Optimal. Leaf size=298 \[ -\frac {2 d^2 (b c-a d) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi ^2}-\frac {3 d^3 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2}-\frac {(b c-a d)^4 \text {FresnelC}(a+b x)}{4 b^4 d}+\frac {3 d^3 \text {FresnelC}(a+b x)}{4 b^4 \pi ^2}+\frac {(c+d x)^4 \text {FresnelC}(a+b x)}{4 d}+\frac {3 d (b c-a d)^2 S(a+b x)}{2 b^4 \pi }-\frac {(b c-a d)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }-\frac {3 d (b c-a d)^2 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }-\frac {d^2 (b c-a d) (a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }-\frac {d^3 (a+b x)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi } \]
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Rubi [A]
time = 0.26, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps
used = 14, number of rules used = 10, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {6564, 3515,
3433, 3461, 2717, 3467, 3432, 3377, 2718, 3466} \begin {gather*} -\frac {d^2 (a+b x)^2 (b c-a d) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^4}-\frac {2 d^2 (b c-a d) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi ^2 b^4}-\frac {(b c-a d)^4 \text {FresnelC}(a+b x)}{4 b^4 d}+\frac {3 d (b c-a d)^2 S(a+b x)}{2 \pi b^4}-\frac {(b c-a d)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^4}-\frac {3 d (a+b x) (b c-a d)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 \pi b^4}+\frac {3 d^3 \text {FresnelC}(a+b x)}{4 \pi ^2 b^4}-\frac {d^3 (a+b x)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 \pi b^4}-\frac {3 d^3 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 \pi ^2 b^4}+\frac {(c+d x)^4 \text {FresnelC}(a+b x)}{4 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 2717
Rule 2718
Rule 3377
Rule 3432
Rule 3433
Rule 3461
Rule 3466
Rule 3467
Rule 3515
Rule 6564
Rubi steps
\begin {align*} \int (c+d x)^3 C(a+b x) \, dx &=\frac {(c+d x)^4 C(a+b x)}{4 d}-\frac {b \int (c+d x)^4 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) \, dx}{4 d}\\ &=\frac {(c+d x)^4 C(a+b x)}{4 d}-\frac {\text {Subst}\left (\int \left (b^4 c^4 \left (1+\frac {a d \left (-4 b^3 c^3+6 a b^2 c^2 d-4 a^2 b c d^2+a^3 d^3\right )}{b^4 c^4}\right ) \cos \left (\frac {\pi x^2}{2}\right )+4 b^3 c^3 d \left (1-\frac {a d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right )}{b^3 c^3}\right ) x \cos \left (\frac {\pi x^2}{2}\right )+6 b^2 c^2 d^2 \left (1+\frac {a d (-2 b c+a d)}{b^2 c^2}\right ) x^2 \cos \left (\frac {\pi x^2}{2}\right )+4 b c d^3 \left (1-\frac {a d}{b c}\right ) x^3 \cos \left (\frac {\pi x^2}{2}\right )+d^4 x^4 \cos \left (\frac {\pi x^2}{2}\right )\right ) \, dx,x,a+b x\right )}{4 b^4 d}\\ &=\frac {(c+d x)^4 C(a+b x)}{4 d}-\frac {d^3 \text {Subst}\left (\int x^4 \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4}-\frac {\left (d^2 (b c-a d)\right ) \text {Subst}\left (\int x^3 \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^4}-\frac {\left (3 d (b c-a d)^2\right ) \text {Subst}\left (\int x^2 \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{2 b^4}-\frac {(b c-a d)^3 \text {Subst}\left (\int x \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^4}-\frac {(b c-a d)^4 \text {Subst}\left (\int \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4 d}\\ &=-\frac {(b c-a d)^4 C(a+b x)}{4 b^4 d}+\frac {(c+d x)^4 C(a+b x)}{4 d}-\frac {3 d (b c-a d)^2 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }-\frac {d^3 (a+b x)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }-\frac {\left (d^2 (b c-a d)\right ) \text {Subst}\left (\int x \cos \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^4}-\frac {(b c-a d)^3 \text {Subst}\left (\int \cos \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^4}+\frac {\left (3 d^3\right ) \text {Subst}\left (\int x^2 \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4 \pi }+\frac {\left (3 d (b c-a d)^2\right ) \text {Subst}\left (\int \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{2 b^4 \pi }\\ &=-\frac {3 d^3 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2}-\frac {(b c-a d)^4 C(a+b x)}{4 b^4 d}+\frac {(c+d x)^4 C(a+b x)}{4 d}+\frac {3 d (b c-a d)^2 S(a+b x)}{2 b^4 \pi }-\frac {(b c-a d)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }-\frac {3 d (b c-a d)^2 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }-\frac {d^2 (b c-a d) (a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }-\frac {d^3 (a+b x)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }+\frac {\left (3 d^3\right ) \text {Subst}\left (\int \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{4 b^4 \pi ^2}+\frac {\left (d^2 (b c-a d)\right ) \text {Subst}\left (\int \sin \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{b^4 \pi }\\ &=-\frac {2 d^2 (b c-a d) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi ^2}-\frac {3 d^3 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2}-\frac {(b c-a d)^4 C(a+b x)}{4 b^4 d}+\frac {3 d^3 C(a+b x)}{4 b^4 \pi ^2}+\frac {(c+d x)^4 C(a+b x)}{4 d}+\frac {3 d (b c-a d)^2 S(a+b x)}{2 b^4 \pi }-\frac {(b c-a d)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }-\frac {3 d (b c-a d)^2 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }-\frac {d^2 (b c-a d) (a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }-\frac {d^3 (a+b x)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }\\ \end {align*}
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Mathematica [A]
time = 0.53, size = 424, normalized size = 1.42 \begin {gather*} \frac {-8 b c d^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+5 a d^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-3 b d^3 x \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+\left (4 b^3 c^3 \pi ^2 (a+b x)+6 b^2 c^2 d \pi ^2 \left (-a^2+b^2 x^2\right )+4 b c d^2 \pi ^2 \left (a^3+b^3 x^3\right )+d^3 \left (3-a^4 \pi ^2+b^4 \pi ^2 x^4\right )\right ) \text {FresnelC}(a+b x)+6 d (b c-a d)^2 \pi S(a+b x)-4 b^3 c^3 \pi \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+6 a b^2 c^2 d \pi \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-4 a^2 b c d^2 \pi \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+a^3 d^3 \pi \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-6 b^3 c^2 d \pi x \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+4 a b^2 c d^2 \pi x \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-a^2 b d^3 \pi x \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-4 b^3 c d^2 \pi x^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+a b^2 d^3 \pi x^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-b^3 d^3 \pi x^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.49, size = 398, normalized size = 1.34
method | result | size |
derivativedivides | \(\frac {\frac {\FresnelC \left (b x +a \right ) \left (a d -c b -d \left (b x +a \right )\right )^{4}}{4 b^{3} d}-\frac {\frac {d^{4} \left (b x +a \right )^{3} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {3 d^{4} \left (-\frac {\left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {\FresnelC \left (b x +a \right )}{\pi }\right )}{\pi }+\frac {\left (-4 a \,d^{4}+4 b c \,d^{3}\right ) \left (b x +a \right )^{2} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {2 \left (-4 a \,d^{4}+4 b c \,d^{3}\right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi ^{2}}+\frac {\left (6 a^{2} d^{4}-12 a b c \,d^{3}+6 b^{2} c^{2} d^{2}\right ) \left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {\left (6 a^{2} d^{4}-12 a b c \,d^{3}+6 b^{2} c^{2} d^{2}\right ) \mathrm {S}\left (b x +a \right )}{\pi }+\frac {\left (-4 a^{3} d^{4}+12 a^{2} b c \,d^{3}-12 a \,b^{2} c^{2} d^{2}+4 b^{3} c^{3} d \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+a^{4} d^{4} \FresnelC \left (b x +a \right )-4 a^{3} b c \,d^{3} \FresnelC \left (b x +a \right )+6 a^{2} b^{2} c^{2} d^{2} \FresnelC \left (b x +a \right )-4 a \,b^{3} c^{3} d \FresnelC \left (b x +a \right )+b^{4} c^{4} \FresnelC \left (b x +a \right )}{4 b^{3} d}}{b}\) | \(398\) |
default | \(\frac {\frac {\FresnelC \left (b x +a \right ) \left (a d -c b -d \left (b x +a \right )\right )^{4}}{4 b^{3} d}-\frac {\frac {d^{4} \left (b x +a \right )^{3} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {3 d^{4} \left (-\frac {\left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {\FresnelC \left (b x +a \right )}{\pi }\right )}{\pi }+\frac {\left (-4 a \,d^{4}+4 b c \,d^{3}\right ) \left (b x +a \right )^{2} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {2 \left (-4 a \,d^{4}+4 b c \,d^{3}\right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi ^{2}}+\frac {\left (6 a^{2} d^{4}-12 a b c \,d^{3}+6 b^{2} c^{2} d^{2}\right ) \left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {\left (6 a^{2} d^{4}-12 a b c \,d^{3}+6 b^{2} c^{2} d^{2}\right ) \mathrm {S}\left (b x +a \right )}{\pi }+\frac {\left (-4 a^{3} d^{4}+12 a^{2} b c \,d^{3}-12 a \,b^{2} c^{2} d^{2}+4 b^{3} c^{3} d \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+a^{4} d^{4} \FresnelC \left (b x +a \right )-4 a^{3} b c \,d^{3} \FresnelC \left (b x +a \right )+6 a^{2} b^{2} c^{2} d^{2} \FresnelC \left (b x +a \right )-4 a \,b^{3} c^{3} d \FresnelC \left (b x +a \right )+b^{4} c^{4} \FresnelC \left (b x +a \right )}{4 b^{3} d}}{b}\) | \(398\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 375, normalized size = 1.26 \begin {gather*} \frac {6 \, \pi {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) + {\left (\pi ^{2} {\left (4 \, a b^{3} c^{3} - 6 \, a^{2} b^{2} c^{2} d + 4 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} + 3 \, d^{3}\right )} \sqrt {b^{2}} \operatorname {C}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) - {\left (3 \, b^{2} d^{3} x + 8 \, b^{2} c d^{2} - 5 \, a b d^{3}\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) + {\left (\pi ^{2} b^{5} d^{3} x^{4} + 4 \, \pi ^{2} b^{5} c d^{2} x^{3} + 6 \, \pi ^{2} b^{5} c^{2} d x^{2} + 4 \, \pi ^{2} b^{5} c^{3} x\right )} \operatorname {C}\left (b x + a\right ) - {\left (\pi b^{4} d^{3} x^{3} + \pi {\left (4 \, b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} + \pi {\left (6 \, b^{4} c^{2} d - 4 \, a b^{3} c d^{2} + a^{2} b^{2} d^{3}\right )} x + \pi {\left (4 \, b^{4} c^{3} - 6 \, a b^{3} c^{2} d + 4 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )}\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right )}{4 \, \pi ^{2} b^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (c + d x\right )^{3} C\left (a + b x\right )\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \mathrm {FresnelC}\left (a+b\,x\right )\,{\left (c+d\,x\right )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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