3.4.90 \(\int \sec (c+d x) (a+b \sin (c+d x))^2 \, dx\) [390]

Optimal. Leaf size=61 \[ -\frac {(a+b)^2 \log (1-\sin (c+d x))}{2 d}+\frac {(a-b)^2 \log (1+\sin (c+d x))}{2 d}-\frac {b^2 \sin (c+d x)}{d} \]

[Out]

-1/2*(a+b)^2*ln(1-sin(d*x+c))/d+1/2*(a-b)^2*ln(1+sin(d*x+c))/d-b^2*sin(d*x+c)/d

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Rubi [A]
time = 0.06, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2747, 716, 647, 31} \begin {gather*} \frac {(a-b)^2 \log (\sin (c+d x)+1)}{2 d}-\frac {(a+b)^2 \log (1-\sin (c+d x))}{2 d}-\frac {b^2 \sin (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((a + b)^2*Log[1 - Sin[c + d*x]])/d + ((a - b)^2*Log[1 + Sin[c + d*x]])/(2*d) - (b^2*Sin[c + d*x])/d

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 716

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \sec (c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac {b \text {Subst}\left (\int \frac {(a+x)^2}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {b \text {Subst}\left (\int \left (-1+\frac {a^2+b^2+2 a x}{b^2-x^2}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac {b^2 \sin (c+d x)}{d}+\frac {b \text {Subst}\left (\int \frac {a^2+b^2+2 a x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac {b^2 \sin (c+d x)}{d}-\frac {(a-b)^2 \text {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{2 d}+\frac {(a+b)^2 \text {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{2 d}\\ &=-\frac {(a+b)^2 \log (1-\sin (c+d x))}{2 d}+\frac {(a-b)^2 \log (1+\sin (c+d x))}{2 d}-\frac {b^2 \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 54, normalized size = 0.89 \begin {gather*} \frac {-(a+b)^2 \log (1-\sin (c+d x))+(a-b)^2 \log (1+\sin (c+d x))-2 b^2 \sin (c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a + b)^2*Log[1 - Sin[c + d*x]]) + (a - b)^2*Log[1 + Sin[c + d*x]] - 2*b^2*Sin[c + d*x])/(2*d)

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Maple [A]
time = 0.28, size = 62, normalized size = 1.02

method	result	size
derivativedivides	\(\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-2 a b \ln \left (\cos \left (d x +c \right )\right )+b^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\)	\(62\)
default	\(\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-2 a b \ln \left (\cos \left (d x +c \right )\right )+b^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\)	\(62\)
norman	\(\frac {-\frac {2 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 b^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (a^{2}-2 a b +b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 a b \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(131\)
risch	\(2 i a b x +\frac {i b^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i b^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {4 i a b c}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{d}\)	\(175\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^2*ln(sec(d*x+c)+tan(d*x+c))-2*a*b*ln(cos(d*x+c))+b^2*(-sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c))))

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Maxima [A]
time = 0.28, size = 60, normalized size = 0.98 \begin {gather*} -\frac {2 \, b^{2} \sin \left (d x + c\right ) - {\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b^2*sin(d*x + c) - (a^2 - 2*a*b + b^2)*log(sin(d*x + c) + 1) + (a^2 + 2*a*b + b^2)*log(sin(d*x + c) -
1))/d

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Fricas [A]
time = 0.41, size = 62, normalized size = 1.02 \begin {gather*} -\frac {2 \, b^{2} \sin \left (d x + c\right ) - {\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b^2*sin(d*x + c) - (a^2 - 2*a*b + b^2)*log(sin(d*x + c) + 1) + (a^2 + 2*a*b + b^2)*log(-sin(d*x + c) +
1))/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \sec {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*sin(d*x+c))**2,x)

[Out]

Integral((a + b*sin(c + d*x))**2*sec(c + d*x), x)

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Giac [A]
time = 4.14, size = 62, normalized size = 1.02 \begin {gather*} -\frac {2 \, b^{2} \sin \left (d x + c\right ) - {\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b^2*sin(d*x + c) - (a^2 - 2*a*b + b^2)*log(abs(sin(d*x + c) + 1)) + (a^2 + 2*a*b + b^2)*log(abs(sin(d*
x + c) - 1)))/d

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Mupad [B]
time = 5.16, size = 50, normalized size = 0.82 \begin {gather*} -\frac {\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,{\left (a+b\right )}^2}{2}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,{\left (a-b\right )}^2}{2}+b^2\,\sin \left (c+d\,x\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((log(sin(c + d*x) - 1)*(a + b)^2)/2 - (log(sin(c + d*x) + 1)*(a - b)^2)/2 + b^2*sin(c + d*x))/d

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