3.4.34 \(\int \cos ^3(e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\) [334]

Optimal. Leaf size=157 \[ \frac {a^2 (a+6 b) \tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{16 b^{3/2} f}+\frac {a (a+6 b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{16 b f}+\frac {(a+6 b) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 b f}-\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 b f} \]

[Out]

1/16*a^2*(a+6*b)*arctanh(sin(f*x+e)*b^(1/2)/(a+b*sin(f*x+e)^2)^(1/2))/b^(3/2)/f+1/24*(a+6*b)*sin(f*x+e)*(a+b*s
in(f*x+e)^2)^(3/2)/b/f-1/6*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(5/2)/b/f+1/16*a*(a+6*b)*sin(f*x+e)*(a+b*sin(f*x+e)^2
)^(1/2)/b/f

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Rubi [A]
time = 0.09, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3269, 396, 201, 223, 212} \begin {gather*} \frac {a^2 (a+6 b) \tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{16 b^{3/2} f}-\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 b f}+\frac {(a+6 b) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 b f}+\frac {a (a+6 b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{16 b f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(a + 6*b)*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(16*b^(3/2)*f) + (a*(a + 6*b)*Sin[e
+ f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(16*b*f) + ((a + 6*b)*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(3/2))/(24*b*f)
- (Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(5/2))/(6*b*f)

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 3269

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \cos ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx &=\frac {\text {Subst}\left (\int \left (1-x^2\right ) \left (a+b x^2\right )^{3/2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=-\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 b f}+\frac {(a+6 b) \text {Subst}\left (\int \left (a+b x^2\right )^{3/2} \, dx,x,\sin (e+f x)\right )}{6 b f}\\ &=\frac {(a+6 b) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 b f}-\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 b f}+\frac {(a (a+6 b)) \text {Subst}\left (\int \sqrt {a+b x^2} \, dx,x,\sin (e+f x)\right )}{8 b f}\\ &=\frac {a (a+6 b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{16 b f}+\frac {(a+6 b) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 b f}-\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 b f}+\frac {\left (a^2 (a+6 b)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{16 b f}\\ &=\frac {a (a+6 b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{16 b f}+\frac {(a+6 b) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 b f}-\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 b f}+\frac {\left (a^2 (a+6 b)\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{16 b f}\\ &=\frac {a^2 (a+6 b) \tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{16 b^{3/2} f}+\frac {a (a+6 b) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{16 b f}+\frac {(a+6 b) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 b f}-\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 b f}\\ \end {align*}

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Mathematica [A]
time = 0.58, size = 149, normalized size = 0.95 \begin {gather*} \frac {\sqrt {a+b \sin ^2(e+f x)} \left (3 a^{3/2} (a+6 b) \sinh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a}}\right )+\sqrt {b} \sin (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}} \left (-3 a (a-10 b)-2 (7 a-6 b) b \sin ^2(e+f x)-8 b^2 \sin ^4(e+f x)\right )\right )}{48 b^{3/2} f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*Sin[e + f*x]^2]*(3*a^(3/2)*(a + 6*b)*ArcSinh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a]] + Sqrt[b]*Sin[e + f*x
]*Sqrt[1 + (b*Sin[e + f*x]^2)/a]*(-3*a*(a - 10*b) - 2*(7*a - 6*b)*b*Sin[e + f*x]^2 - 8*b^2*Sin[e + f*x]^4)))/(
48*b^(3/2)*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a])

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Maple [A]
time = 7.60, size = 198, normalized size = 1.26

method	result	size
default	\(\frac {-8 \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, b^{\frac {7}{2}} \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )+2 \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, b^{\frac {5}{2}} \left (2 b +7 a \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, b^{\frac {3}{2}} \left (-3 a^{2}+16 a b +4 b^{2}\right ) \sin \left (f x +e \right )+3 a^{3} \ln \left (\sin \left (f x +e \right ) \sqrt {b}+\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\right ) b +18 a^{2} \ln \left (\sin \left (f x +e \right ) \sqrt {b}+\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\right ) b^{2}}{48 b^{\frac {5}{2}} f}\)	\(198\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(-8*(a+b-b*cos(f*x+e)^2)^(1/2)*b^(7/2)*sin(f*x+e)*cos(f*x+e)^4+2*(a+b-b*cos(f*x+e)^2)^(1/2)*b^(5/2)*(2*b+
7*a)*cos(f*x+e)^2*sin(f*x+e)+(a+b-b*cos(f*x+e)^2)^(1/2)*b^(3/2)*(-3*a^2+16*a*b+4*b^2)*sin(f*x+e)+3*a^3*ln(sin(
f*x+e)*b^(1/2)+(a+b-b*cos(f*x+e)^2)^(1/2))*b+18*a^2*ln(sin(f*x+e)*b^(1/2)+(a+b-b*cos(f*x+e)^2)^(1/2))*b^2)/b^(
5/2)/f

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Maxima [A]
time = 0.29, size = 186, normalized size = 1.18 \begin {gather*} \frac {\frac {3 \, a^{3} \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} + \frac {18 \, a^{2} \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {b}} + 12 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right ) + 18 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} a \sin \left (f x + e\right ) - \frac {8 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}} \sin \left (f x + e\right )}{b} + \frac {2 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \sin \left (f x + e\right )}{b} + \frac {3 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2} \sin \left (f x + e\right )}{b}}{48 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*a^3*arcsinh(b*sin(f*x + e)/sqrt(a*b))/b^(3/2) + 18*a^2*arcsinh(b*sin(f*x + e)/sqrt(a*b))/sqrt(b) + 12*
(b*sin(f*x + e)^2 + a)^(3/2)*sin(f*x + e) + 18*sqrt(b*sin(f*x + e)^2 + a)*a*sin(f*x + e) - 8*(b*sin(f*x + e)^2
+ a)^(5/2)*sin(f*x + e)/b + 2*(b*sin(f*x + e)^2 + a)^(3/2)*a*sin(f*x + e)/b + 3*sqrt(b*sin(f*x + e)^2 + a)*a^
2*sin(f*x + e)/b)/f

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Fricas [A]
time = 1.45, size = 577, normalized size = 3.68 \begin {gather*} \left [\frac {3 \, {\left (a^{3} + 6 \, a^{2} b\right )} \sqrt {b} \log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{2} b^{2} + 24 \, a b^{3} + 24 \, b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 32 \, a^{3} b + 160 \, a^{2} b^{2} + 256 \, a b^{3} + 128 \, b^{4} - 32 \, {\left (a^{3} b + 10 \, a^{2} b^{2} + 24 \, a b^{3} + 16 \, b^{4}\right )} \cos \left (f x + e\right )^{2} - 8 \, {\left (16 \, b^{3} \cos \left (f x + e\right )^{6} - 24 \, {\left (a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{3} - 10 \, a^{2} b - 24 \, a b^{2} - 16 \, b^{3} + 2 \, {\left (5 \, a^{2} b + 24 \, a b^{2} + 24 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {b} \sin \left (f x + e\right )\right ) - 8 \, {\left (8 \, b^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b - 16 \, a b^{2} - 4 \, b^{3} - 2 \, {\left (7 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{384 \, b^{2} f}, -\frac {3 \, {\left (a^{3} + 6 \, a^{2} b\right )} \sqrt {-b} \arctan \left (\frac {{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \, {\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 8 \, a b + 8 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-b}}{4 \, {\left (2 \, b^{3} \cos \left (f x + e\right )^{4} + a^{2} b + 3 \, a b^{2} + 2 \, b^{3} - {\left (3 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) + 4 \, {\left (8 \, b^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b - 16 \, a b^{2} - 4 \, b^{3} - 2 \, {\left (7 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{192 \, b^{2} f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/384*(3*(a^3 + 6*a^2*b)*sqrt(b)*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + 2*b^4)*cos(f*x + e)^6 + 32*(5*a^2*
b^2 + 24*a*b^3 + 24*b^4)*cos(f*x + e)^4 + a^4 + 32*a^3*b + 160*a^2*b^2 + 256*a*b^3 + 128*b^4 - 32*(a^3*b + 10*
a^2*b^2 + 24*a*b^3 + 16*b^4)*cos(f*x + e)^2 - 8*(16*b^3*cos(f*x + e)^6 - 24*(a*b^2 + 2*b^3)*cos(f*x + e)^4 - a
^3 - 10*a^2*b - 24*a*b^2 - 16*b^3 + 2*(5*a^2*b + 24*a*b^2 + 24*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a
+ b)*sqrt(b)*sin(f*x + e)) - 8*(8*b^3*cos(f*x + e)^4 + 3*a^2*b - 16*a*b^2 - 4*b^3 - 2*(7*a*b^2 + 2*b^3)*cos(f
*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/(b^2*f), -1/192*(3*(a^3 + 6*a^2*b)*sqrt(-b)*arctan(1/
4*(8*b^2*cos(f*x + e)^4 - 8*(a*b + 2*b^2)*cos(f*x + e)^2 + a^2 + 8*a*b + 8*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b
)*sqrt(-b)/((2*b^3*cos(f*x + e)^4 + a^2*b + 3*a*b^2 + 2*b^3 - (3*a*b^2 + 4*b^3)*cos(f*x + e)^2)*sin(f*x + e)))
+ 4*(8*b^3*cos(f*x + e)^4 + 3*a^2*b - 16*a*b^2 - 4*b^3 - 2*(7*a*b^2 + 2*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x
+ e)^2 + a + b)*sin(f*x + e))/(b^2*f)]

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**3*(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4846 deep

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Giac [A]
time = 0.62, size = 134, normalized size = 0.85 \begin {gather*} -\frac {{\left (2 \, {\left (4 \, b \sin \left (f x + e\right )^{2} + \frac {7 \, a b^{4} - 6 \, b^{5}}{b^{4}}\right )} \sin \left (f x + e\right )^{2} + \frac {3 \, {\left (a^{2} b^{3} - 10 \, a b^{4}\right )}}{b^{4}}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a} \sin \left (f x + e\right ) + \frac {3 \, {\left (a^{3} + 6 \, a^{2} b\right )} \log \left ({\left | -\sqrt {b} \sin \left (f x + e\right ) + \sqrt {b \sin \left (f x + e\right )^{2} + a} \right |}\right )}{b^{\frac {3}{2}}}}{48 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*((2*(4*b*sin(f*x + e)^2 + (7*a*b^4 - 6*b^5)/b^4)*sin(f*x + e)^2 + 3*(a^2*b^3 - 10*a*b^4)/b^4)*sqrt(b*sin
(f*x + e)^2 + a)*sin(f*x + e) + 3*(a^3 + 6*a^2*b)*log(abs(-sqrt(b)*sin(f*x + e) + sqrt(b*sin(f*x + e)^2 + a)))
/b^(3/2))/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (e+f\,x\right )}^3\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(e + f*x)^3*(a + b*sin(e + f*x)^2)^(3/2),x)

[Out]

int(cos(e + f*x)^3*(a + b*sin(e + f*x)^2)^(3/2), x)

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