) \int \frac {1}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}} \, dx\\ &=-\frac {10 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{21 d}-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d}+\frac {\left (5 \left (a^2+b^2\right )^2 \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}\right ) \int \frac {1}{\sqrt {\cos \left (c+d x-\tan ^{-1}(a,b)\right )}} \, dx}{21 \sqrt {a \cos (c+d x)+b \sin (c+d x)}}\\ &=-\frac {10 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{21 d}-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d}+\frac {10 \left (a^2+b^2\right )^2 F\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right ) \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}}{21 d \sqrt {a \cos (c+d x)+b \sin (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 1.82, size = 205, normalized size = 1.10 \[ \frac {\frac {20 \left (a^2+b^2\right )^2 \tan \left (\tan ^{-1}\left (\frac {a}{b}\right )+c+d x\right ) \sqrt {\cos ^2\left (\tan ^{-1}\left (\frac {a}{b}\right )+c+d x\right )} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (c+d x+\tan ^{-1}\left (\frac {a}{b}\right )\right )\right )}{\sqrt {b \sqrt {\frac {a^2}{b^2}+1} \sin \left (\tan ^{-1}\left (\frac {a}{b}\right )+c+d x\right )}}+\sqrt {a \cos (c+d x)+b \sin (c+d x)} \left (\left (3 b^3-9 a^2 b\right ) \cos (3 (c+d x))-23 b \left (a^2+b^2\right ) \cos (c+d x)+2 a \sin (c+d x) \left (3 \left (a^2-3 b^2\right ) \cos (2 (c+d x))+13 a^2+7 b^2\right )\right )}{42 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*Cos[c + d*x] + b*Sin[c + d*x]]*(-23*b*(a^2 + b^2)*Cos[c + d*x] + (-9*a^2*b + 3*b^3)*Cos[3*(c + d*x)] +
2*a*(13*a^2 + 7*b^2 + 3*(a^2 - 3*b^2)*Cos[2*(c + d*x)])*Sin[c + d*x]) + (20*(a^2 + b^2)^2*Sqrt[Cos[c + d*x +
ArcTan[a/b]]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[c + d*x + ArcTan[a/b]]^2]*Tan[c + d*x + ArcTan[a/b]])
/Sqrt[Sqrt[1 + a^2/b^2]*b*Sin[c + d*x + ArcTan[a/b]]])/(42*d)

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fricas [F]  time = 1.10, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (3 \, a b^{2} \cos \left (d x + c\right ) + {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (b^{3} + {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((3*a*b^2*cos(d*x + c) + (a^3 - 3*a*b^2)*cos(d*x + c)^3 + (b^3 + (3*a^2*b - b^3)*cos(d*x + c)^2)*sin(d
*x + c))*sqrt(a*cos(d*x + c) + b*sin(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.44, size = 183, normalized size = 0.98 \[ \frac {\left (a^{2}+b^{2}\right )^{2} \left (6 \left (\sin ^{5}\left (d x +c -\arctan \left (-a , b\right )\right )\right )+5 \sqrt {1+\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \sqrt {-2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )+2}\, \sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \EllipticF \left (\sqrt {1+\sin \left (d x +c -\arctan \left (-a , b\right )\right )}, \frac {\sqrt {2}}{2}\right )+4 \left (\sin ^{3}\left (d x +c -\arctan \left (-a , b\right )\right )\right )-10 \sin \left (d x +c -\arctan \left (-a , b\right )\right )\right )}{21 \cos \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {a^{2}+b^{2}}}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21*(a^2+b^2)^2*(6*sin(d*x+c-arctan(-a,b))^5+5*(1+sin(d*x+c-arctan(-a,b)))^(1/2)*(-2*sin(d*x+c-arctan(-a,b))+
2)^(1/2)*(-sin(d*x+c-arctan(-a,b)))^(1/2)*EllipticF((1+sin(d*x+c-arctan(-a,b)))^(1/2),1/2*2^(1/2))+4*sin(d*x+c
-arctan(-a,b))^3-10*sin(d*x+c-arctan(-a,b)))/cos(d*x+c-arctan(-a,b))/(sin(d*x+c-arctan(-a,b))*(a^2+b^2)^(1/2))
^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a\,\cos \left (c+d\,x\right )+b\,\sin \left (c+d\,x\right )\right )}^{7/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*cos(d*x+c)+b*sin(d*x+c))**(7/2),x)

[Out]

Timed out

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